The post Three-Phase Electricity Explained appeared first on Electrical A2Z.

]]>**Figure 1** A single circuit consisting of a source and a load.

**Figure 2** illustrates three of these assumed AC circuits near each other. Intentionally, the generators are shown forming a triangle, but at this time it is only a graphic layout.

**Figure 2** Three single-phase systems shown together.

For ease of reference, these systems are named A, B, and C. Suppose now that these three single-phase electric circuits are such that we can connect together the points that are near each other; that is, point A to B′, B to C′, and C to A′, as shown in **Figure 3**. If this can be done, then we may use only three wires for connection to the loads, instead of 6, provided that the loads permit this. This is shown in **Figure 3**, thus saving wire cost.

**Figure 3** Saving in the number of wires used.

The arrangement on the right-hand side of **Figure 3** can be rearranged to look as shown in **Figure 4b**, which is one of the ways a three-phase load can be connected to a three-phase source. **The necessary condition** to be able to connect the associated points in this circuit (A to B′, and so on) is that these points must have the same electric potential (the same voltage with respect to a reference) at any instant of time.

**Figure 4** Common representation of a three-phase system. The three loads in (a) are shown as in (b).

If there is one winding, the generated electricity is a single phase but, if there are three windings as shown in **Figure 5**, then the generated electricity is three phase.

As you see in **Figure 5**, the three windings are placed at 120° from each other. This causes a phase difference between the three voltages that are generated. The reason is obvious; one of the windings sees the magnetic field before the other two, and again the second winding sees it before the third before everything is repeated.

**Figure 5** A simple representation of a three-phase generator, indicating the reason for the phase difference between the voltages in the three windings.

**In other words,** what happens to one winding in a given time happens to the other two windings with a delay that corresponds to 120° and 240° physical displacement.

As a result of the rotation of the magnetic field, the generated voltage in each winding has a sinusoidal form. For one cycle variation of the magnetic field, the generated voltages in each winding are illustrated in **Figure 6**.

**Figure 6** Individual waveform variation of the voltage in each winding for one cycle period.

When putting together, the three sinusoidal waveforms of a three-phase electricity system are shown in **Figure 7**. The three phases are addressed as phases A, B, and C, or phases 1, 2, and 3.

**Figure 7** Waveforms of a three-phase system shown together on a common axis.

No matter what the three phases are called, their order is very important and must be respected. Observe that in **Figure 7 **phase A reaches its maximum before B and C. B reaches its maximum after 120°, and phase C reaches its maximum 120° after B (240° after A).

Also, notice that the vertical axis has no title because it can represent voltage or current. In **Figure 7**, line MN corresponds to the time for one full cycle and the variations of each phase are those shown in **Figure 6**.

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]]>The post Parallel RLC Circuit: Analysis & Example Problems appeared first on Electrical A2Z.

]]>For a parallel *RLC* circuit, the voltage is common for all the three types of components because it is the **same voltage** that is applied to each component. Nevertheless, the **currents in the three branches are not in phase with each other**. This means that the currents in the three branches do not simultaneously reach their peak values or zero values.

Hence, the total current cannot be determined by algebraically adding the individual values of the currents in the resistor, inductor, and capacitor.

A parallel *RLC* circuit is shown in **Figure 1**. As in the case of series *RLC* circuits, we need to find the total current and the power consumption for the whole circuit or for each individual branch.

**Figure 1** Schematic of parallel *RLC* circuits.

For this circuit the voltage applied to each component in each branch is the same. Therefore, the current in each component can be found from dividing the voltage by the branch impedance. Then the currents can be added together.

**However**, because the currents in the three components are not in phase with each other (they do not reach their maximum and minimum values at the same time), they cannot be algebraically added together and must be added in vector form.

- You May Also Read: Series RLC Circuit: Analysis & Example Problems

**Figure 1** illustrates the vector representation of the three currents in a typical parallel *RLC* circuit. It shows that the current in the resistor is in phase with the applied voltage, the current in the capacitor leads the applied voltage (remember *ICE*) and the current in the inductor lags the voltage (remember *ELI*).

**Furthermore**, note that for this vector representation of the currents and voltage in a parallel *RLC* circuit, because the voltage is the common variable for all branches, you start by drawing the vector for the voltage as the reference vector. (In series *RLC* circuit you started this process by drawing the vector for the current.)

To find the total current in a parallel *RLC* circuit, one needs to find the vector sum of the currents in *R, L*, and *C*.

Because the current in the inductor and the current in the capacitor are 180° out of phase, in adding them together their values are subtracted from each other. Thus, the relationship for the total current of the circuit, *I*, and the individual component currents *I _{R}*,

$\begin{matrix} I={{\sqrt{I_{R}^{2}+\left( {{I}_{L}}-{{I}_{C}} \right)}}^{2}} & {} & \left( 1 \right) \\\end{matrix}$

**Figure 2** Vectors for the voltage and the three different currents in the *RLC* parallel circuit.

In the circuit shown in **Figure 3** the current is 1.8 A. If the current through the capacitor is 1.5 A, find the applied voltage and the resistance of the resistor.

**Figure 3** Circuit corresponding to **Example 1**.

**Solution**

For 60 Hz frequency, the reactance of the capacitor is

\[{{X}_{C}}=\frac{1}{2*3.14*60*0.000045}=59\Omega \]

Thus, the applied voltage is

$59*1.5=88.5V$

Because this circuit has no inductor, the value of *L* in **Equation 1** is set to zero and the result is

$I=\sqrt{I_{R}^{2}+I_{C}^{2}}$

Which leads to

\[{{I}_{R}}=\sqrt{{{1.8}^{2}}-{{1.5}^{2}}}=0.995=1A\]

And the resistance of the resistor is

$88.5\div 1=88.5\Omega $

If in **Equation 1**, the values for *I _{R}*,

\[\frac{V}{Z}=\sqrt{{{\left( \frac{V}{R} \right)}^{2}}+{{\left( \frac{V}{{{X}_{L}}} \right)}^{2}}}\]

By omitting *V* from both sides the relationship between *Z* and *R, L*, and *C* can be found then as

\[\begin{matrix} \frac{1}{Z}=\sqrt{{{\left( \frac{1}{R} \right)}^{2}}+{{\left( \frac{1}{{{X}_{L}}}-\frac{1}{{{X}_{C}}} \right)}^{2}}} & {} & \left( 2 \right) \\\end{matrix}\]

**Equation 2** can be used to find the equivalent impedance of the three components in parallel. The circuit current can also be found this way by dividing the applied voltage by *Z* or by directly multiplying $\frac{1}{Z}$ by the applied voltage.

In the circuit shown in **Figure 4**, *R* = 55 Ω, *L* = 0.08 H, and *C* = 1 μF, find the impedance of the circuit and the applied voltage.

**Figure 4 **Circuit for **Example 2**.

**Solution**

$\begin{align} & {{X}_{L}}=2*3.14*60*0.08=30.16\Omega \\ & {{X}_{C}}=\frac{1}{2*3.14*60*0.000001}=26.5\Omega \\ & \frac{1}{Z}=\sqrt{{{\left( \frac{1}{55} \right)}^{2}}+{{\left( \frac{1}{55}-\frac{1}{26.5} \right)}^{2}}}=\frac{1}{53.33} \\ & Z=53.33\Omega \\\end{align}$

Applied voltage = *V* = *ZI* = (53.33) (1.8) = **96 V.**

**Equation 2** also implies that the value for *Z* is smaller than *R* for parallel *RLC* circuits. A vector representation of *I _{R}*,

**Figure 5** Vectors for (a) currents and (b) powers in parallel *RLC* circuits.

Reactive power is the vector sum of the inductive and capacitive powers. Depending on if inductive power (*Q _{L}*) or the capacitive power (

Because in practice the majority of applications (including home and industrial circuits) are parallel circuits, any circuit is categorized to be leading or lagging. If in a circuit the current leads the voltage, the circuit is said to be leading; if the current lags the voltage, the circuit is said to be lagging.

**Figure 5** shows a lagging circuit. In practice, most of the circuits are lagging because of the presence of electric motors, unless the effects of electric motors are compensated by inserting capacitors that introduce capacitive power to a circuit (see power factor correction). The power factor in a parallel *RLC* circuit is determined from

\[\begin{matrix} pf=\frac{Z}{R}=\frac{{{I}_{R}}}{I}=\frac{Active\text{ }Power}{Apparent\text{ }Power} & {} & \left( 3 \right) \\\end{matrix}\]

Note that the power factor by itself is not sufficient to describe a circuit. It has to be accompanied by the statement for leading or lagging. A circuit may have the same power factor in two cases, either leading or lagging. Sometimes the leading or lagging is attributed to the power factor. For example, one may say a circuit has a leading power factor of 0.90.

In the circuit shown in Figure 6, the total current is 150 mA and the current through the inductor is 100 mA. Determine what the applied voltage is. Also, knowing that the frequency is 50 Hz, find the value of *L*.

