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]]>Similarly, connecting a very low resistance **ammeter** across terminals A and B shows that the short-circuit current is about 2.0 mA. Hence, the black box appears to have an internal resistance of

\[{R}_{Th}=\frac{{{E}_{oc}}}{{{I}_{sc}}}=\frac{80V}{2mA}=40k\Omega \]

**Figure 1** (a) Original voltage source; (b) Thevenin-equivalent source

As far as we can tell without opening it, the box contains an 80-V source, as shown in Figure 1(b). Any load connected to terminals A and B of the equivalent circuit of Figure 1(b) draws exactly the same current with exactly the same voltage drop as the load would if it were connected to the network of Figure 1(a).

In fact, all that we can determine from measurements at the output terminals of any network containing one or more voltage sources is that the network is equivalent to a simple constant-voltage source with a single internal resistance in series with it. The **French engineer Leon Charles Thevenin (1857–1927) stated the principle** of this equivalence as a theorem:

Any two-terminal network of fixed resistances and voltage sources may be replaced by a single voltage source that has

- An equivalent voltage equal to the open-circuit voltage at the terminals of the original network, and
- An internal resistance equal to the resistance looking back into the network from the two terminals with all the voltage sources replaced by their internal resistances.

We can apply Thevenin’s theorem to any of the resistance networks by treating one branch of the network as a load and the remainder of the network as a two-terminal network containing one or more voltage sources.

Having decided which branch of the original network to treat as a load, we remove it from the original network and place it in a Thevenin-equivalent circuit. Then, we apply Thevenin’s theorem to determine the rest of the equivalent circuit.

Note that Thevenin uses an “**ohmmeter**” approach to determine the internal resistance “looking back into” the open-circuit terminals of the source network.

What current does a 10-kV resistor draw when it is connected to a 100-V source through the T-network shown in Figure 1(a)?

**Solution**

**Step 1**

Connect the 10-kV resistor to a Thévenin-equivalent source consisting of a constant voltage source E_{Th} with a series internal resistance R_{Th}, as shown in **Figure 2(b)**.

According to Thevenin’s theorem, E_{Th} equals the open circuit voltage between terminals A and B of Figure 1(a).

Since an open circuit draws no current, there is no voltage drop across the 20-kV resistor. Hence, the open-circuit terminal voltage is the same as the voltage drop across the 100-kV resistor.

When terminal A is open-circuit, the 25-kV and the 100-kV resistors form a simple series circuit. Then, the **voltage-divider principle** provides us with the voltage drop across the 100-kV resistor. This voltage drop is the open-circuit voltage between terminals A and B, which is E_{Th} in the Thévenin-equivalent circuit.

\[{{E}_{Th}}=\frac{100}{25+100}\times 100V=80V\]

**Figure 2 ** Thevenin’s Equivalent circuit for Step 2 of Example 1

**Step 2**

Replace the actual source in Figure 1(a) by its internal resistance, which is zero ohms. The original circuit of Figure 1(a) then becomes the **series-parallel resistance network** of Figure 2(a). The equivalent resistance for this network is

\[{{R}_{Th}}=20k\Omega +\frac{25k\Omega \times 100k\Omega }{25k\Omega +100k\Omega }=40k\Omega \]

**Step 3**

Solve for the current through the 10-kV resistor in the Thevenin- equivalent circuit of Figure 2(b).

\[I=\frac{{{E}_{Th}}}{{{R}_{T}}}=\frac{80V}{40k\Omega +10k\Omega }=1.6mA\]

A resistor passing a 20-mA current is in parallel with a 5.0-kV resistor. This combination is in series with another 5.0-kV resistor, and the whole network is connected to a 500-V source. Find the resistance of the resistor that is passing the 20-mA current.

**Solution**

**Step 1**

We select the unknown resistance passing the 20-mA current as the load, remove it from the original circuit in Figure 3(a), and place it in the Thevenin- equivalent circuit of Figure 3(b).

The voltage-divider principle then gives the open-circuit terminal voltage of the remaining circuit in Figure 3(a):

\[{{E}_{Th}}=\frac{5}{5+5}\times 500V=250V\]

**Figure 3** Circuit diagram for Step 1 of Example 2

**Step 2**

If we replace the original voltage source with its internal resistance of zero ohms, the circuit of Figure 3(a) becomes the simple parallel resistance network of Figure 4(a), and

\[{{R}_{Th}}=20k\Omega +\frac{5k\Omega \times 5k\Omega }{5k\Omega +5k\Omega }=2.5k\Omega \]

**Figure 4** Thevenin’s Equivalent circuit for Step 2 of Example 2

Often the load in a circuit varies while the other circuit elements are fixed. For example, various appliances may be connected to a household outlet, thus changing the load on that branch of the house wiring.

When we are analyzing a circuit that has a varying load, Thevenin’s theorem can save us some tedious calculations by letting us replace the unchanged portion of the circuit with a simple equivalent circuit, as shown in Figure 5.

However, we can apply Thevenin’s theorem only to linear components (components with values that do not vary with voltage or current).

**Figure 5** Thevenin’s Theorem Application

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]]>The post Nodal Analysis with Solved Examples appeared first on Electrical A2Z.

]]>Nodal analysis is a circuit-analysis format that combines Kirchhoff’s current- law equations with the **source transformation**. Converting all voltage sources to equivalent constant-current sources allows us to standardize the way we write the **Kirchhoff’s current-law** equations.

For nodal analysis, we consider source currents to flow into a node. If the arrows in a circuit diagram showing that a source current actually flows out of a particular node, the current is a negative quantity for that node.

**Similarly**, we consider all resistor currents to flow out of a node. We are thus restating Kirchhoff’s current law in the form:

At any independent node, the algebraic sum of the resistor currents leaving the node equals the algebraic sum of the source currents entering the node.

The Kirchhoff’s current-law equations then have the format

$\begin{matrix} {{I}_{R1}}+{{I}_{R2}}+\cdots ={{I}_{S1}}+{{I}_{S2}}+\cdots & {} & \left( 1 \right) \\\end{matrix}$

The next step is to replace the resistor currents using I = VG.

The voltage across each resistor equals the difference between the voltages, relative to the reference node, of the nodes at each end of the resistor. We always subtract the voltage of the adjacent node from the voltage of the node for which we are writing the equation. For node 1 in

**Figure 1**, the current through R_{1} equals V_{1}G_{1}. The voltage across R_{3} is V_{1 }– V_{2}. Hence the Kirchhoff’s current- law equation for node 1 becomes

${{V}_{1}}{{G}_{1}}+\left( {{V}_{1}}-{{V}_{2}} \right){{G}_{2}}={{I}_{S1}}$

Collecting the voltage terms gives

\[\begin{matrix} \left( {{G}_{1}}+{{G}_{3}} \right){{V}_{1}}-{{G}_{3}}{{V}_{2}}={{I}_{S1}} & {} & \left( 2 \right) \\\end{matrix}\]

**Figure 1** circuit diagram for writing nodal equations

Equation 2 illustrates the standard format for writing a nodal equation. We can use determinants to solve for the voltages.

We can write **Equation 2** directly by noting that the positive voltage term on the left-hand side of the equation is the unknown voltage for the node in question multiplied by the sum of the conductance connected to that node. **From this positive term**, we must subtract a term for the node voltage at every adjacent node connected by a resistor to the node in question. This term is the adjacent node voltage multiplied by the conductance between the two nodes. Using this procedure to write the nodal equation for node 2 gives

\[\left( {{G}_{2}}+{{G}_{3}} \right){{V}_{2}}-{{G}_{3}}{{V}_{1}}=-{{I}_{S2}}\]

Nodal analysis is particularly useful for networks where a common portion of the network is fed from several sources in parallel.

An automobile generator with an internal resistance of 0.20 V develops an open-circuit voltage of 16.0 V. the storage battery has an internal resistance of 0.10 V and an open-circuit voltage of 12.8 V. Both sources are connected in parallel to a 1.0-V load. Determine the load current.

**Solution**

**Step 1**

First, we draw the equivalent circuit with the voltage sources converted to constant-current sources, as shown in **Figure 2**.

**Figure 2** Circuit diagram for Example 1

The short-circuit current for the battery is

\[{{I}_{bat}}=\frac{{{E}_{bat}}}{{{R}_{bat}}}=\frac{12.8V}{0.10\Omega }=128A\]

Similarly, the short-circuit current of the generator is

\[{{I}_{gen}}=\frac{{{E}_{gen}}}{{{R}_{gen}}}=\frac{16V}{0.20\Omega }=80A\]

With a bit of mental arithmetic, we can fill in the required data on the circuit diagram of Figure 2.

**Step 2**

Next, we find the conductance of each branch:

$\begin{align} & {{G}_{gen}}=\frac{1}{{{R}_{gen}}}=\frac{1}{0.20\Omega }=5S \\ & {{G}_{bat}}=\frac{1}{{{R}_{bat}}}=\frac{1}{0.10\Omega }=10S \\ & and \\ & {{G}_{L}}=\frac{1}{{{R}_{L}}}=1S \\\end{align}$

**Step 3**

At first glance, Figure 2 appears to show six junction points, but there are only two because the same voltage appears across all parallel branches. For our nodal analysis, we can redraw the circuit in the form shown in **Figure 3**.

If we label the lower node as the reference node, the upper node is the only independent node. Consequently, we require only one **Kirchhoff’s current-law** equation for the circuit of Figure 3.

**Figure 3** Equivalent circuit for Example 1

At the single independent node, Equation 1 becomes

${{I}_{5}}+{{I}_{1}}+{{I}_{10}}={{I}_{gen}}+{{I}_{bat}}$

Or, we can go directly to the format of Equation 2,

$\begin{align} & \left( {{G}_{5}}+{{G}_{1}}+{{G}_{10}} \right)V={{I}_{gen}}+{{I}_{bat}} \\ & V=\frac{80+128}{5+1+10}=13V \\ & {{I}_{L}}={{G}_{L}}{{V}_{L}}=13V\times 1S=13A \\\end{align}$

The nodal analysis allowed us to solve Example 1 with not much more than mental arithmetic since the circuit has only a single independent node. **However**, we do need to solve simultaneous equations when the network has more than one independent node.

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]]>The post Mesh Current Analysis with Solved Problems appeared first on Electrical A2Z.

]]>Mesh Current Analysis is a technique that simplifies and speeds up writing the simultaneous equations for solving various resistance networks. The format for mesh equations is straightforward, but it cannot handle some of the networks that we can solve with the loop procedure.

A mesh is a closed loop that does not enclose any circuit elements. Therefore, the circuit diagram cannot contain any conductors that cross without being joined at the point of crossing. In other words, the network diagram must be strictly two-dimensional or planar.

The mesh format requires all sources to be voltage sources. We must convert any current sources to equivalent voltage sources before we use the mesh format. Instead of choosing convenient tracing loops, we draw a current loop for each mesh.

All mesh currents must have the same direction, either clockwise or counterclockwise. The more common choice is clockwise. These rules provide a standard format for every mesh equation.

In the loop procedure, we write equations for voltage drops around the current loops and then rewrite the Kirchhoff’s voltage-law equations to collect the current terms. In the mesh format, we go directly to this second step.

We can use the network shown in **Figure 1** to illustrate the mesh format. The network is clearly planar and the four meshes have no internal electric components.

**Figure 1** Mesh network for Example 1

For any particular mesh current, all other currents through common circuit elements are in the opposite direction. For example, I_{A} and I_{B} flow through the 30-V resistor in opposite directions. The mesh equation for mesh A has the form

${{R}_{T}}{{I}_{A}}-{{R}_{AB}}{{I}_{B}}-{{R}_{AC}}{{I}_{C}}=\sum{E}$

Where R_{T} is the total resistance around mesh A, R_{AB} is the resistance that is common to mesh A and mesh B, R_{AC} is the resistance that is common to mesh A and mesh C, and ΣE is the algebraic sum of the voltage sources around the mesh.

Where I_{A} passes inside a voltage source from its negative terminal to its positive terminal, E is positive. If the direction of I_{A} through a voltage source is from the positive terminal to the negative terminal, E is negative.

Find the magnitudes and polarities of the voltage drops across the 30-Ω, 50- Ω, and 60- Ω resistors in the network shown in Figure 1.

**Solution**

Using the mesh format, write a Kirchhoff’s voltage-law equation for each of the four meshes in Figure 1. In this example, the resistances are in ohms and the potential differences are in volts, so the currents will be in amperes.

$\begin{align} & \left( 10+30+50 \right){{I}_{A}}-30{{I}_{B}}-50{{I}_{C}}=20 \\ & \left( 30+20+40+60 \right){{I}_{B}}-30{{I}_{A}}-60{{I}_{D}}=0 \\ & \left( 80+70+50 \right){{I}_{C}}-50{{I}_{A}}=-30 \\ & \left( 90+60 \right){{I}_{D}}-60{{I}_{B}}=30-40 \\\end{align}$

Collecting like terms in each equation gives

\[\begin{align} & 90{{I}_{A}}-30{{I}_{B}}-50{{I}_{C}}=20\cdots \left( 1 \right) \\ & -30{{I}_{A}}+150{{I}_{B}}-60{{I}_{D}}=0\cdots \left( 2 \right) \\ & -50{{I}_{A}}+200{{I}_{C}}=-30\cdots \left( 3 \right) \\ & -60{{I}_{B}}+150{{I}_{D}}=-10\cdots \left( 4 \right) \\\end{align}\]

Unfortunately, the calculations for solving fourth-order determinants are not as simple as those for second- and third-order determinants. We can get around this problem by eliminating one of the unknowns. From Equation 4,

\[{{I}_{D}}=\frac{60{{I}_{B}}-10}{150}\]

Substituting for I_{D} in Equation 2 gives

$-30{{I}_{A}}+150{{I}_{B}}-\left( 24{{I}_{B}}-4 \right)=0$

We now have three simultaneous equations:

\[\begin{align} & \begin{matrix} 90{{I}_{A}}-30{{I}_{B}}-50{{I}_{C}}=20 & {} & \left( 1 \right) \\\end{matrix} \\ & \begin{matrix} -30{{I}_{A}}-126{{I}_{B}}=-4 & {} & \left( 5 \right) \\\end{matrix} \\ & \begin{matrix} -50{{I}_{A}}+200{{I}_{C}}=-30 & {} & \left( 3 \right) \\\end{matrix} \\\end{align}\]

\[{{I}_{A}}=\frac{\left| \begin{matrix} 20 & -30 & -50 \\ -4 & 126 & 0 \\ -30 & 0 & 200 \\\end{matrix} \right|}{\left| \begin{matrix} 90 & -30 & -50 \\ -30 & 126 & 0 \\ -50 & 0 & 200 \\\end{matrix} \right|}=\frac{291,000}{1,773,000}=0.164A\]

\[{{I}_{B}}=\frac{\left| \begin{matrix} 90 & 20 & -50 \\ -30 & -4 & 0 \\ -50 & -30 & 200 \\\end{matrix} \right|}{D}=\frac{13,000}{1,773,000}=0.0073A\]

\[{{I}_{C}}=\frac{\left| \begin{matrix} 90 & -30 & 20 \\ -30 & 126 & -4 \\ -50 & 0 & -30 \\\end{matrix} \right|}{D}=\frac{-193,200}{1,773,000}=-0.109A\]

\[{{I}_{C}}=\frac{60\times 7.33\times {{10}^{-3}}-10}{150}=-0.064A\]

$\begin{align} & {{I}_{30}}={{I}_{A}}-{{I}_{B}}=0.164-0.0073=157mA \\ & {{V}_{30}}=30\Omega \times 157mA=4.7V\left[ positive\text{ }at\text{ }top \right] \\ & {{I}_{50}}={{I}_{A}}-{{I}_{C}}=0.164+0.109=273mA \\ & {{V}_{50}}=50\Omega \times 273mA=14V\left[ positive\text{ }on\text{ }the\text{ }right \right] \\ & {{I}_{60}}={{I}_{B}}-{{I}_{D}}=0.0073-0.064=71mA \\ & {{V}_{60}}=60\Omega \times 71mA=4.3V\left[ positive\text{ }on\text{ }the\text{ }right \right] \\\end{align}$

It is important to understand that Mesh analysis can be applied to either AC resistive circuit or DC resistive circuit. In any type of network, the numbers of linear equations are dependent on the number of loop currents.

- You May Also Read: Mesh Analysis using Matlab

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]]>The post Source Transformation Example Problems with Solutions appeared first on Electrical A2Z.

]]>Source transformation is a circuit analysis technique in which we convert voltage source in series with resistor into a current source in parallel with the resistor and vice versa.

For a given constant-voltage source, R_{int} in the equivalent constant-current source has the same value but appears in parallel with the ideal current source, as shown by the example in **Figure 1**.

**Similarly**, for a given constant-current source, R_{int} in the equivalent constant-voltage source has the same value but appears in series with the ideal voltage source.

In order for us to replace a constant-voltage source in a network with its constant-current equivalent, there must be some resistance R_{x} in series with an ideal voltage source.

R_{x} may be an internal resistance of the source, or it may be one of the network resistors in series with the voltage source.

**Similarly**, there must be some form of resistance in parallel with an ideal current source before we can replace it with its constant-voltage equivalent.