**Figure 6** Circuit of **Example 3**.

**Solution**

The applied voltage can be found by multiplying the resistor current by 100 Ω. Having only a resistor and an inductor in this circuit **Equation 1** leads to

$\begin{align} & {{I}_{R}}=\sqrt{{{I}^{2}}-I_{L}^{2}}=\sqrt{{{150}^{2}}-{{100}^{2}}}=0.1118A \\ & V=100*0.0008=11.18V \\ & {{X}_{L}}=11.18\div 0.100=111.8\Omega \\ & L=\frac{{{X}_{L}}}{2\pi f}=\frac{111.8}{2\pi *50}=35.6mH \\\end{align}$

In a parallel AC circuit, if the current leads the voltage, the circuit is said to be leading; if the current lags, the voltage the circuit is said to be lagging.

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]]>The post Series RLC Circuit: Analysis & Example Problems appeared first on Electrical A2Z.

]]>If any of the components is absent (usually, the inductor or the capacitor, not the resistor), then in calculations, the corresponding value for that element can be set to zero or the associated term deleted from the formulae.

**Figure 1** illustrates a series *RLC* circuit. Note that the order the three components are shown is not important. Thus, in **Figure 2** the two circuits are equivalent.

**Figure 1** A series *RLC* circuit.

**Figure 2** Two equivalent Series RLC circuits.

In tackling the circuit at hand, we need to know the relationship between the applied voltage and the current and the power consumption of the circuit, using all the knowledge that has been gained so far.

The simplest question with a series *RLC* circuits is finding the current in the circuit if the particulars of the loads and the applied voltage are given.

In the above circuit (Figure 1) *V* is the applied voltage, *I* is the common current for all the three elements, *f* is the frequency, and *R, L*, and *C* represent the values for resistance, inductance, and capacitance, respectively, of the three components in the circuit.

- You May Also Read: Parallel RLC Circuit: Analysis & Example Problems

The applied voltage in this circuit is divided between the three components. In this regard, the corresponding voltages across *R, L*, and C are denoted by *V _{R}*,

The voltage across the resistor is in phase with the current, the voltage across the capacitor is lagging the current by 90° and the voltage across the inductor leading the current by 90°.

We now try to show these variables in the vector form. Because the three components have the same current, the most appropriate reference entity for showing the vectors is the circuit current *I*.

**Figure 3** shows these vectors, irrespective of their numerical values (because we do not have any number of values given yet) but with the correct orientation. Nevertheless, remember that the scale for current and voltage can be different, whereas the scale for all identical values (voltages here) must be the same.

As can be seen in **Figure 3**, the voltage across *R* is in phase with the current, the voltage across *L* is leading the current, and the voltage across *C* is lagging the current. **This implies** that *R, X _{C}*, and

**Figure 3** Vectors for the current and the three different voltages in the *RLC* series circuit.

Because a capacitor and an inductor have opposite effects, their corresponding vectors are opposite to each other, and, thus, their sum is represented by a number smaller than the larger value (*X _{C}* –

**Figure 4** illustrates two cases, one when *X _{C}* >

**Figure 4** Relationship between R, X_{C}, X_{L}, and Z.** (a)** Positive phase angle**. (b)** Negative phase angle.

Using the Pythagorean Theorem the value of the impedance Z can be written as

\[\begin{matrix} Z=\sqrt{{{R}^{2}}+{{\left( {{X}_{L}}-{{X}_{C}} \right)}^{2}}} & {} & \left( 1 \right) \\\end{matrix}\]

(Note that it does not matter if one enters *X _{L}* –

In a series *RLC*, circuit *R* = 30 Ω, *L* = 15 mH, and *C* = 51 μF. If the source voltage and frequency are 12 V and 60 Hz, respectively, what is the current in the circuit?

**Solution**

$\begin{align} & {{X}_{L}}=2*3.14*60*0.015=5.655\Omega \\ & {{X}_{C}}=\frac{1}{2*3.14*60*0.000051}=5.655\Omega \\ & Z=\sqrt{{{30}^{2}}+{{\left( 52-5.655 \right)}^{2}}}=55.21\Omega \\ & I=\frac{12}{55.21}=217mA \\\end{align}$

**Figure 5a** is the same as **Figure 4a**. Because for the three components in series the current is the same, if all the values represented by the three sides of the triangle are multiplied by the current, then a similar triangle, as shown in **Figure 5b**, results. The sides of this triangle represent the voltages, since *RI* = V* _{R}*,

Furthermore, if the sides of this new triangle are multiplied by I one more time, again a similar triangle results, which reflect the values for the various powers, as shown in **Figure 5c**.

**Figure 5** Similar triangles showing powers, voltages, and their relationships in a series *RLC *circuit. (a) Circuit resistance versus circuit impedance. (b) The voltage across resistance versus the applied voltage. (c) Circuit active power versus apparent power.

As can be understood from **Figure 5b**, the three voltage values *V _{R}*,

$\begin{matrix} V=\sqrt{V_{R}^{2}+{{\left( {{V}_{L}}-{{V}_{C}} \right)}^{2}}} & {} & \left( 2 \right) \\\end{matrix}$

Where *V* is the applied voltage. The second term under the radical is, in fact, the sum of the voltage across the inductor and the voltage across the capacitor, but, because these two voltages are 180° out of phase, their sum appears as a difference in values.

**Figure 5a**–**c** clearly shows that for a series *RLC* circuit the power factor can be found from one of the following relationships:

\[\begin{matrix} pf=\frac{R}{Z}=\frac{{{V}_{R}}}{V}=\frac{Active\text{ }Power}{Apparent\text{ }Power} & {} & \left( 3 \right) \\\end{matrix}\]

It is also easy to understand if the circuit is more inductive or more capacitive. This is reflected in the vector sum *X _{L}* –

A series *RLC* circuit consists of a 20 Ω resistor, a 51 μF capacitor, and a 25 mH inductor. If the source frequency is 50 Hz, and the circuit current is 350 mA, what is the applied voltage?

**Solution**

$\begin{align} & R=20\Omega \\ & {{X}_{L}}=2*3.14*50*0.025=7.85\Omega \\ & {{X}_{C}}=\frac{1}{2*3.14*50*0.000051}=62.445\Omega \\ & Z=\sqrt{{{20}^{2}}+{{\left( 7.85-62.445 \right)}^{2}}}=58.15\Omega \\ & V=IZ=58.15*0.35=20V \\\end{align}$

In a series *RLC* circuit the voltages across the three components are not in phase with each other.

If the applied voltage to the circuit of **Example 2** is 12 V, what is the voltage across the capacitor?

**Solution**

In **Example 2** the applied voltage was 20 V. The distribution of this voltage among the three components is as follows:

$\begin{align} & {{V}_{R}}=20*0.35=7V \\ & {{V}_{C}}=62.445*0.35=21.85V \\ & {{V}_{L}}=7.85*0.35=2.75V \\\end{align}$

In the current case, when the applied voltage is 12 V (i.e., 0.6× the previous case) because nothing has changed in the circuit, the current will be accordingly smaller by 0.6×. As a result, all voltages will be 0.6× smaller. The voltage across the capacitor, therefore, is

${{V}_{C}}=21.85*0.6=13.1V$

You see that it is not necessary always to do repeat all the calculation. Also, note that the voltage across the capacitor is larger than the applied voltage of 12 V. This is always possible, and we do not expect all the voltages to be smaller than the applied voltage.

In the circuit of **Example 2**, what is the phase angle between the voltage and the current?

**Solution**

The answer can be found by first finding the power factor from any of the relationships in **Equation 3**. The values of *R* and *Z* are readily available and are the best choice.

$\begin{align} & pf=\frac{R}{Z}=\frac{20}{58.15}=0.344 \\ & Phase\text{ }Angle={{\cos }^{-1}}\left( 0.344 \right)={{70}^{o}} \\\end{align}$

Knowing only the value of the phase angle to be 70° is not sufficient, and we need to express if the circuit is leading or lagging (i.e., if the current leads the voltage or it lags the voltage). This can be judged based on the values of *X _{C}* and

When a series *RLC* circuit is subject to 48 V, *V _{R}* is 15 V, and

**Solution**

According to **Equation 2**,

$V=\sqrt{V_{R}^{2}+{{\left( {{V}_{L}}-{{V}_{C}} \right)}^{2}}}=48V$

Because ${{V}_{R}}=15V,{{\left( {{V}_{L}}-{{V}_{C}} \right)}^{2}}$ is determined as (directly from **Equation 2**)

${{\left( {{V}_{L}}-{{V}_{C}} \right)}^{2}}={{V}^{2}}-V_{R}^{2}={{48}^{2}}-{{15}^{2}}=2079={{45.6}^{2}}$

Attention must be paid here in calculating (*V _{L}* –

$\pm \left( {{V}_{L}}-{{V}_{C}} \right)=45.6V$

Having *V _{L}* = 22 V leads to only one acceptable value. (The other value is negative and is not acceptable.) Hence, the value for

${{V}_{C}}=67.6V$

One of the characteristics of a series *RLC* circuit is that the voltage between two points can be higher than the applied voltage.

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]]>The post Capacitors in AC Circuits appeared first on Electrical A2Z.

]]>During the time that charging takes place a current flows in the circuit (wires connecting the capacitor to the power source). This current is due to the electrons moving in the wires and not in the capacitor itself, as shown in **Figure 1**. This current is not long lasting and decays to zero.