Applying Equation 1 to the constant-voltage source of Figure 1(a) shows that I_{x} for the equivalent constant-current source (Figure 1(b)) is the short-circuit current:

\[\begin{matrix} {{\text{R}}_{\text{int}}}\text{=}\frac{\text{open circuit terminal voltage}}{\text{short circuit terminal current}} & {} & \left( \text{1} \right) \\\end{matrix}\]

\[\begin{matrix} {{I}_{X}}=\frac{{{E}_{oc}}}{{{R}_{X}}} & {} & \left( 2 \right) \\\end{matrix}\]

Similarly, given the constant-current source of Figure 1(c), E_{x} for the equivalent constant-voltage source of Figure 1(d) is the open-circuit voltage:

$\begin{matrix} {{E}_{X}}={{I}_{sc}}{{R}_{X}} & {} & \left( 3 \right) \\\end{matrix}$

**Figure 1** Source transformation

(a) Determine the equivalent constant-current source for a voltage source with an open-circuit voltage of 120 V and an internal resistance of 6.0 V.

(b) Determine the equivalent constant-voltage source for a current source with a constant current of 40 A and an internal resistance of 0.80 V.

(c) Check that the equivalent sources in part (a) produce the same results when connected to a 12-V load.

**Solution**

- The internal resistance for the constant-current source is the same as for the voltage source: R
_{int}5 6.0 V

\[{{I}_{X}}=\frac{{{E}_{oc}}}{{{R}_{\operatorname{int}}}}=\frac{120V}{6\Omega }=20A\]

- R
_{int}for the constant-voltage equivalent source is the same as for the current source:

$\begin{align} & {{R}_{\operatorname{int}}}=0.80\Omega \\ & {{E}_{X}}={{I}_{sc}}{{R}_{\operatorname{int}}}=40A\times 0.80\Omega =32V \\\end{align}$

- When we connect a 12-V load to the voltage source of Figure 1(a),

\[{{I}_{L}}=\frac{{{E}_{X}}}{{{R}_{X}}+{{R}_{L}}}=\frac{120V}{6\Omega +12\Omega }=6.7V\]

And

${{V}_{L}}=6.7A\times 12\Omega =80V$

When we connect a 12-V load to the current source of Figure 1(b),

$\begin{align} & {{V}_{L}}={{I}_{X}}\times \frac{{{R}_{X}}\times {{R}_{L}}}{{{R}_{X}}+{{R}_{L}}}=20A\times \frac{6\times 12}{6+12}=80V \\ & {{I}_{L}}=\frac{{{V}_{L}}}{{{R}_{L}}}=\frac{80V}{12\Omega }=6.7A \\\end{align}$

Once we are familiar with the concept, we can convert voltage sources to equivalent current sources (and vice versa) for network analysis purposes with just a bit of mental arithmetic. Analysis of transistor circuits often involves resistance networks containing both voltage and current sources. Such networks are usually easier to solve if we convert the sources so that they are either all voltage sources or all current sources.

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]]>The post Series-Parallel Circuit: Definition & Examples | Series-Parallel Resistors appeared first on Electrical A2Z.

]]>As measured from the terminals of the voltage source, the simplified circuit of **Figure 2** is equivalent to the original circuit of Figure 1. In Figure 2, R_{1} is in series with the equivalent resistance of R_{2} and R_{3} in parallel. Therefore, we can solve the circuit of Figure 2 as a simple series circuit.

**Figure 1** Simple series-parallel circuit

**Figure 2** Equivalent circuit for Figure 1

In the circuit shown in **Figure 3(a)**, R_{2} and R_{3} have the same current through them and are therefore in series. We can replace them with an equivalent resistor, as shown in** Figure 3(b)**. We can now solve the simplified circuit of Figure 3(b) as a simple parallel circuit.

**Figure 3** (a) Series-parallel circuit; (b) Equivalent circuit

We define a series-parallel circuit as one in which some portions of the circuit have the characteristics of simple series circuits while the other portions have the characteristics of simple parallel circuits.

- You May Also Read: Series Circuit: Definition & Examples | Resistors in Series

**Series-Parallel Circuit Solution using Equivalent-Circuit Method**

We can solve series-parallel circuits by substituting the equivalent resistances for various portions of the circuit until the original circuit is reduced to either a simple series or a simple parallel circuit.

**Example 1**

Find the voltage drop, current, and power for each resistor in the circuit diagram of Figure 4.

**Solution**

**Step 1**

Draw a fully labeled schematic diagram for this particular circuit.

**Figure 4** Circuit diagram for Example 1

**Step 2**

Since R_{2} and R_{3} are in parallel,

\[{{\operatorname{R}}_{eq}}=\frac{{{R}_{3}}\times {{R}_{3}}}{{{R}_{2}}+{{R}_{3}}}=\frac{10\times 40}{10+40}=8\Omega \]

The total resistance of the circuit is

${{R}_{T}}={{R}_{1}}+{{\operatorname{R}}_{eq}}=12+8=20\Omega $

**Step 3**

From Ohm’s law,

\[{{I}_{T}}=\frac{E}{{{R}_{T}}}=\frac{100V}{20\Omega }=5A\]

**Step 4 **

Since R_{1} is in series with the source,

${{I}_{1}}={{I}_{T}}=5A$

**Step 5 **

Applying Ohm’s law to R_{1},

${{V}_{1}}={{I}_{1}}{{R}_{1}}=5A\times 12\Omega =60V$

**Step 6**

From Kirchhoff’s voltage law,

${{V}_{eq}}=E-{{V}_{1}}=100-60=40V$

Returning now to the original circuit,

${{V}_{2}}={{V}_{3}}={{V}_{eq}}=40V$

**Step 7 **

Applying Ohm’s law to R_{2} and R_{3},

$\begin{align} & {{I}_{2}}=\frac{{{V}_{2}}}{{{R}_{2}}}=\frac{40V}{10\Omega }=4A \\ & {{I}_{3}}=\frac{{{V}_{3}}}{{{R}_{3}}}=\frac{40V}{40\Omega }=1A \\\end{align}$

**To verify our calculations, we can check that the currents we calculated satisfy Kirchhoff’s current law for this circuit: **

**${{I}_{1}}={{I}_{2}}+{{I}_{3}}$ **

**Step 8**

Since,

$P=VI$

So,

$\begin{align} & {{P}_{1}}={{V}_{1}}{{I}_{1}}=60V\times 5A=0.30kW \\ & {{P}_{2}}={{V}_{2}}{{I}_{2}}=40V\times 4A=0.16kW \\ & {{P}_{1}}={{V}_{1}}{{I}_{1}}=40V\times 1A=40W \\ & {{P}_{1}}={{V}_{1}}{{I}_{1}}=100V\times 5A=0.50kW \\\end{align}$

**We can verify our calculations by checking that:**

**${{P}_{T}}={{P}_{1}}+{{P}_{2}}+{{P}_{3}}$ **

For the series circuit of Figure 5, Kirchhoff’s voltage law states that E = V_{1} + V_{2} + V_{3}. In other words, the total applied voltage is divided among the three resistors. We can connect a voltmeter across six possible combinations of the terminals A, B, C, and D, as shown in Figure 5. Thus, the series combination of R_{1}, R_{2}, and R_{3} acts as a voltage divider.

**Figure 5** Voltage-divider principle

To solve for the six voltages in Figure 5, we could calculate R_{T}, find I from Ohm’s law, and then calculate the voltage drop across each resistance from V_{1} = IR_{1}, V_{2} = IR_{2}, and so on. However, we can calculate each voltage in a single step. A property of **series circuits** that is often called the voltage-divider principle:

In a

series circuit, the ratio between any two voltage drops equals the ratio of the two resistances across which these voltage drops occur.

Hence, for any resistor R_{n} in a voltage divider,

\[\begin{align} & \frac{{{V}_{n}}}{E}=\frac{{{R}_{n}}}{{{R}_{T}}} \\ & and \\ & \begin{matrix} {{V}_{n}}=E\times \frac{{{R}_{n}}}{{{R}_{T}}} & {} & \left( 1 \right) \\\end{matrix} \\\end{align}\]

**Example 2**

What is the voltage between terminals B and D in the circuit of Figure 5?

**Solution**

\[\begin{align} & {{V}_{BD}}=E\times \frac{{{R}_{2}}+{{R}_{3}}}{{{R}_{1}}+{{R}_{2}}+{{R}_{3}}} \\ & =350V\times \frac{\left( 5k\Omega +10k\Omega \right)}{\left( 20k\Omega +5k\Omega +10k\Omega \right)} \\ & =150V \\\end{align}\]

Using a variable resistor as a voltage divider provides a continuously variable terminal voltage, as shown in Figure 6.

**Figure 6** Potentiometer circuit

Voltage dividers are widely used in power supplies for electronic circuits so that a single voltage source can supply all the various voltages required by a piece of equipment, reducing the cost, size, and weight of the equipment.