**Figure 1** Capacitor charging instants (a) before and after (b) switch is closed.

When the voltage across the capacitor is the same as that of the power supply, the current is zero. The change in the current in the circuit containing the capacitor is shown in **Figure 2**. At the first instance, the connection is established the current has its maximum value.

**Figure 2** Current change in the circuit containing a charging capacitor.

The same argument is true when a charged capacitor discharges through a circuit. A current in the circuit flows and continues until the capacitor is fully discharged.

At the first instant, the capacitor is connected to the circuit, the current has its highest magnitude, which gradually decays to zero. **Figure 3** shows the current change in a discharging capacitor. (The changing pattern is based on a decaying exponential curve.)

- Consider a circuit as shown in
**Figure 3a**in which a capacitor is in a DC circuit with two resistors*R*_{1}and*R*_{2}. When the switch is turned on and the resistors are powered, the capacitor charges and the voltage across it is the same as that of the battery. - In
**Figure 3b**the charge is shown as 1, which implies 100 percent of the battery voltage. Suppose now that at this time the switch is turned off. This instant is the beginning of discharge for the capacitor. - The discharge takes place through the resistors
*R*_{1}and*R*_{2}, which form a closed loop with the capacitor. The voltage across the capacitor changes from the initial value to zero in a smooth way and depending on the time constant defined by*C*and resistors*R*_{1}and*R*_{2}. - In
**Figure 3b**,*T*represents the time constant (as you remember, it takes approximately**5× the time constant**for the circuit to settle to a new value after a change in the voltage has occurred).

**Figure 3** Current change in the circuit containing a discharging capacitor. (a) The DC circuit containing a capacitor. (b) After the switch is turned off following being on, the discharge current starts from some value and decays to zero.

- You May Also Read: Inductors in AC Circuits

For a better understanding of what happens in an AC circuit containing a capacitor, we first assume a square wave AC signal. When the connection is made, the capacitor starts charging, but after it is charged (or before it is fully charged, depending on the capacitance), the half cycle terminates and the polarity changes.

The charged capacitor now must discharge and start charging in the opposite polarity. This process is shown in **Figure 4**. Because this charging and discharging happens continuously while the circuit is switched on, there is always some current flowing in the wires connecting the source to the capacitor, but not inside the capacitor. **Thus**, the circuit has a current, which can be measured.

**Associated with this current**, according to **Ohm’s law**, an ohmic value (that is measured in ohms) can be determined. The latter represents a current limiting entity, while it is not a resistance as in a resistor. It does not turn the electricity into heat.

**Figure 4** Current due to a capacitor in a circuit with a square wave voltage.

Similar to what happens with the inductors, the associated current limiting entity is called reactance, and because it stems from a capacitor, it is called **capacitive reactance**.

**Capacitive reactance:** Reactive effect of a capacitor when connected to AC electricity, measured in ohms. Capacitive reactance determines the current in a circuit containing a capacitor.

**Figure 5** Effect of capacitance and frequency on the current in AC circuits with a capacitor. (a) **Lower frequency**: there is enough time for charging and discharging currents to fall to zero. (b) **Higher frequency**: there is not enough time for current during charging and discharging to become zero; **result**: more current in the case of higher frequency.

**Figure 5**shows two different possible cases for the current in a circuit with a square wave power source and a capacitor.- In
**Figure 5a**the capacitor is fully charged before it starts to discharge; in**Figure 5b**the capacitor does not get sufficient time to charge during a half cycle. - In
**Figure 5a**, by the end of each half cycle, the current has dropped to zero, while in**Figure 5b**the current has a nonzero value. We can say that the average magnitude of the current in**Figure 5b**is larger than the current in**Figure 5a**because there is less duration of zero value current. - The difference between the current values is due to the capacitance of a capacitor. It can be seen that a larger capacitance leads to a larger current.
**Figure 5**also reflects the effect of frequency. The larger the frequency is, the less a capacitor has time to completely charge and discharge. Thus, the average current is larger because there are no or smaller periods of zero current.**Figure 5**also shows that the maximum current (peak value of current) corresponds to the instance where the applied voltage is zero (the voltage changes sign; it passes through zero). This is more evident for a sinusoidal signal, as shown in**Figure 6**because the changeover from charge to discharge (and this is when the current has its maximum absolute value) starts at the time the voltage changes sign.

Moreover, for the case of a pure capacitor connected to an AC source (sinusoidal waveform) the circuit current reaches its maximum (minimum) value at 90° before the applied voltage reaches its maximum (minimum). This is the opposite of the case for an inductor, as shown in **Figure 6**.

**Figure 6** Current due to a capacitor in a sinusoidal wave AC circuit.

An observation of the current in **Figure 5** reveals that the line frequency and the capacitance of a capacitor in an AC circuit have a direct effect on the **magnitude of the current**.

A larger capacitor (higher capacitance) takes more time to charge (or discharge), and a smaller capacitor takes less time to charge (or discharge). More time for charging and discharging implies **higher current** because current does drop to zero after charging.

A higher frequency implies smaller cycle time and, thus, less available time for charging and discharging.

Less available time implies that the current does not get sufficient time to reach zero; thus, more nonzero current (higher current). Therefore, the current in a circuit containing a capacitor is proportional to both the capacitor capacitance and the source frequency.

In consonance with what was said about an **inductor in an AC circuit**, the effect of a capacitor is exhibited by its capacitive reactance.

Capacitive reactance of a capacitor is determined from

\[\begin{matrix} {{X}_{C}}=\frac{1}{2\pi fC} & {} & \left( 1 \right) \\\end{matrix}\]

and is measured in ohms. In **Equation 1**, *f* is the line frequency in Hz and *C* is the capacitance in farads. It implies that as the frequency increases the value of *X _{C}* decreases. Also, as

When the frequency of an AC source connected to a circuit containing a capacitor increases, the capacitive reactance of the circuit decreases and circuit current increases.

What is the circuit current when a 12 V, 60 Hz electricity source is connected to a 51 μF capacitor?

**Solution**

Capacitive reactance is calculated from **Equation 1** as

$\begin{align} & {{X}_{C}}=\frac{1}{2\pi fC}=\frac{1}{2\pi *60*0.000051}=52\Omega \\ & I=\frac{12}{52}=231mA \\\end{align}$

What is the reactance of the capacitor in **Example 1** if it is connected to a signal with 10,000 Hz frequency?

**Solution**

Using the new value of frequency in **Equation 1** gives

\[{{X}_{C}}=\frac{1}{2\pi fC}=\frac{1}{2\pi *10,000*0.000051}=0.321\Omega \]

From this example, you notice better a fact about capacitors in AC circuits. Whereas for 60 Hz frequency the reactance of a capacitor is 52 Ω, this value reduces to 0.3 Ω when the frequency is 10 KHz. For radio and TV signals that have a much higher frequency this value is almost zero. This means that a capacitor behaves like a solid connection.

At very high frequencies a capacitor behaves as a solid connection.

When a capacitor is connected to a 6 V, 50 Hz power source, the current is 500 mA. What is the capacitance of the capacitor?

**Solution**

$\begin{align} & XC=\frac{6}{0.5}=12\Omega \\ & \frac{1}{2\pi *50*C}=12 \\ & C=\frac{1}{2\pi *50*12}=265\mu F \\ \end{align}$

Capacitors in series are shown in **Figure 7**. We need to find the one equivalent capacitor that replaces those in series. The rule for series capacitors is the opposite of the rule for resistors and inductors. For series capacitors,

**Figure 7** Capacitors in series.

\[\begin{matrix} \frac{1}{{{C}_{eq}}}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}+\frac{1}{{{C}_{3}}}+\cdots & {} & \left( 2 \right) \\\end{matrix}\]

If both sides of the above equation are multiplied by 1/2π*f*, then we get

\[\frac{1}{2\pi f{{C}_{eq}}}=\frac{1}{2\pi f{{C}_{1}}}+\frac{1}{2\pi f{{C}_{2}}}+\frac{1}{2\pi f{{C}_{3}}}+\cdots \]

Which is

\[\begin{matrix} {{X}_{eq}}={{X}_{1}}+{{X}_{2}}+{{X}_{3}}+\cdots & {} & \left( 3 \right) \\\end{matrix}\]

Thus, for finding the equivalent capacitor for series capacitors, **Equation 2** must be used; however, if the reactance values are employed, they add together.

In a previous repair job in a circuit, three capacitors had been put in series with each other instead or the damaged capacitor. If these capacitors are 47, 68, and 100 μF, find the value of the capacitor in the original circuit.

**Solution**

Using **Equation 2**, we find

\[\frac{1}{{{C}_{eq}}}=\frac{1}{47}+\frac{1}{68}+\frac{1}{100}=\frac{1}{21.74}\]

Thus, the original capacitor has been a 22 μF (the nearest standard value capacitor).

We may need to put capacitors in parallel to obtain a capacitance value that cannot be found among standard capacitors. Capacitors in parallel are shown in **Figure 8**.

The rule for capacitors in parallel is easier. You add their capacitances together. That is,

$\begin{matrix} {{C}_{eq}}={{C}_{1}}+{{C}_{2}}+{{C}_{3}}+\cdots & {} & \left( 4 \right) \\\end{matrix}$

To find the equivalent reactance, however, they must be treated as resistors in parallel. This stems from multiplying the two sides of **Equation 4** by 2πf. Thus,

$2\pi {{C}_{eq}}=2\pi {{C}_{1}}+2\pi {{C}_{2}}+2\pi {{C}_{3}}+\cdots $

Or

\[\begin{matrix} \frac{1}{{{X}_{C}}}=\frac{1}{{{X}_{C1}}}+\frac{1}{{{X}_{C2}}}+\frac{1}{{{X}_{C3}}}+\cdots & {} & \left( 5 \right) \\\end{matrix}\]

A 33, 47, and 10 μF capacitor are put in parallel together. How much is the resulting capacitor?

**Figure 8** Capacitors in parallel.

**Solution**

It is sufficient just to add the three values together

$C=33+47+10=90\mu F$

Note that 90 μF can also be obtained from putting two standard 68 and 22 μF capacitors in parallel.

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]]>The post Inductors in AC Circuits appeared first on Electrical A2Z.

]]>The phenomenon just discussed is called **Lenz’s law**.

Before a statement for Lenz’s law is made, let’s define the term **induction**. When an electromagnetic force (emf) is generated in an inductor, a more common statement employed is that “an emf is induced in the inductor.” Thus, induction is used instead of generation, which goes well with inductor and inductance.

**Lenz’s law states** that an induced emf always opposes its cause, meaning that the direction of the current owing to the induced voltage is opposite to the direction of the current that made the induction. As a result of this opposition, the current in a DC circuit containing an inductor after a switch is turned on is not an abrupt change and, instead, is a gradual growth from zero to a steady value. This is illustrated in **Figure 2**.

**Induction:** Generation of electricity in a wire when the magnetic flux is cut by the wire (e.g., when the wire moves with respect to a magnetic field or the strength and/or direction of the magnetic field varies).

**Figure 1:** Moving a magnet near a coiled wire induces electricity in the coil.

**Figure 2** Gradual change of current from zero to 1.5 A in a circuit containing a coil.

As can be observed, it takes a short time for the current to settle to its steady value. The steady value of the current depends on the total resistance in the circuit.

The length of time that takes for the current to reach its final value depends on (1) the value of inductance of the coil (measured in henries) and (2) the total resistance of the circuit (measured in ohm).

For the sake of better understanding, first, let’s assume a square wave AC signal provided by a source and connected to an inductor, shown in **Figure 3**. Furthermore, we assume that the resistance of the coil is negligible and can be ignored.

When the circuit containing an inductor is turned on, a current starts to develop in the circuit, starting from zero and reaching a high value, according to the curve as shown in **Figure 2**.

But, before the current reaches its maximum value the voltage has a sudden change and the polarity changes. Consequently, a current in the opposite direction starts to develop in the same manner.

On the basis of the gradual development of current and the fact that the AC voltage switches back and forth from negative to positive and vice versa, it is easy to understand that for the circuit of **Figure 3** the current looks as illustrated in **Figure 4**. It can be further understood from **Figure 4** that

- There is always a current flowing in the circuit.
- Current changes between positive and negative, i.e., it is AC.
- Current has its maximum (and minimum) value at the time that the voltage is zero (points A, A′, A″, etc., and points B, B′, B″, etc.).

**Figure 3** Inductor connected to an AC source with a square waveform.

**Figure 4** Current in an inductor connected to an AC source with a square waveform.

**Figure 5** Current in an inductor connected to an AC source with a sinusoidal waveform.

If instead of a square wave, we have a sinusoidal AC waveform, which is the common case, the current is also sinusoidal and looks like **Figure 5**. Its peaks correspond to the instants the voltage value is zero.

Note that, as depicted in **Figure 5**, in an AC circuit containing an inductor (only) the current is 90° behind the applied voltage. This means that the current reaches its maximum (minimum) after the voltage has reached its maximum (minimum), and the difference between the time that voltage reaches its peak and the time that current reaches its peak is 1/4 of a cycle or 90°. This is always the case for such a circuit.

We established that when an inductor is connected to an AC source, a current exists in the circuit. This implies that, on the basis of **Ohm’s law**, there is some resistance (to electric flow) in the circuit associated with this current. This so-called resistance (it is not really resistance because there is zero ohm value in the circuit, as assumed) can be determined.

But, first, let’s establish another fact that the current in this circuit changes with the frequency of the AC electricity. This can be easily understood from the comparison of currents shown for two cases in **Figure 6**.

Again, for better understanding, a square waveform is assumed. The waveform in **Figure 6b** has 2× the frequency as in **Figure 6a**. The current in **Figure 6b** is smaller than the current in **Figure 6a**. The reason for the difference in the value of current is that in **Figure 6b** there is not sufficient time for reaching the peak value before the polarity switching occurs. The same is true for a sinusoidal waveform.

**Figure 6** Comparison of currents in two AC square wave signals when frequency doubles. (a) Lower frequency: larger peak current. (b) Higher frequency: smaller peak current.

It can be followed from **Figure 6** that the current through an inductor has an inverse effect with the frequency of the AC line.

The effect of the insertion of an inductor in an AC circuit is exhibited in the form of impedance to the current, but because it is not a resistance (with ohm value that can be measured), it is called **reactance**.

So, a reactance in an AC circuit is what exhibits a resistance to the flow of current, but it is not because of a resistive element that converts the electrical energy to heat. Reactance is measured in ohms. To specify that reactance is due to an inductor, when necessary, it is more particularly addressed as **inductive reactance**.

The reactance of an inductor, denoted by *X _{L}*, depends on the frequency and the inductance and can be found from

**Reactance:** The apparent resistance (measured in ohms) that a capacitor or an inductor when connected in an AC circuit exhibits to the flow of electricity. Reactance depends on AC frequency and unlike a resistor, the energy involved does not convert to heat.

**Inductive reactance:** The apparent resistance to the flow of electricity exhibited by an inductor in an AC circuit. It is measured in ohm and determines the current in the inductor based on the applied voltage.

$\begin{matrix} {{X}_{L}}=2\pi fL & {} & \left( 1 \right) \\\end{matrix}$

Where π is a constant (π = 3.14159265), *f* is the frequency of the AC electricity measured in Hz, and *L* is the inductance measured in henries.

What is the inductive reactance of an inductor when connected to 60 Hz electricity? The inductor has an inductance of 0.05 H.

**Solution**

It follows directly from **Equation 1** that

${{X}_{L}}=2\pi fL=2\pi *60*0.05=18.85\Omega $

Find the current for an inductor when connected to a 9 V, 50 Hz AC power source, if its inductance is 20 mH.

**Solution**

$\begin{align} & {{X}_{L}}=2\pi L=2\pi *50*0.020=6.28\Omega \\ & I={}^{9}/{}_{6.28}=1.43A \\\end{align}$

What is the current in the inductor in **Example 2** if the frequency of the power source is 60 Hz?

**Solution**

$\begin{align} & {{X}_{L}}=2\times 3.14\times 60\times 0.020=7.64\Omega \\ & I={}^{9}/{}_{7.54}=1.19A \\\end{align}$

You can observe from the above examples that when the frequency goes up, the reactance of an inductor goes up proportionally to it and the current in the inductor decreases.

When the frequency of electricity connected to an inductor goes up, the reactance of the inductor increases proportionally to the frequency and the current through the inductor decreases.

In the same way that resistors could be combined in series and in parallel with each other, inductors can be put in series or in parallel, when necessary. **Figure 7** illustrates three inductors in series. The relationships for a number of inductors put in series with each other are

\[\begin{matrix} L={{L}_{1}}+{{L}_{2}}+{{L}_{3}}+\cdots & {} & \left( 2 \right) \\\end{matrix}\]

\[\begin{matrix} {{X}_{L}}={{X}_{{{L}_{1}}}}+{{X}_{{{L}_{2}}}}+{{X}_{{{L}_{3}}}}+\cdots & {} & \left( 3 \right) \\\end{matrix}\]

*L* is the equivalent inductor for all the inductors in series, and *X _{L}* is the equivalent inductive reactance of them.

**Figure 7** Inductors in series.

When inductors are connected in series, all of them share the same current. Accordingly, similar to the case for resistors, the total voltage divides among them, proportional to their reactance values.

What is the current in three inductors in series, when the applied voltage is 48 V? The frequency of the power supply is 60 Hz, and the inductors are 5, 10, and 15 mH.

**Solution**

For this problem, either **Equation 2** or 3 can be used. Using the former takes less calculation because similar calculations must be repeated for each inductor to find its reactance.

$\begin{align} & L=5+10+15=30mH \\ & {{X}_{L}}=2\pi *60*0.030=11.31\Omega \\ & I={}^{48}/{}_{11.31}=4.24A \\\end{align}$

If one needs to connect inductors in parallel with each other, as shown in **Figure 8**, then the rule of parallel inductors can be used. You have noticed that the rule for inductors in series is the same as that used for resistors in series. The same is true for inductors in parallel with each other. Either of the following equations can be used.

\[\begin{matrix} \frac{1}{L}=\frac{1}{{{L}_{1}}}+\frac{1}{{{L}_{2}}}+\frac{1}{{{L}_{3}}}+\cdots & {} & \left( 4 \right) \\\end{matrix}\]

\[\begin{matrix} \frac{1}{{{X}_{L}}}=\frac{1}{{{X}_{{{L}_{1}}}}}+\frac{1}{{{X}_{{{L}_{2}}}}}+\frac{1}{{{X}_{{{L}_{3}}}}}+\cdots & {} & \left( 5 \right) \\\end{matrix}\]

Three inductors with inductance 5, 10, and 15 mH are put in parallel with each other and connected to 120 V line. If the frequency of the line is 60 Hz, what is the current in the line?