The series-dropping resistor of Figure 7 provides the simplest method of obtaining the required voltage drop across a particular circuit element.

**Figure 7** Series-dropping resistor

**Example 3**

A portion of an electronic circuit requires an operating voltage of 15 V and a current of 20 mA. If the supply terminal voltage is 25 V, what value of the series-dropping resistor is required?

**Solution**

From Ohm’s law, we can represent this load circuit by a resistor, as in Figure 7.

\[{{R}_{L}}=\frac{{{V}_{L}}}{{{I}_{L}}}=\frac{15V}{20mA}=750\Omega \]

From Kirchhoff’s voltage law, the voltage drop across the series- dropping resistor must be

${{V}_{D}}=E-{{V}_{L}}=25-15=10V$

Since the dropping resistor and the load are in series,

$\begin{align} & {{I}_{D}}={{I}_{L}}=20mA \\ & {{R}_{S}}=\frac{{{V}_{D}}}{{{I}_{D}}}=\frac{10V}{20mA}=500\Omega \\\end{align}$

To complete the design, we find the minimum power rating for R_{S}:

$P=VI=10V\times 20mA=0.20W$

A 0.5-W resistor would be a good choice since it will operate at a lower temperature than a 0.25-W resistor and thus be less likely to fail.

The **advantage of the simple series-dropping resistor** is that the only drain on the power supply is the current through the load. But this circuit has the **disadvantage** that any change in load resistance changes the current through the series-dropping resistor, which, in turn, changes the voltage drop across the resistor and the voltage supplied to the load.

In circuits where the load is a transistor, R_{L} can become very high. Under such circumstances, the voltage drop across a series-dropping resistor is almost zero and the voltage across the load is close to the full applied voltage, which may be high enough to damage the transistor.

To prevent **such damage**, some voltage-divider designs include a bleeder resistor in parallel with the load, as in Figure 8. The bleeder resistor ensures there is always enough current through the series-dropping resistor to maintain an appreciable voltage drop across it. Hence, the load voltage cannot rise to the full applied voltage.

**Figure 8** Voltage divider with a bleeder resistor

In power supplies for electronic equipment, a bleeder current of 10–25% of the total current drawn from the source provides sufficient protection against excessive load voltage when the load resistance increases.

The greater the bleeder current, the fewer variations in load current affect the load voltage. Thus, improved **voltage regulation** is achieved at the expense of extra current drain from the source and extra heat produced in the voltage-divider resistors.

In designing voltage dividers for loads consisting of a single transistor, voltage regulation is more significant than a few extra milliamperes of bleeder current.

In circuits that obtain several different voltages from one power supply, it is convenient to have a common reference point for all voltage measurements.

Often the circuit’s metal frame or chassis is connected to the circuit and used as the reference point. The chassis may also be connected to the earth, usually through the ground wire of the electrical system.

- You May Also Read: Parallel Circuit: Definition & Examples | Resistors in Parallel

The symbol at the bottom of **Figure 9** indicates that the reference point is grounded. Hence, we can say that the voltage at point A is 1250 V with respect to ground. The symbol in the middle of **Figure 10** indicates the point where the chassis is electrically connected to the circuit.

With respect to chassis, the voltage at point A in this circuit is 112 V and the voltage at point B is 212 V.

**Figure 9** Voltage divider with multiple output voltages

Transistor circuits can require positive or negative voltages relative to a common chassis connection, and many circuits require both. We can obtain these positive and negative voltages from one power supply by connecting a tap on a voltage divider to the chassis as shown in **Figure 10**.

**Figure 10** Power-supply voltage divider for a transistor amplifier

**Example 5**

At full load, the power supply in Figure 10 has a terminal voltage of 24 V DC. Design a voltage divider to provide 112 V and 212 V with respect to the chassis when the current drains are 400 mA at 112 V and 200 mA at 212 V. The total current drain on the power supply is 500 mA.

**Solution**

Since R1 is in parallel with load 1, V1 5 12 V. Applying Kirchhoff’s current law to junction A,

$\begin{align} & I1=200mA-400mA=100mA \\ & {{R}_{1}}=\frac{{{V}_{1}}}{{{I}_{1}}}=\frac{12V}{100mA}=120\Omega \\ & P1={{V}_{1}}{{I}_{1}}=12V\times 100mA=1.2W \\\end{align}$

Similarly,

$\begin{align} & {{V}_{2}}=12V, \\ & and \\ & {{I}_{2}}=500mA-200mA=300mA \\ & {{R}_{2}}=\frac{12V}{300mA}=40\Omega \\ & {{P}_{2}}=12V\times 300mA=3.6W \\\end{align}$

The current-divider principle can be described as:

In a parallel circuit, the ratio between any two branch currents equals the ratio of the two conductance through which these currents flow.

Since the voltage is common for resistors in parallel,

\[V=\frac{{{I}_{1}}}{{{G}_{1}}}=\frac{{{I}_{2}}}{{{G}_{2}}}=\frac{{{I}_{n}}}{{{G}_{n}}}=\frac{{{I}_{T}}}{{{G}_{T}}}\]

And

\[\begin{matrix} {{I}_{n}}={{I}_{T}}\times \frac{{{G}_{n}}}{{{G}_{T}}} & {} & \left( 2 \right) \\\end{matrix}\]

Substituting G_{n} = 1/R_{n}, and GT = 1/R_{eq }in Equation 2 gives

\[\begin{matrix} {{I}_{n}}={{I}_{T}}\times \frac{{{\operatorname{R}}_{eq}}}{{{R}_{n}}} & {} & \left( 3 \right) \\\end{matrix}\]

Hence, we can restate the current-divider principle:

In a parallel circuit, the ratio between any two branch currents is the inverse of the ratio of the branch resistances.

For two resistors in parallel,

$\begin{align} & {{\operatorname{R}}_{eq}}=\frac{{{R}_{1}}\times {{R}_{2}}}{{{R}_{1}}+{{R}_{2}}} \\ & {{I}_{1}}={{I}_{T}}\times \frac{{{R}_{1}}\times {{R}_{2}}}{{{R}_{1}}\left( {{R}_{1}}+{{R}_{2}} \right)} \\\end{align}$

And

\[\begin{align} & \begin{matrix} {{I}_{1}}={{I}_{T}}\times \frac{{{R}_{2}}}{\left( {{R}_{1}}+{{R}_{2}} \right)} & {} & \left( 4 \right) \\\end{matrix} \\ & similarly, \\ & \begin{matrix} {{I}_{1}}={{I}_{T}}\times \frac{{{R}_{1}}}{\left( {{R}_{1}}+{{R}_{2}} \right)} & {} & \left( 5 \right) \\\end{matrix} \\\end{align}\]

**Example 6**

Calculate the bleeder current in the voltage divider if the load resistance is 3.0 kΩ, as shown in Figure 11.

**Figure 11** Circuit diagram for Example 6

**Solution**

The equivalent resistance of the bleeder resistor and the load resistance in parallel is

$\begin{align} & {{\operatorname{R}}_{eq}}=\frac{{{R}_{1}}\times {{R}_{2}}}{{{R}_{1}}+{{R}_{2}}}=\frac{300\Omega \times 3k\Omega }{300\Omega +3k\Omega }=273\Omega \\ & {{R}_{T}}={{\operatorname{R}}_{eq}}+{{R}_{S}}=143+273=416\Omega \\ & {{I}_{T}}=\frac{E}{{{R}_{T}}}=\frac{25V}{416\Omega }=60.1mA \\ & {I}_{B}={{I}_{T}}\times \frac{{{R}_{L}}}{{{R}_{B}}+{{R}_{L}}}=60.1mA\times \frac{3k\Omega }{3.3k\Omega }=55mA \\\end{align}$

A series-parallel battery connection, as shown in Figure 12, provides both greater voltage and greater capacity than the individual cells. The cells must be identical to avoid circulating currents within the battery. The total capacity of the battery is determined by the number of rows connected in parallel.

**Figure 12** Equivalent circuit of a series-parallel connected battery

For a battery consisting of P parallel rows with S cells in series in each row, the EMF of the battery is the EMF of one series row, the same as for a series battery connection:

${{E}_{bat}}=S{{E}_{cell}}$

The total internal resistance is the combination of P identical parallel rows of S resistors in series, with each resistor having a resistance of R_{cell}, so

\[\begin{matrix} {{R}_{bat}}=\frac{resistance\text{ }of\text{ }each\text{ }row}{number\text{ }of\text{ }rows}=\frac{S{{R}_{cell}}}{P} & {} & \left( 6 \right) \\\end{matrix}\]

**Example 7**

Twenty identical cells are connected in four parallel rows, each with five cells in series. Each cell has an internal resistance of 0.20 V. The battery delivers a current of 5.0 A to a 1.25-V resistor as shown in Figure 13. Calculate the EMF of each cell.