**Figure 8** Inductors in parallel.

**Solution**

For this problem, it is easier to use **Equation 4**.

\[\frac{1}{L}=\frac{1}{5}+\frac{1}{10}+\frac{1}{15}=\frac{11}{30}\]

The equivalent inductance for the three inductors is

$L=2.727mH$

and the reactance of them for 60 Hz frequency is

${{X}_{L}}=2\pi *60*0.0002727=1.08\Omega $

Thus, the current in the line is about 120 A (120 V ÷ 1 Ω).

Note that the resulting inductance is smaller than that of the smallest (5 mH) inductor.

Understand that the current in the line when the three inductors with inductance 5, 10, and 15 mH are connected in parallel is very high (120 A). Suppose that this is a case in reality when you had to replace a damaged inductor but that you did not have the exact value and you had decided to make the equivalent inductor by combining the available ones.

Suppose that you had made a mistake and instead of connecting these inductors in series you had connected them in parallel. How many times the current calculated in **Example 5** is compared with the expected current? (The line is 120 V and 60 Hz.)

**Solution**

When these three inductors are in series, their equivalent inductance is

$L=5+10+15=30mH$

And their reactance is

${{X}_{L}}=2\pi *60*0.030=11.3\Omega $

And

$\begin{align} & I=\frac{120V}{11.3\Omega }=10.6A \\ & Since, \\ & \frac{120A}{10.6A}=11.32 \\\end{align}$

Thus, the current is 11.32× larger than expected. This can be very damaging in a real situation.

The three parallel inductors in **Example 5** (with inductance 5, 10, and 15 mH) were, in fact, taken from a device that worked with 400 Hz. Find what the current was through each individual inductor and what the total current was for 48 V applied voltage.

**Solution**

First, we need to find the reactance for each inductor.

$\begin{align} & {{X}_{L1}}=2\times 3.14\times 60\times 0.005=12.566\Omega \\ & {{X}_{L2}}=2\times 3.14\times 60\times 0.010=25.133\Omega \\ & {{X}_{L3}}=2\times 3.14\times 60\times 0.015=37.699\Omega \\\end{align}$

The three currents for each inductor are, respectively,

$\begin{align} & {{I}_{1}}=\frac{48}{12.566}=3.82A \\ & {{I}_{2}}=\frac{48}{25.133}=1.91A \\ & {{I}_{3}}=\frac{48}{37.66}=1.27A \\\end{align}$

And the total current is the sum of the three currents

$I=3.82+1.91+1.27=7A$

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]]>The post Inductors in DC Circuits appeared first on Electrical A2Z.

]]>The effect of an inductor in an electric circuit is always to oppose a change in the circuit current. For instance, at the instant the switch is closed, current tends to increase from zero to a value that depends on the voltage and resistance in the circuit.

An inductor wants to keep the current unchanged. Eventually, after a short period of change, the current settles at a value and remains constant for this DC circuit.

- Adding an inductor in an electric circuit is analogous to adding a flexible (stretchable) tank (reservoir) to a hydraulic pipeline. When the tank stretches, its volume changes.
- When the valve in the pipeline is opened to let water flow, the flowing water must fill the flexible reservoir before it can continue its normal flow. Thus, part of the flow is used to fill the reservoir.
- Once the reservoir is filled, water flow continues in the pipeline at its normal rate. In other words, the flow of water is initially resisted and limited for a short period of time but afterward will have a continuous and constant current.
- The effect of an inductor in an electric circuit is to oppose a change in the circuit current (keeping the current unchanged).
- Here again, it is interesting to know how long it takes for the electric current in
**Figure 1b**to reach its constant value or for the water in**Figure 2b**to reach its normal flow.

Associated with the circuit resistance *R* and inductance *L* values, we may define a time constant (similar to the case for a capacitor) as follows:

\[\begin{matrix} \tau =\frac{L}{R} & {} & \left( 1 \right) \\\end{matrix}\]

When *L* is defined in **henries** and *R* in **ohms**, τ will be in **seconds**. This equation implies that for smaller values of *R* the duration of the effect of the inductor is longer. It takes approximately **5× the time constant** for the current to reach its constant value.

**Figure** **1**: Adding an inductor in series with a resistor in a DC circuit. **(a)** Original circuit. **(b)** Circuit with an inductor.

**Figure 2:** Analogy of a hydraulic system to an electric circuit when an inductor is added. **(a)** A circuit without an inductor. **(b)** Circuit with an inductor.

What is the time constant in a circuit containing a resistor and a coil in series? The resistor has a value of 51 Ω, and the coil has an inductance of 11 mH and a resistance of 4 Ω.

**Solution**

Substituting in **Equation 1** gives

\[\tau =\frac{L}{R}=\frac{0.011}{51+4}=0.0002\sec \]

It takes approximately 5× the time constant for the effect of an inductor in a DC electric circuit to disappear.

In the same way that a capacitor causes a delay for the voltage change in a circuit, because of its intrinsic electricity storage property, an inductor in a circuit causes a delay. However, the delay caused by an inductor is because of the property of an inductor to resists a change in the current in the circuit it is a part.

The time constant for a circuit containing an inductor and a resistor is defined by

\[\tau =\frac{L}{R}\]

When *L* is henries and *R* is in ohms the calculated value of the time constant is in seconds.

All that was said about the time delay for changing voltage in a circuit **containing capacitors and resistors** and the corresponding graphs is true for a circuit having an inductor in it. However, in this case, the values and the percentage of change shown at the end of periods *T*, 2*T*, 3*T*, etc., corresponds to **changes of current in a circuit** and not voltage.

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]]>The post Magnetism, Electromagnetism & Magnetic Materials appeared first on Electrical A2Z.

]]>Certain metals of the iron family can become magnetized if their atoms align such that this property is enhanced. This alignment can be due to contact with another magnet, electricity influence, being placed in a magnetic field (being in the vicinity of a magnet), or by natural events in nature.

As a result of magnetism, magnetic media can attract or reject other media by inserting attraction force or rejection forces on them.

In nature, the metallic form and salts (compounds) of the ferromagnetic materials can be found as magnets. This is the way magnetism was first recognized. **Ferromagnetic materials** are those that can relatively easily become magnets, and they can retain their magnetism.

Some other materials, called **diamagnetic materials**, can never become a magnet.

A third category is called **paramagnetic materials**, which can exhibit a very small degree of magnetism, but they can never retain it.

**Ferromagnetic:** Type of material from the iron family that is suitable for magnetization.

**Diamagnetic:** Materials without any significant magnetism property, such as wood and clay. They can never become a magnet.

**Paramagnetic:** Metals with very little (negligible) magnetic property that never can retain any magnetism after being in a magnetic field, as opposed to ferromagnetic materials.

Examples of ferromagnetic material are iron, steel, chromium, and nickel. Examples of diamagnetic material are copper, lead, mercury, silver, tin, and graphite (carbon). Examples of paramagnetic material are tungsten, aluminum, magnesium, sodium, and titanium.

A magnetic field is a part of the space around a magnet that the magnetic influence can be felt and measured. For example, there is a magnetic field between the two poles of a horseshoe magnet, as shown in Figure 1. Also, the Earth has a magnetic field, between the North Pole and the South Pole.

To elaborate on the meaning of field, consider the Earth also having a gravity field. As one moves away from the Earth, the gravity field becomes weaker until it becomes so small that it can be ignored. At that point, everything becomes weightless.

- You May Also Read: Magnetic Flux Definition & Unit

A field can be assumed to consist of a number of imaginary lines that show the orientation of the field. For example, for a magnetic field we always consider the direction to be from the North Pole to the South Pole, and the gravity field around the Earth, it is always toward the center of Earth. For a magnetic field, these lines are called flux lines.

**Magnetic field:** Limited part of the space around a magnet where the magnetic effect exists.

**Flux lines:** Imaginary lines around a magnet indicating the direction of the magnetic effect. Stronger magnetism effect implies more flux lines through the same area.

As well as orientation, a field has strength. This strength is not constant. A magnetic field can be strong, or it can be weak at a certain point. As one moves away from the magnetic field, it becomes weaker.

For instance, the magnetic field in Figure 1 is stronger at point A than at point B. A stronger field can be assumed to have more lines (magnetic flux lines) per area; that is, the flux lines are denser.

**Figure 1** Magnetic field.

**Figure 2** Direction of a magnetic field inside and outside the magnet body.

As you know, a magnet can be in various forms, a yoke or a bar, for instance. No matter what the shape of a magnet is, the magnetic field direction is always from the North Pole toward the South Pole outside the magnet body along the flux lines and from south to north inside the magnet. The magnetic flux lines form closed curves. This is as shown in Figure 2.

A magnetic field can be created by electricity. The core body of such a magnet loses its magnetic property when electricity is turned off. However, many magnets do retain their magnetism. These are called permanent magnets.

Today, strong permanent magnets can be made from alloys of rare metals and iron. They are employed in many motors and recently in some small **generators** such as those in some **wind turbines.**

**Permanent magnet:** Magnet with the permanent magnetic property that cannot be turned on and off or altered.

The magnetic property of a permanent magnet is due to its atomic structure. Thus, if a bar magnet is broken in the middle, each piece by itself becomes a perfect magnet having two poles, one north, and one south.

In the same way, if each of the two broken magnets is divided again into pieces, each piece becomes a magnet. Inversely, two magnets can be attached together only when their opposite poles are put together. This is why opposite poles attract and similar poles repel each other.

Heat is not good for a permanent magnet. A magnet can lose it magnetism by being heated. This changes the atomic structure of the metal.

So, if you want to destroy a magnet, you can heat it. Alternatively, if a magnet is struck hard by a hammer, it will lose its magnetism.

A ferromagnetic material can easily become a magnet.

One of the properties of electricity is **magnetism**. When an electric current passes through a conductor, an electric field is generated around the conductor. This can be checked by placing a compass near the current carrying wire.

The compass needle aligns itself with the field generated by the electric current. This is shown in **Figure 3**. Each piece of the wire creates a magnetic field around it. The field lines are circles perpendicular to the wire segment; thus, depending on the point on this circle, the direction of the magnetic field changes.

The needle of a compass always stays tangent to the magnetic lines. This magnetic field is all around the wire; so, for a straight wire, the magnetic field forms a cylinder.

**Figure 3** Magnetic field of a current carrying wire.

**Figure 4** Right-hand rule to find the direction of the magnetic field around a wire.

Note that in Figure 3 the current direction is from positive to negative, using the conventional direction of the current. The direction of the magnetic field generated this way can always be found by the right-hand rule, as shown in **Figure 4**. **That is,** if the thumb in the right hand is aligned with the current direction, the fingers show the magnetic field direction.

The **strength of this magnetic field** depends on:

**(1)** The current in the wire and;

**(2)** The distance from the wire. The higher the current is, the stronger the field is.

**(3)** Also, the farther from the wire, the weaker the field is.

An inductor is a wound wire: when connected to electricity, it carries a current. In the previous section, we learned that if a straight wire carries a current, then a cylindrical magnetic field is created around it.

In the case of a coil (an inductor) the wire is circular, and normally there are a large number of loops that carry the same current. In this sense, there are many magnetic fields created because of the electric current, and they interact with each other owing to the loops in the wire. Figure 5 shows what happens in such a case.

**Figure 5** Magnetic field of a coil. (a) Field direction for each individual loop. (b) Resultant magnetic field of all loops.

For each loop in the coil, the **magnetic fields** form a torus around the loop. Magnetic lines at two points are shown in **Figure** **5a**.

At some points (such as A and A′), magnetic lines are in the opposite direction and cancel each other, but at some other points (such as B and B′) they combine with each other and their effects add together. The resultant magnetic field for each loop is along a line perpendicular to the loop (along with the loop axis).

Magnetic fields of all the loops, then, add together to construct the magnetic field of the coil. This field is along the coil axis, as shown in **Figure 5b**.

**In conclusion**, when a coil carries electric current, a magnetic field is generated. The direction of this magnetic field depends on the direction of the current.

**Figure 6** shows a coil and the direction of its magnetic field based on the electric current direction. As noted before, the direction of the magnetic field is from north to south outside of the magnet generated by the current flowing in the coil (from + toward −).

**Figure 6** Direction of the magnetic field of a coil.

**Figure 7** Right-hand rule to find the direction of the magnetic field of a coil.

**Moreover**, the field strength depends on the current as well as other factors, such as the physical dimensions of the coil and the material of its core.

To enhance the magnetic field of a coil, a core made of a ferromagnetic material is inserted inside the coil.

The right-hand rule again can be helpful here to determine which side is the north and which side is the south, knowing the current direction. This is shown in Figure 7.

This rule **states that** if your right-hand fingers are in the direction of the current, then your thumb indicates the direction of the magnetic field (the thumb shows the north pole of the resulting magnetic field). Study this figure together with Figure 6. Also, note the difference between Figures 4 and 7.

The right-hand rule states that if your right-hand fingers are in the direction of the current in a coil of wires (from + toward −), then your thumb indicates the north pole of the coil magnetic field.

The magnet created by the coil in Figure 6 is not a permanent magnet. As soon as the current is turned off, the magnetic field vanishes and the magnetic effect disappears. This arrangement (a coil and a core in it) is called an **electromagnet**, and it has many industrial applications.

As you might imagine, it can be used for collection of (ferromagnetic) metals and for exerting force, in addition to many other applications such as opening and turning a switch or a valve on and off.

**Electromagnet:** A (not permanent) magnet made by a wire coil wrapped around a ferromagnetic core when carrying an electric current. The magnet can be turned on and off or its strength can be adjusted.

The post Magnetism, Electromagnetism & Magnetic Materials appeared first on Electrical A2Z.

]]>The post Analog & Digital Multimeter appeared first on Electrical A2Z.

]]>A multimeter is a must-have tool for someone working with electricity and electronics. It is always necessary to verify values, to measure voltage, current, and resistance values in a circuit for diagnostic purposes, or to identify a damaged component.

Analog multimeters work based on the deflection of a needle from its zero position point. During a measurement, one needs to read the correct value just under the needle.

The main part of an analog meter is a galvanometer that consists of a small winding. The winding can rotate about a pin, and a needle is attached to it. It also has a spring to return the needle to the home (zero) position. Depending on the intensity of the current through the winding, it deflects from its rest position. **Thus**, a galvanometer is a current-sensitive device.

By making some changes to a galvanometer and adding more components, it can be used as an **ammeter**, a **voltmeter**, an **ohmmeter**, and other measurement devices. The deflection of the needle in an analog meter is always from left to right.

**Galvanometer:** A device consisting of a needle attached to a coil that can rotate around a pin shaft as a result of an electric current flowing through the coil. The coil behaves also as a spring, limiting the motion of the needle. A galvanometer is used to measure electric current, but the needle position can be graduated for other electric entities, like voltage.

**Ohmmeter:** Device for measurement of electric resistance.

**Figure 1** shows a typical analog multimeter. It has a selector rotary switch by which one selects the category and range of values to be measured.

Not all the meters are built the same way and have the same range of values and the same number of terminals. More details of the switches and terminals in the multimeter of Figure 1 are illustrated in **Figure 2**.

In addition to the main selector switch for changing from volt to ohm or amp and selecting their range of values (e.g., 100 mA, 500 mA, 25 V, and 500 V), there is another switch for AC, –DC, and +DC.

There are two main terminals for connecting the black (common) and red leads, but additional terminals are used for 10 A current (current larger than 500 mA and up to 10 A), 500 V and 1000 V (higher voltages).

For measuring these relatively high values the black lead still goes to the common terminal, but the red lead must be inserted into the appropriate hole.

**Figure 1** Typical analog multimeter.

**Figure 2** Switches and terminals on an analog multimeter.

**Figure 3** Various graduations on a typical analog multimeter.

**Figure 3** illustrates the set of various readings for the multimeter shown in Figure 1. As can be understood, these graduations are different, and for one position of the needle, there are various numbers.

Each graduation corresponds to one or more selector switch positions. For instance, for a selector switch range of 2.5, 25, and 250 (see Figure 2) one must use the graduations ending in one of these numbers, whereas for other selections the numbers are between 0 and 10 (which must be multiplied by a power of 10, accordingly).

Also, in the meter shown, the AC values are shown in red, whereas the DC values are in black.

Note that the zero value for ohm measurement (top scale) is to the extreme right (all the other zero values are on the extreme left). This is because a higher resistance value leads to a lower current value (see Ohm’s law in this chapter) and vice versa.

A typical digital multimeter (DMM) is depicted in **Figure 4**. It has a rubber casing that protects it against shocks and scratches. Normally, it is easier to work with a digital multimeter because it directly gives the measured values in numbers.

A DMM does not have any moving parts and works based on converting an analog reading to a digital value.

**Digital multimeter (DMM):** A device to measure resistance, current, voltage, and other electrical parameters, in which the reading is automatically adjusted and displayed by 3 or 4 digits, as compared to analog multimeters in which the position of a rotating needle represents a measured value.

Similar to an analog multimeter, a DMM has a common terminal to which the black lead is connected and the red lead goes into one of the other terminals, depending on the design of the DMM.

In the DMM shown in Figures 4 and 5 the red terminal is inserted in the hole to the right of the common (com) terminal, but for measuring current (based on its value if it is in the order of amps or much smaller) the other two holes are used.

It is always safer to use the higher current terminal first (the one on the left) before using the middle terminal for higher precision. In this way, you do not subject the fine meter to a current much larger than its capacity, which can damage the meter (or blow its fuse).

**Figure 4** Typical digital multimeter (DMM).

**Figure 5** Selector switch and the display of a typical DMM.

In the multimeter of Figure 4 the switch can be positioned to off, when the meter is not in use, or it can be set for AC voltage, DC voltage, small DC voltage (mV), resistance (ohm, Ω), and current (Ampere A, milliamp mA, and μA micro-amp). There is one more selection for the diode (test).