**Solution**

$\begin{align} & {{R}_{bat}}=\frac{S{{R}_{cell}}}{P}=\frac{5\times 0.20\Omega }{4}=0.25\Omega \\ & {{E}_{bat}}={{I}_{L}}{{R}_{T}}=5A\times \left( 0.25\Omega +1.25\Omega \right)=7.5V \\ & {{E}_{cell}}=\frac{{{E}_{bat}}}{S}=\frac{7.5V}{5}=1.5V \\\end{align}$

**Figure 13** Equivalent circuit for Example 7

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]]>**Figure** **1** Simple parallel circuit: (a) Customary configuration, (b) Equivalent configuration

We can identify a parallel circuit by connections among the components. In Figure 1, E, R_{1}, R_{2}, and R_{3} are all in parallel because they are all connected between the same two points, A and B. Since each of the resistors is connected directly across the voltage source, V_{1} = V_{2} = V_{3} = E. We can omit the subscripts since the voltage is common to all the components connected in parallel.

Two or more electric components are in parallel in an electric circuit if a common voltage appears across all of the components.

For a **series circuit**, we determined the total resistance in order to find the current in the circuit. For a parallel circuit, we find the total current first and use it to determine the resistance of the circuit.

Given the applied voltage and the values of each resistance in the circuit of Figure 1, we can solve for the current in each branch by using **Ohm’s law**.

If we think of the current in each branch in terms of electrons flowing through the branch, it is apparent that the current through the source must be the sum of the branch currents.

In a simple parallel circuit, the total current is the sum of all the branch currents:

$\begin{matrix} {{I}_{T}}={{I}_{1}}+{{I}_{2}}+{{I}_{3}}+\cdots & {} & \left( 1 \right) \\\end{matrix}$

For the circuit in Figure 1, assume that R_{1} is 40 Ω, R_{2} is 30 Ω, R_{3} is 20 Ω, and E is 120 V. What single resistance would draw the same current from the source?

**Solution**

\[\begin{matrix} {{I}_{1}}=\frac{{{V}_{1}}}{{{R}_{1}}}=\frac{120V}{40\Omega }=3A \\ {{I}_{2}}=\frac{{{V}_{2}}}{{{R}_{2}}}=\frac{120V}{30\Omega }=4A \\ {{I}_{3}}=\frac{{{V}_{3}}}{{{R}_{3}}}=\frac{120V}{20\Omega }=6A \\\end{matrix}\]

${{I}_{T}}={{I}_{1}}+{{I}_{2}}+{{I}_{3}}=3+4+6=13A$

\[{{\operatorname{R}}_{eq}}=\frac{E}{{{I}_{T}}}=\frac{120V}{13A}=9.23\Omega \]

For a series circuit, the total resistance is greater than the resistance of any individual resistor. However, the resistance in a parallel circuit (e.g., the combination of resistors in Example 1) is less than the resistance of any individual resistor. Hence, when dealing with parallel circuits, we do not speak of total resistance. Instead, we speak of the equivalent resistance, R_{eq}, of two or more resistors in parallel.

Parallel circuits have characteristics that, in many respects, are similar but opposite to those of series circuits.

Kirchhoff extended the total-current principle of Equation 1 to apply to all electric circuits.

**Kirchhoff’s current law:** At any junction in a circuit, the algebraic sum of the currents entering the junction equals the algebraic sum of the currents leaving the junction.

Find the current in the R_{2} branch of the circuit of **Figure 2**.

**Figure 2** Circuit diagram for Example 2

**Solution**

\[{{I}_{1}}=\frac{V}{{{R}_{1}}}=\frac{50V}{10\Omega }=5A\]

The current flowing into junction X is I_{T} and the currents flowing away from junction X are I_{1} and I_{2}. Therefore,

$\begin{align} & {{I}_{T}}={{I}_{1}}+{{I}_{2}} \\ & {{I}_{2}}={{I}_{T}}-{{I}_{1}}=12-5=7A \\\end{align}$

As Examples 1 and 2 demonstrate, the total current is always greater than the current through any branch of a parallel circuit. Therefore, the equivalent resistance is always less than the smallest of the branch resistances. The more resistors we connect in parallel, the smaller the equivalent resistance becomes.

**In other words**, the more resistors we connect in parallel, the more readily the circuit can pass current since there are more parallel branches for current to flow through.

Conductance is a measure of the ability of an electric circuit to pass current. The letter symbol for conductance is G.

The Siemens(S) is the SI unit of conductance: $1S=\frac{1}{1\Omega }$

Conductance is the reciprocal of resistance:

\[\begin{matrix} G=\frac{1}{R} & {} & \left( 2 \right) \\\end{matrix}\]

Where G is the conductance of a circuit in Siemens and R is the resistance of the same circuit in ohms.

Dividing both sides of Equation 1 by E (or V, since they are the same for simple parallel circuits), we get

\[\frac{{{I}_{T}}}{E}=\frac{{{I}_{1}}}{V}+\frac{{{I}_{2}}}{V}+\frac{{{I}_{3}}}{V}+\cdots \]

Substituting Ohm’s law into Equation 2 gives

\[G=\frac{1}{R}=\frac{1}{{V}/{I}\;}=\frac{I}{V}\]

Therefore,

$\begin{matrix} {{G}_{T}}={{G}_{1}}+{{G}_{2}}+{{G}_{3}}+\cdots & {} & \left( 3 \right) \\\end{matrix}$

In parallel circuits, the total conductance is equal to the sum of the conductance of all the individual branches.

The equivalent resistance is simply R_{eq} = 1/G_{T}.

We can use conductance to find the equivalent resistance for a parallel circuit without calculating the total current.

**Example 1A**

For the circuit in Figure 1, assume that R_{1} is 40 Ω, R_{2} is 30 Ω, R_{3} is 20 Ω, and E is 120 V. What single resistance would draw the same current from the source?

**Solution**

$\begin{align} & {{G}_{T}}={{G}_{1}}+{{G}_{2}}+{{G}_{3}}=\frac{1}{40\Omega }+\frac{1}{30\Omega }+\frac{1}{20\Omega }=0.1083S \\ & {{\operatorname{R}}_{eq}}=\frac{1}{{{G}_{T}}}=\frac{1}{0.1083S}=9.23\Omega \\\end{align}$

When only two resistors are connected in parallel, Equation 3 becomes

\[{{G}_{T}}={{G}_{1}}+{{G}_{2}}=\frac{1}{R1}+\frac{1}{{{R}_{2}}}=\frac{{{R}_{1}}+{{R}_{2}}}{{{R}_{1}}\times {{R}_{2}}}\]

and

\[\begin{matrix} {{\operatorname{R}}_{eq}}=\frac{{{R}_{1}}\times {{R}_{2}}}{{{R}_{1}}+{{R}_{2}}} & {} & \left( 4 \right) \\\end{matrix}\]

For two resistors in parallel, the equivalent resistance equals their product over their sum.

We can rearrange Equation 4 to find the resistance R_{2} that we must connect in parallel with a given resistance R_{1} to obtain a desired equivalent resistance:

\[\begin{matrix} {{\operatorname{R}}_{2}}=\frac{{{R}_{1}}\times {{R}_{eq}}}{{{R}_{1}}-{{R}_{eq}}} & {} & \left( 5 \right) \\\end{matrix}\]

Now we can use Ohm’s law and Kirchhoff’s current law to find the current through resistor R_{2}:

\[\begin{align} & {{V}_{1}}={{V}_{2}}={{I}_{1}}{{R}_{1}}={{I}_{2}}{{R}_{2}} \\ & {{I}_{1}}{{R}_{1}}=\left( {{I}_{T}}-{{I}_{2}} \right){{R}_{1}} \\\end{align}\]

\[\begin{matrix} {{I}_{2}}={{I}_{T}}\times \left( \frac{{{R}_{1}}}{{{R}_{1}}+{{R}_{2}}} \right) & {} & \left( 6 \right) \\\end{matrix}\]

When N equal resistors are connected in parallel, we can show that

\[\begin{matrix} {{\operatorname{R}}_{eq}}=\frac{R}{N} & {} & \left( 7 \right) \\\end{matrix}\]

Where R is the resistance of each of the parallel resistors and N is the number of resistors.

**Example 3**

What is the equivalent resistance of a 1-kV and a 4-kV resistor in parallel?

Solution

\[{{\operatorname{R}}_{eq}}=\frac{{{R}_{1}}\times {{R}_{2}}}{{{R}_{1}}+{{R}_{2}}}=\frac{1k\Omega \times 4k\Omega }{\left( 1k\Omega +4k\Omega \right)}=0.8k\Omega \]

We defined resistivity as the resistance of a unit length and cross-section of a material. Conductivity is defined in a similar way.