Better DMMs have four digits, as shown in Figure 5. Some cheaper DMMs have only three digits. Normally, on the basis of the measured values, the scale is automatically adjusted; for instance, in Figure 5 the scale is automatically set to kΩ (appearing on the screen).

A resistor’s resistance or the resistance of the part of a circuit can be measured by an **ohmmeter** or a multimeter. An ohmmeter or the ohmmeter part of a multimeter works based on measuring the current in a circuit where the item to be measured constitutes the main load of that circuit.

Nevertheless, the graduations are made in terms of resistance values, not current values. In this sense, the following differences exist between an ohmmeter and a voltmeter or ammeter:

- An ohmmeter needs a battery to power its circuit, whereas for measuring current or voltage, no battery is needed because there is already a current (when the switch is closed).
- Because current is measured, using Ohm’s law, for a smaller resistance, there is a large current and for a larger resistance, the current is smaller. This necessitates that the zero reading for resistance (in an analog meter) be at the extreme right side of the scale (the highest current) and the larger resistance values are on the left side.
- Because the ohmmeter battery loses its strength with time, an ohmmeter has a zero adjustment knob that is used to bring the needle to zero reading. It is always necessary to adjust the zero reading before any measurement is made with an analog ohmmeter. This can be done by directly putting together the ends of the two leads, then turn the adjustment knob to bring the needle to zero.
- Measuring resistance must always be performed when the power to a circuit (the resistance of a part of which is to be measured) is turned off. Otherwise, the reading will be erroneous.

An ohmmeter can be used for measuring and verifying the value of any resistor or resistive element. In addition, it can be used for checking the continuity of a circuit to see if any part of a circuit is open or there is any short (two points unnecessarily contacting) in the circuit.

When measuring the resistance of the part of a circuit, the power to the circuit must be off.

An ohmmeter (usually, only analog meters) can also be used to check if a capacitor is good or damaged. If a capacitor is damaged, it either becomes short (when the two plates of the capacitor contact each other) or open (contacts are lost).

**To check a capacitor**, it is connected to an ohmmeter. If it is short, then it shows a high current (near zero resistance) and the meter needle stays at the same point.

If it is open, then it shows a very high resistance (the needle stays in the very left side of the meter). If the capacitor is good, the needle of the meter quickly moves to the right and slowly goes back to the left.

In a similar way, we can use an ohmmeter to see if an inductor is fine or if it is damaged. A digital or an analog multimeter can be used.

Note that in a DMM when the value to be measured is beyond the range of values selected an overload sign will appear on the screen by the letters “OL” as shown in Figure 6.

Specifically, when measuring resistance, an open circuit (equivalent to a very high resistance measured) is realized if the meter shows **OL.**

**An inductor** can be shorted or can be open. If after connecting its two ends to an ohmmeter high resistance is noticed, then it is open. A short inductor shows zero resistance. A good inductor shows a small value of resistance but not zero.

**Figure 6** Very high-value resistance or an open circuit.

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]]>The post Capacitors in DC Circuits appeared first on Electrical A2Z.

]]>**Figure 1a** is a simple DC circuit with a resistive load, to which a capacitor is added, as in **Figure 1b**. An analogous equivalent hydraulic circuit, for better tangibility, is shown in **Figure 2**.

Adding a capacitor to an electric circuit is equivalent to adding a closed tank with limited capacity or a tall tank (with walls that are tall enough and comparable with the main reservoir) at the end of a pipeline.

Water flows in the pipeline in **Figure 2a** as the electricity flows in the circuit of **Figure 1a**, as long as the reservoir (battery) is not emptied.

Nevertheless, in the case of **Figure 2b**, a current of water flows as long as the small tank is not full. That is, for a short period of time, there is a current in the pipeline, but after a while, water stops flowing.

Moreover, this flow of water is not uniform during the period. At the beginning (when the tank is empty), more water flows but toward the end the flow of water becomes less and less, approaching zero. This is exactly what happens in the electric circuit containing a capacitor, too.

**Figure 1:** Adding a capacitor to a DC circuit. (a) A circuit without capacitor; normal current after the switch is closed. (b) Circuit with a capacitor; current only for a short time no current afterward.

**Figure 2:** Adding a small tank to the end of a pipeline. (a) Hydraulic system analogous to **Figure 1a**. (b) Hydraulic system analogous to the DC circuit with a capacitor, in **Figure 1b**.

When a capacitor is inserted inside a DC circuit, for a short period of time after the switch is turned on, current flows in the circuit. In the beginning, this current is higher but gradually becomes smaller and smaller until it diminishes. This is when the capacitor has charged, and it does not accept an electric charge anymore.

**At this time and afterward**, there is no current flowing in the circuit. Thus, except for a short period, in the beginning, a capacitor in a DC circuit blocks the circuit and does not allow any current.

A charged capacitor contains electricity and behaves like live electricity. It can be dangerous if the voltage is sufficiently high. The voltage level to which a capacitor charge is the same as that of the battery (power supply) if the capacitor becomes fully charged.

If the circuit is broken (the switch is turned off) while the capacitor is charging, it only partially charges and to a voltage less than the applied voltage.

A capacitor in a DC circuit blocks the current, except for only a short period following a change such as after a switch is closed (or opened if already closed).

It is interesting to know how long it takes for a capacitor to charge. In fact, in many timing devices, including electronic watches, this time period is used for timekeeping.

The time for charging or discharging a capacitor has a direct correspondence with the problem for a water tank to fill (under a given water pressure) or empty if already full (based on the height of water and the resistance of the pipeline from which water discharges).

Here a simple rule is given for the calculation of the time for charging a capacitor. Referring to **Figure 1b** the circuit consists of a resistor with resistance *R* and a capacitor with capacitance *C* in series with each other connected to the battery (power supply) of voltage *V*. We define a **time constant** τ (the Greek letter corresponding to English t, and pronounced tow).

**Time**** constant:** Duration of time for a circuit containing a capacitor or inductor and resistors to reach 63 percent of a new value after a change has happened to the circuit (e.g., its power has been turned on or off).

$\tau =RC$

When *R* is defined in ohm and *C* in farad, τ is automatically expressed in seconds. As can be seen, the time constant τ depends on the values of *R* and *C*. It takes approximately 5× the time constant τ for the capacitor to fully charge to voltage *V*. This time starts from the instant that the switch is closed. But, if during this time interval the switch is opened, the capacitor does not fully charge.

The time constant is sometimes shown by

T(uppercase) instead of τ.

A 51 Ω resistor and a 10 mF (milli-farad) capacitor in series are connected to a 12 V battery. How long does it take for the capacitor to charge?

**Solution**

The time constant is calculated first:

$\tau =RC=51*\left( {}^{10}/{}_{1000} \right)=0.51\sec $

The required time for charging the capacitor = 5τ = (5)(0.51) = 2.55 sec

Except for a very short period, in the beginning, a capacitor in a DC circuit behaves as an open circuit and does not allow any current. It takes approximately 5× the time constant for a capacitor to either charge or discharge.

A capacitor has a storage capability for electricity. When it is part of a DC circuit, it exhibits an apparent opposition to a change in the circuit voltage.

When a switch turns on or off the power to an electric circuit, it introduces a change of voltage to the circuit. During turning the power on in a DC circuit containing a capacitor, first, the capacitor is charged. Similarly, at power off, first, a capacitor in the circuit releases its charge to the circuit. Consequently, a delay is associated with both turning on and turning off a DC circuit.

**As a result of this delay**, it takes a short while before the circuit containing a capacitor reaches its equilibrium and comes to its final condition when a change in voltage, either increasing or decreasing, occurs. This delay depends on the capacitance of the capacitor and the resistance in the circuit.

The delay just mentioned is approximately 5× the time constant of the circuit.

The time constant for a circuit containing capacitors with an equivalent capacitance *C* and resistors with an equivalent resistance *R* is

$\tau =RC$

When *R* is in ohm and *C* is in Farad, the time constant is determined in second.

**Figure 3** below shows the **two cases** of increasing and decreasing the voltage in a DC circuit. Because this change in voltage can be any value, it is normalized and its magnitude is shown as 1, which percentage-wise represents 100 percent.

- The figure illustrates the gradual change of voltage during the time it takes for this change. The time constant is depicted as
*T*, and the percentage change after the elapse of each time constant*T*is also shown. **For an increase in voltage**(like turning on a switch), part (a) in figure 3 illustrates that subsequent to the initiation of an increase in voltage, after*T*seconds the voltage increases by 63 percent of the difference between the new and the old values.- In the figure, the old value is zero, but it can be a nonzero value. After 2
*T*sec the voltage is augmented to 86 percent of the change. After 5*T*, 99 percent of the change is reached, which is almost the total change to occur. **Part (b)**of figure 3 illustrates**a decrease in voltage**. If the change is caused by turning the power off, so that the final voltage is zero, 63 percent of the decrease in voltage takes place after one time constant. That is, the circuit voltage drops to 37 percent of its initial value.- Accordingly, after 2
*T*the voltage drops to 14 percent and after 5*T*the decrease is 99 percent and the voltage in the circuit is 1 percent of its initial magnitude. After this time the voltage change is practically zero.

**Figure 3:** Capacitor Charging and Discharging

The delay introduced by a capacitor has many applications, particularly in electronic circuits used in radio, TV, communications, and so on, as well as in electrical applications that need control and regulation.

By changing the time constant, throughRandC, the desired delay can be obtained for any application.

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]]>The post What is Electric Power | Definition, Formula & Examples appeared first on Electrical A2Z.

]]>A good comprehension of the relationships for ** electric power**, unit of power measurement, and the rules that determine power is very important.

You must always bear in mind that for an electric power source (e.g., generator and battery), power rating implies the maximum power that the device can deliver.

For a consumer, (a load) power rating implies the electric power requirement by the device so that within the desired operating conditions (voltage and current) it can function with good efficiency and with minimum risk of damage.

The required electric power must be available to the device so that it functions properly. If that much power is not provided (by the source, the line, or the power supply), some shortcoming can happen that can lead to damage and failures.

**Electric power:** Power in the form of electricity and measured by electrical units (power is the amount of work in 1 sec).

If the current does not change but the voltage doubles, the electric power doubles, and if the voltage does not change but the current doubles, the power doubles. This relationship is

\[\begin{matrix} P=VI & {} & \left( 1 \right) \\\end{matrix}\]

Where *P* is in watts, *V* is in volts, and *I* is in amps.

** Equation 1** clearly shows that for the electric power to be constant (this situation often occurs in practice), if the voltage decreases, then the current must increase or if the current increases, the voltage must decrease. Current increase beyond a rated value is equivalent to overload, which is not desirable.

Measured current in the filament of a lightbulb, when connected to a 12 V battery, is 5 A. What is the electric power consumption of the lightbulb?

**Solution**

Power can be directly found by the product of the voltage and the current:

$P=VI=12*5=60W$

“Energy” is the potential to do *work*, and it can be in different forms, such as electrical energy, nuclear energy, thermal energy (heat), wind energy, and solar energy. “Work” here implies a mechanical or another type of work.

Mechanical work is, for instance, when a weight is lifted or when an engine drives a car. Mechanical work is more tangible compared to other types of work. **For instance**, a motor can do mechanical work, but a battery does not directly do mechanical work. However, it can run a motor that does mechanical work; thus, a battery has the potential to work. It has electrical energy.

Energy can be converted from one form to another. For instance, consider a steam turbine that can do mechanical work. That is the conversion of heat to mechanical energy. In this sense, when an electric kettle heats water, it performs work. It consumes energy and converts it to work.

Energy can be measured, like any other entity, in its appropriate unit(s). Suppose that a machine or device has energy. How much energy does it have? Energy can be measured in terms of heat units like calorie and BTU (British thermal unit), work units, or energy units.

The **unit for energy is joule**, and units for work can be foot-pound and newton-meter. One joule is, in fact, one newton-meter.

In conjunction with the energy, we have *power*. We may always ask the question: “If a machine can do a certain amount of work, how long does it take to do it?” For example, how long does it take for a kettle to boil the water in it? The answer to this question stands in *power*, which is also used for comparison between various energy sources.

Power is the amount of work done in 1 sec by a device that can do work. This is the measure of the strength of energy sources. For instance, a smaller motor has less power than a larger motor. That is, it can do less work than the larger motor in the same duration of time or, in doing the same work for the smaller motor, it takes more time.

**Similarly**, it takes more time for a smaller (less powerful) kettle to boil the same amount of water compared to a larger kettle.

Thus, electric power is determined from energy divided by time.

\[Power=\frac{Energy}{Time}\]

Electric power is normally measured in **watt and kilowatt**. Energy is sometimes (especially in the case of electrical energy) measured in terms of the unit of power multiplied by the unit of time; that is, watt-second and kilo-watt-hour are used as units of energy.

For almost any device, including light bulbs, the power is written on the device itself or on a nameplate. In addition, the operating voltage is always shown. For example, on a lightbulb, you may see “100 W, 120 V.” This implies that

- Operating voltage of the light bulb is 120 V. You should not connect this bulb to a voltage that is considerably higher than 120 V; whereas it is generally acceptable if you connect it to 125 V, but if you apply 180 V to this bulb, you will definitely burn it out.
- If connected to 120 V, the power consumed by the light bulb is 100 W. Moreover, if you connect it to a higher voltage, the power will be more than 100 W, and if the voltage used for the bulb is lower, then the power is lower than 100 W.

Note that physically the lightbulb does not change and it has only a filament in it with a certain resistance. The resistance of the filament when it is cold is less than when it is lighted.

What is the resistance of the filament in a 100 W, 120 V lightbulb?

**Solution**

The current in the light bulb filament can be found from ** Equation 1**:

\[I=\frac{P}{V}=\frac{100W}{120A}=0.83A\]

From V=IR,

\[R=\frac{V}{I}=\frac{120}{0.83}=144\Omega \]

What is the power consumption of a 100 W, 120 V lightbulb if it is connected to 110 V?

**Solution**

No matter to what voltage the lightbulb is connected, its filament does not change. When the light bulb is connected to 110 V, its filament still gets hot and with good precision, we can assume that its resistance stays at 144 Ω (see ** Example 2**).

The current due to connecting this lightbulb to 110 V is

\[I=\frac{V}{R}=\frac{110V}{144\Omega }=0.764A\]

Thus the power of lightbulb is

$P=VI=110*0.764=84W$

It can be clearly seen that if the voltage applied to a lightbulb (or any other device with the resistive element) is not the same as its rated voltage, then one cannot expect to obtain the rated electric power.

Note that if the applied voltage is much smaller than the rated value, then the lightbulb filament is not heated enough and its resistance cannot be assumed to stay the same. Its resistance becomes slightly smaller.

As can be seen from the **Ohm’s law **and from the relationship for an electric power, **there are four interrelated electrical entities for any resistive load**: resistance, voltage, current, and power. When any two of these values are known, the other two can be obtained.

The following equations are quite useful in obtaining the unknown quantities.

**Electric Power**

$\begin{align} & P={{I}^{2}}R \\ & P=\frac{{{V}^{2}}}{R} \\\end{align}$

**Current**

$\begin{align} & I=\frac{P}{V} \\ & I=\frac{V}{R} \\\end{align}$

**Voltage**

$\begin{align} & V=IR \\ & V=\frac{P}{I} \\\end{align}$

**Resistance**

$\begin{align} & R=\frac{V}{I} \\ & R=\frac{{{V}^{2}}}{P} \\ & R=\frac{P}{{{I}^{2}}} \\\end{align}$

All of the equations can be put together in the form of a chart shown in **Figure 1**.

**Figure 1** Summary chart for the relationships among voltage, current, resistance, and electric power.

From the fact that the electric power is the product of voltage and current, it is obvious that a **watt-meter** to measure power must work based on a needle (if analog) or a display (if digital) whose displacement or value is proportional to the values for both the voltage and the current in a circuit.

While a watt-meter can be used in panels of **electric generators** to indicate the instantaneous power, it is normally not used by a technician, and it is not included in the functions of a **multimeter.**

However, measuring energy is very common for the calculation of electric energy cost. Each house and building is equipped with an electricity meter that measures the consumed electric energy. The basis of such a meter is a circular disk that rotates and causes a series of gears to rotate as in a clock.

The speed of rotation of this disk is proportional to the power consumed in a circuit, and the number of revolutions during a period represents the energy consumed during that period.

The unit for measuring electric power is watt.

A kilowatt (1000 watts) is a larger unit, and for even larger measures of energy production, megawatt (1,000,000 watts) is a more common unit of power.

Because energy is the product of power and time, we may write

$\begin{matrix} E=Pt & {} & \left( 2 \right) \\\end{matrix}$

If *P* is the electric power in watt and *T* is the time in second, then *E* determines energy in joules. It is very common in electricity that a larger unit of energy is employed.

If the power is given in kW and time in hr (3600 sec), then the unit of energy is kW-hr (kilowatt-hour). Kilowatt-hour is a unit of energy equivalent to (1000)(3600) = 3,600,000 joule.

A better understanding of kW-hr is possible in the following manner. One kW-hr is the energy consumption of one 100 W lightbulb in 10 hours, or of two 100 W lightbulbs in 5 hours, or ten 100 W lightbulbs in 1 hour.

A 100 W lightbulb is turned on 5 hours per day. Calculate the amount of energy consumed by the light bulb in a year.

**Solution**

$\begin{align} & Number\text{ }of\text{ }hours\text{ }the\text{ }light\text{ }is\text{ }on\text{ }in\text{ }a\text{ }year=365*5=1825\text{ }Hr. \\ & Energy=Pt=100*1825=182,500W-hr=182.5kW-hr \\\end{align}$

If the cost of electricity is 8 cents per kW-hr, what is the cost of electricity used by the lightbulb in ** Example 4**?

**Solution**

$AnnualCost=182.5*0.08=\$14.60$

If the average daily electricity consumption in a household is 4 kW-hr during a given season, what is the monthly cost of electricity if the rate of electricity is 7.5 cents per kW-hr?

**Solution**

$\begin{align} & Average\text{ }Monthly\text{ }Consumption=30*4=120\text{ }kW-hr \\ & Cost=120*0.75=\text{ }\$90\\\end{align}$

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