The conductivity of a material is the conductance of a unit length and cross-section of that material. The letter symbol for conductivity is the Greek letter σ (sigma). Conductivity is measured in Siemens per meter.

Since conductance is the reciprocal of resistance, conductivity is the reciprocal of resistivity:

\[\begin{matrix} \sigma =\frac{1}{\rho } & {} & \left( 8 \right) \\\end{matrix}\]

- You May Also Read: Series Circuit: Definition & Examples | Resistors in Series

Since V = IR and R = 1/G, V = I/G. The voltage is the same across all components connected in parallel, so

\[\begin{matrix} V=\frac{{{I}_{T}}}{{{G}_{T}}}=\frac{{{I}_{1}}}{{{G}_{1}}}=\frac{{{I}_{2}}}{{{G}_{2}}}=\frac{{{I}_{3}}}{{{G}_{3}}}=\cdots \begin{matrix} {} & {} & {} \\\end{matrix} & {} & \left( 9 \right) \\\end{matrix}\]

By transposing variables in each pair of equal terms, we get

\[\begin{matrix} \frac{{{I}_{1}}}{{{I}_{2}}}=\frac{{{G}_{1}}}{{{G}_{2}}} & , & \begin{matrix} \frac{{{I}_{2}}}{{{I}_{3}}}=\frac{{{G}_{2}}}{{{G}_{3}}}=\frac{{{R}_{3}}}{{{R}_{2}}} & , & \text{and so on} \\\end{matrix} \\\end{matrix}\]

In a parallel circuit, the ratio between any two branch currents equals the ratio of their conductance or the inverse of the ratio of their resistances.

**Example 4**

The total current drawn by a 12.5-kV resistor and a 50-kV resistor in parallel is 15 mA. Find the current through the 50-kV resistor.

**Solution**

From the circuit diagram of **Figure 3**,

${{I}_{1}}+{{I}_{2}}=15mA$$\begin{align} & \frac{{{I}_{1}}}{{{I}_{2}}}=\frac{{{R}_{2}}}{{{R}_{1}}}=\frac{50k\Omega }{12.5k\Omega }=4 \\ & {{I}_{1}}=4{{I}_{2}} \\\end{align}$

Substituting for I_{1} in the first equation gives

$\begin{align} & 4{{I}_{2}}+{{I}_{2}}=15mA \\ & {{I}_{2}}=\frac{15mA}{5}=3mA \\\end{align}$

**Figure 3 **Circuit diagram for Example 4

When the **internal resistance** of the source is negligible, altering the resistance of one branch of a parallel circuit does not affect the voltage across, or the current through, the other branches. **T****herefore**, changes in one branch of a parallel circuit have a negligible effect on the other branches.

In-house wiring, lighting circuits are connected in parallel so that switching one circuit on or off does not affect the operation of the other circuits (see **Figure 4**).

**Figure 4** Parallel connection of loads in house wiring

The **following characteristics** can help us recognize parallel circuits:

- The voltage is the same across all components.
- The total conductance is the sum of all the individual branch conductance:

${{G}_{T}}={{G}_{1}}+{{G}_{2}}+{{G}_{3}}+\cdots $

- The total current is the sum of all the individual branch currents:

${{I}_{T}}={{I}_{1}}+{{I}_{2}}+{{I}_{3}}+\cdots $

- The ratio between branch currents is the same as the conductance ratio and the inverse of the resistance ratio.
- Each branch is independent of any changes in the other branches, providing the voltage across the parallel circuit is constant.

A battery of identical cells connected in parallel (see **Figure 5**) has the same EMF as a single cell but can deliver a greater maximum current.

The total capacity of the battery is proportional to the number of cells. If the cells do not all have the same EMF, currents will flow between them since the cells with lower EMFs will act as loads for the cells with higher EMFs.

Standard AA-size cells are sometimes connected in parallel to make a compact battery with enough current capacity to run motors in devices such as cordless electric razors.

**Figure 5** Equivalent circuit of a parallel connected battery

In Figure 5, P is the number of cells connected in parallel, E_{cell} is the EMF of each cell, and R_{cell} is the internal resistance of each cell. The battery EMF is

E_{bat} = E

and the total internal resistance of the battery is

\[\begin{matrix} {{R}_{bat}}=\frac{{{R}_{cell}}}{P} & {} & \left( 10 \right) \\\end{matrix}\]

**Example 5**

How much current flows through an external resistance of 4.00 V connected to a battery consisting of ten cells connected in parallel, with each cell having an EMF of 1.50 V and an internal resistance of 0.20 V? Find the terminal voltage of the battery.

**Solution**

$\begin{align} & {{E}_{bat}}=1.5V \\ & {{R}_{bat}}=\frac{{{R}_{cell}}}{P}=\frac{0.20\Omega }{10}=0.020\Omega \\ & I=\frac{E}{{{R}_{T}}}=\frac{1.5}{4\Omega +0.020\Omega }=1.49V \\ & VT=I{{R}_{L}}=0.373A\times 4\Omega =1.49V \\\end{align}$

**Figure 6** shows an equivalent circuit for the battery.

**Figure 6** Equivalent circuit for Example 5

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]]>A series circuit can be identified by the connection between components or by the current through them. **For example**, in the circuit of **Figure 1**, R_{1} and R_{2} are connected in series because no other component or branch is connected to the junction of R_{1} and R_{2}. None of the junctions in this circuit have a second branch, so all of the components are in series. Thus, a series circuit has only one path for current.

**Figure 1** Simple series circuit

Since resistors cannot store charge, the flow of electrons into R_{1} is equal to the flow of electrons out of R_{1}, which in turn is equal to the flow of electrons into R_{2}.

Similarly, the flow of electrons through the source is equal to the flow of electrons through any other part of the series circuit. Georg Ohm was the first person to recognize this key property of series circuits.

The current is the same in all parts of a simple series circuit. Conversely, two or more electric components are in series if a common current flows through them.

Since the current in a series circuit is common to all components, it is not necessary to use a subscript with I to distinguish the current through the various components.

In the circuit of Figure 1, I represents the current through R_{1}, R_{2}, and R_{3}, as well as the current through the source and the connecting wires.

Suppose that resistor R_{1} is made with 2 m of Nichrome wire, while resistor R_{2} contains 1 m and R_{3} contains 3 m of the same wire.

An electron flowing through the circuit passes through a total of 6 m of Nichrome wire. Since the resistance of an electric conductor is directly proportional to its length, the total resistance of this circuit is the sum of the individual resistances. Generalizing,

The total resistance of a series circuit equals the sum of all the individual resistances in the circuit:

$\begin{matrix} {{R}_{T}}={{R}_{1}}+{{R}_{2}}+{{R}_{3}}+\cdots & {} & \left( 1 \right) \\\end{matrix}$

Once we know the total resistance of a series circuit, we can use Ohm’s law to find the common current:

\[I=\frac{E}{{{R}_{T}}}\]

This current produces a total voltage drop V_{T}, equal to the applied voltage E. Since the current through all the resistors is equal, the voltage drops across equal resistances are also equal.

What current will flow in a series circuit consisting of a 45-Ω source, a 20- Ω, a 10- Ω, and a 30- Ω resistor?

**Solution**

\[\begin{align} & {{R}_{T}}={{R}_{1}}+{{R}_{2}}+{{R}_{3}}=20+10+30=60\Omega \\ & I=\frac{E}{{{R}_{T}}}=\frac{45V}{60\Omega }=0.75A \\\end{align}\]

The current drawn from the source is exactly the same for a single 60-V resistor connected to its terminals as for the 20-V, 10-V, and 30-V resistors connected in series. Therefore, we can think of the resistances in series as being replaced by a single equivalent resistance R_{eq}.

**Figure 2** shows the equivalent circuit for the series circuit in Figure 1. When we analyze more elaborate circuits, we simplify a circuit by replacing two or more components in series with a single equivalent component.

**Figure 2** Equivalent circuit of Figure 1

**Figure 3** shows the simple series circuit of Figure 1 with a **voltmeter** and an **ammeter** added. Although the positions of R_{1} and R_{3} have changed in the circuit diagram, the connections between the components are the same as in Figure 1.

The ammeter must be connected in series with the circuit in order to measure the current through it.

Since the voltmeter is connected between points B and C, it measures the potential difference across R_{2}. The ideal voltmeter draws negligible current and thus does affect the behavior of the circuit.

In the circuit of Figure 3, electrons experience a potential rise inside the voltage source as it uses energy to move them from the positive terminal to the negative terminal.

As electrons flow around the external circuit, they lose the potential energy they gained while moving through the source. Hence, the electrons at point D are at a higher potential than those at point C, and electrons at point B are at a higher potential than those at point A.

Similarly, electrons at point B are at a lower potential than those at point C, and so on.

**Figure 3** Polarity of voltage drops in a series circuit

If negative ions in the voltage source are free to move, they flow in the same direction as the electrons. Any positive charge carriers flow in the opposite direction. Positive charge carriers inside a voltage source experience a potential rise as they move from the negative to the positive terminal.

A voltage source is an active circuit element since it generates electric energy (at the expense of some other form of energy). Passive circuit elements consume electric energy.

Electrons flow from the negative terminal to the positive terminal of a passive component. Thus, the polarity of the voltage drops indicates the direction of current flows in these components.

Note that an ammeter is a passive circuit element. Electrons enter the ammeter through its negative terminal and leave through its positive terminal.

In the circuit of Figure 3, electrons flow counterclockwise around the circuit. We can, therefore, mark the polarities of all voltage drops with 2 and 1 signs. The end of the resistor at which electrons enter is marked 2 and the end of the resistor from which electrons leave is marked 1.

In the circuit of Figure 3, the positive end of R_{3} is connected to the negative end of R_{2}, and so on. So, point C is positive with respect to point D, and point B is even more positive with respect to point D. Similarly, point C is negative with respect to point A, and so on.

For circuit diagrams, the nature of the actual charge carriers is not particularly important, but we often need to keep track of current direction. The direction of both conventional current and electron flow are shown in Figure 3.

The polarities of the voltage drops are exactly the same using either approach, but conventional current “flows” from the positive terminal of a passive circuit element to its negative terminal.

The remainder of the text will show only the direction of conventional current unless the nature of the actual charge carriers is important.

In Figure 3, V_{1} represents the voltage drop across R_{1}. The polarity of V_{1} is obvious since terminal A is positive with respect to terminal B.

Double-subscript notation lets us indicate the polarity of voltage drops. For example, V_{AB} represents the potential at point A with respect to point B. The second subscript normally designates the reference point. In Figure 3, V_{AB} is a positive voltage, V_{BA} is negative, and V_{AB}=-V_{BA}.

**The double-subscript**** notation** is essential for keeping track of voltage polarities and current directions in **three-phase circuits**. For most of the dc circuits, the single-subscript notation is adequate.

Multiplying both sides of Equation 1 by I gives

$I{{R}_{T}}=I{{R}_{1}}+I{{R}_{2}}+I{{R}_{3}}+\cdots $

Since

$\begin{matrix} E=I{{R}_{T}} & and & V=IR \\\end{matrix}$

Hence,

$E\begin{matrix} ={{V}_{1}}+{{V}_{2}}+{{V}_{3}}+\cdots & {} & \left( 2 \right) \\\end{matrix}$

In a series circuit, the sum of all the voltage drops across the individual resistances equals the applied voltage.

The German physicist Gustav Robert Kirchhoff (1824–1887) discovered that a similar relationship applies to any complete electric circuit.

**Kirchhoff’s voltage law:** In any complete electric circuit, the algebraic sum of applied voltages equals the algebraic sum of the voltage drops.

Kirchhoff’s voltage law provides us with another method for solving Example 1.

What current will flow in a series circuit consisting of a 45-V source, a 20-V, a 10-V, and a 30-V resistor?

Solution

$\begin{align} & E=I{{R}_{1}}+I{{R}_{2}}+I{{R}_{3}} \\ & 45=20I+10I+30I=60I \\ & I=\frac{45V}{60\Omega }=0.75A \\\end{align}$

Since the current is the same through all components of a series circuit, applying Ohm’s law to the voltage drops gives

\[I=\frac{{{V}_{1}}}{{{R}_{1}}}=\frac{{{V}_{2}}}{{{R}_{2}}}=\frac{{{V}_{3}}}{{{R}_{3}}}=\frac{{{V}_{T}}}{{{R}_{T}}}\]

By transposing variables in each pair of equal terms, we get

\[\begin{matrix} \begin{matrix} \frac{{{V}_{1}}}{{{V}_{2}}}=\frac{{{R}_{1}}}{{{R}_{2}}} & , \\\end{matrix} & \frac{{{V}_{1}}}{{{V}_{2}}}=\frac{{{R}_{1}}}{{{R}_{2}}} & \begin{matrix} , & and\text{ }so\text{ }on \\\end{matrix} \\\end{matrix}\]

In a series circuit, the ratio of the voltage drops across two resistances equals the ratio of the resistance.

We can use these ratios to save a step when solving for a specific parameter of a series circuit.

A 20-kV resistor and a 15-kV resistor are connected in series to a 140-V source. Find the voltage drop across the 15-kV resistor.

**Solution**

Often, drawing and labeling a schematic diagram, such as **Figure 4**, can help us see the relationships among the circuit parameters.

**Figure 4** Circuit diagram for Example 2

Use ratios to find the voltage drop.

Since,

\[\frac{{{V}_{2}}}{E}=\frac{{{R}_{2}}}{{{R}_{T}}}\]

So,

\[{{V}_{2}}=E\frac{{{R}_{2}}}{{{R}_{T}}}=140V\times \frac{15k\Omega }{35k\Omega }=60V\]

Since the same current flows through all components of a series circuit, changing any one component affects the current through all the components. Therefore, control components such as switches, fuses, and rheostats are connected in series with a load, as shown in **Figure 5**.

**Figure 5** Control components in series with a load

The **following characteristics** can help us recognize a series circuit:

- The current is the same in all parts of a series circuit.
- The total resistance is the sum of all the individual resistances:

${{R}_{T}}={{R}_{1}}+{{R}_{2}}+{{R}_{3}}+\cdots $

- The applied voltage is equal to the sum of all the individual voltage drops:

$E={{V}_{1}}+{{V}_{2}}+{{V}_{3}}+\cdots $

- The ratios of voltage drop equal the ratios of resistances.
- A change to any component of a series circuit affects the current through all components.

Most cells have a nominal voltage between 1.1 V and 1.5 V. The current rating of a cell depends on its size and type.

Identical cells connected in series provide higher voltages than a single cell. The current rating will be the same as for each individual cell since the current in a series circuit is the same in all parts of the circuit.

If cells with different capacities are connected in series, the lower capacity cells will be used up first. Most batteries for portable devices have cells connected in series, like those shown in **Figure 6**.

**Figure 6** Equivalent circuit of a series-connected battery

If S is the number of cells connected in series, E_{cell} is the EMF of each cell, and R_{cell} is the internal resistance of each cell, then total battery EMF is

$\begin{matrix} {{E}_{bat}}=S{{E}_{bat}} & {} & \left( 3 \right) \\\end{matrix}$

and total battery internal resistance is

$\begin{matrix} {{R}_{bat}}=S{{R}_{cell}} & {} & \left( 4 \right) \\\end{matrix}$

How much current flows through an external resistance of 4.0 Ω connected to a battery consisting of ten cells connected in series with each cell having an EMF of 1.5 V and an internal resistance of 0.20 V? Find the terminal voltage of the battery.

**Solution**

\[\begin{align} & {{E}_{bat}}=S{{E}_{bat}}=10\times 1.5V=15V \\ & {{R}_{bat}}=S{{R}_{cell}}=10\times 0.20\Omega =2\Omega \\\end{align}\]

The total resistance in the circuit,

${{R}_{T}}={{R}_{bat}}+{{R}_{L}}=2+4=6\Omega $

Applying Ohm’s law to the whole circuit gives

$\begin{align} & I=\frac{{{E}_{bat}}}{{{R}_{T}}}=\frac{15V}{6\Omega }=2.5A \\ & {{V}_{T}}=I{{R}_{L}}=2.5A\times 4\Omega =10V \\\end{align}$

**Figure 7** shows the equivalent circuit for the battery.

**Figure 7** Equivalent circuit for Example 3

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]]>The post Power & Efficiency: Definition, Unit, Formula, Examples appeared first on Electrical A2Z.

]]>Equation 1 shows the relationship between power and work:

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

Where P is power in watts, W is work in joules, and t is a time in seconds.

When dealing with work and power, keep in mind that joules are a measure of energy while watts are a measure of the rate of transferring energy.

We can use the equations for potential difference and current to relate these quantities to power in a circuit.

Since,

$V=\frac{W}{Q}$

So,

$W=VQ$

Similarly,

\[I=\frac{Q}{t}\]

So, Substituting for W and t in Equation 1 gives

\[\begin{matrix} P=\frac{W}{t}=\frac{VQ}{{}^{Q}/{}_{I}}=V\times I & {} & \left( 2 \right) \\\end{matrix}\]

Where P is power in watts, V is a voltage drop in volts, and I is current in amperes.

We can also substitute V = IR (voltage drop) into Equation 2 to get

$\begin{matrix} P=V\times I=IR\times I={{I}^{2}}R & {} & \left( 3 \right) \\\end{matrix}$

Where P is power in watts, I is current through the resistance in amperes, and R is resistance in ohms.

Similarly, substituting I = V/ R into Equation 2 gives

$\begin{matrix} P=V\times I=\frac{{{V}^{2}}}{R} & {} & \left( 4 \right) \\\end{matrix}$

Where P is power in watts, V is a voltage drop across the resistance in volts, and R is resistance in ohm.

A lamp draws a current of 2.00 A when connected to a 120-V source. How much power is going to the lamp?

**Solution**

$P=E\times I=120V\times 2A=240W$

When designing circuits, we need to specify the power rating of a resistor as well as its resistance. Since a resistor converts all power it consumes into heat, the power rating is the rate at which the resistor can dissipate heat without damage.

Heat dissipation depends on the surface area of the resistor and therefore on its physical size. Standard carbon-composition resistors are manufactured in sizes ranging from 1/8 W to 2 W, as shown in **Figure 1**.

**Figure 1** Approximate sizes of carbon-composition resistors

A 10-kV resistor is connected into a circuit where the current through it is 50 mA. What is the minimum power rating required for this resistor?

**Solution**

$P={{I}^{2}}R={{\left( 50mA \right)}^{2}}\times 10k\Omega =25W$

The resistor should be rated for at least 25 W.

What is the highest voltage that can be applied to a 3.3-kV, 2.0-W resistor without exceeding its heat-dissipating capability?

**Solution**

Since,

\[P=\frac{{{V}^{2}}}{R}\]

And,

$V=E=\sqrt{PR}=\sqrt{2W\times 3.3k\Omega }=81V$

Almost all equipment converts some of its input energy into a form that does not produce useful work. This wasted energy is usually in the form of heat.

The wasted energy increases the cost of making the equipment as well as the cost of running it since the equipment has to be designed to dissipate the wasted energy safely.

The efficiency of a device indicates how much of the input energy, W_{in}, the device converts into useful work, W_{out}.

Efficiency is the ratio of useful output energy to total input energy. The letter symbol for efficiency is the Greek letter η (eta).

\[\begin{matrix} \eta =\frac{{{W}_{out}}}{{{W}_{in}}} & {} & \left( 5 \right) \\\end{matrix}\]

We usually express efficiency as a percentage.

What is the efficiency of an electric hoist if its motor uses 60 KJ to raise a 300-kg mass through 18 m?

**Solution**

\[{{W}_{out}}=300kg\times 9.8{}^{m}/{}_{{{s}^{2}}}\times 18m=52.9kJ\]

\[\eta =\frac{{{W}_{out}}}{{{W}_{in}}}=\frac{52.9kJ}{60kJ}\times 100=88%\]

Since,

\[P=\frac{W}{t}\]

So,

$W=Pt$

Substituting for W_{out }and W_{in} in Equation 5 gives

\[\eta =\frac{{{W}_{out}}}{{{W}_{in}}}=\frac{{{P}_{out}}\times t}{{{P}_{in}}\times t}\]

So,

\[\begin{matrix} \eta =\frac{{{P}_{out}}}{{{P}_{in}}} & {} & \left( 6 \right) \\\end{matrix}\]

For electric motors, P_{out} is mechanical power output, and P_{in} is electric power input. All types of power can be measured in watts. However, in North America, the mechanical power output of motors is often expressed in terms of horsepower (hp), a unit invented by James Watt:

\[\begin{matrix} 1\text{ }hp=746\text{ }W & {} & \left( 7 \right) \\\end{matrix}\]

Find the input power for an electric motor that has an output of 24 hp and an efficiency of 85%.

**Solution**

$\begin{align} & {{P}_{out}}=24hp=24hp\times \frac{746W}{1hp}=17.9kW \\ & {{P}_{in}}=\frac{{{P}_{out}}}{\eta }=\frac{17.9kW}{0.85}=21kW \\\end{align}$

A joule is a relatively small quantity of energy. A typical home uses millions of joules of electrical energy each day. Since power × time = energy, we could use the product of any unit of power with any unit of time as a measure of energy.

In fact, a joule is equal to a watt-second. If we measure power in kilowatts and time in hours, we can use kilowatt hours (kWh) as a unit for energy:

$W=Pt$

Where W is work or energy in kilowatt hours, P is power in kilowatts, and t is a time in hours.

Since

1 kW = 1000 W and 1 h = 3600 s,

$\begin{matrix} 1kWh=1000W\times 3600s=3.6\times {{10}^{6}}J=3.6MJ & {} & \left( 8 \right) \\\end{matrix}$

Many North American electric utilities measure their customers’ energy consumption in kilowatt hours.

Most meters that record electric energy consumption are a form of a motor with an aluminum disk that rotates at a speed directly proportional to both the applied voltage and the current.

Since P = EI, the speed of the disk is proportional to the power used at any given moment, and the number of rotations indicates the energy consumed (power times time).

**Example 6**

At 7¢ per kilowatt-hour, how much will it cost to leave a 60-W lamp burning for five days?

**Solution**

$\begin{align} & W=Pt=60W\times 24h\times 5=7200Wh=7.2kWh \\ & Cost=7.2kWh\times \frac{7c}{kWh}=50.4c \\\end{align}$

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Definition:A resistor that maintains a constant V/I ratio is a linear resistor.

As the current through a resistor increases, more heat is produced in the resistor, raising its temperature. This increase in temperature causes a slight increase in the resistance of most conductor materials.

For the common conductor materials such as copper and aluminum, the change in resistance over the ranges of operating temperatures for most circuits is so small that these materials are usually considered to be linear resistors.

**Figure 1** Current versus voltage Characteristic curve for linear resistors

As the name suggests, **wire-wound resistors** have a metal wire wound on a hollow porcelain tube and sealed in position with a porcelain coating. These resistors are usually made with constantan or other alloys with a temperature coefficient of almost zero.

The larger the size of a resistor, the more readily it can dissipate heat to the surrounding air, and hence the greater the power rating.

Inexpensive, mass-manufactured resistors often have resistances that vary by 10% or more from their nominal values.

While such resistors are adequate for many types of circuits, applications such as measuring instruments require resistors that are accurate to within 1% or less of the nominal resistance. Such **precision resistors** are often made by depositing a thin film of metal or carbon on a small ceramic cylinder, to which leads are attached.

The film may be etched to adjust the resistance to the specified value before being coated with a layer of insulating material.

Resistors used in electronic devices usually have resistances greater than kilohms (a thousand ohms) and pass currents of only a few milliamperes.

A **carbon-composition resistor** is commonly used when the current through the resistor produces less than 2 W of heat. The resistance element consists of finely ground carbon mixed with an insulating binder such as phenolic and pressed into a cylindrical shape with a wire lead embedded in each end. The resistance element is then sealed in a plastic jacket.

The length and width of the cylinder, the proportion of carbon in the mixture, and the way the mixture is compressed to determine the resistance.

Carbon-composition resistors are much cheaper than wire-wound and film resistors, but the resistance of a carbon-composition resistor increases if the temperature varies appreciably from 20°C as shown in **Figure 2**.

For moderate temperature variations, the resistance changes by only a few percents, so we can usually treat carbon-composition resistors as linear resistors.

**Figure 2** Resistance-temperature characteristic curve of carbon-composition resistors

Composition resistors are also made with a ceramic consisting of tin oxide and antimony bound with glass. **Ceramic-composition resistors** are particularly useful for circuits where the resistors must withstand voltage or energy surges.

**Integrated circuits (ICs)** range from simple resistor networks to microprocessors containing millions of microscopic resistors and transistors. ICs start as a wafer of highly purified silicon.

Pure silicon is a poor conductor since it has few free charge carriers. The resistors are made by diffusing tiny regions of the silicon with precisely controlled amounts of elements that supply free electrons, making the regions more conductive.

Definition:The property exhibited by devices whose resistance does not change uniformly with changes in voltage or current.

When an incandescent lamp is switched on, the temperature and the resistance of its tungsten filament increase dramatically.

When white-hot, the filament of an ordinary 60-watt 120-V lamp has a resistance of 240 Ω, but its resistance at room temperature is about 18 Ω.

An incandescent lamp is a **nonlinear resistor.** The inrush current at the instant the lamp is turned on is much greater than its normal operating current:

$\begin{align} & Normal\text{ }I=\frac{V}{R}=\frac{120V}{240\Omega }=0.5A \\ & Inrush\text{ }I=\frac{V}{R}=\frac{120V}{18\Omega }=6.6A \\\end{align}$

Fortunately, the mass of the lamp filament is small enough that it gets white hot in less than a millisecond. Therefore, the current surge is brief, as shown in **Figure 3**.

Nevertheless, switches used with incandescent lamps have to be designed to withstand the inrush current. To avoid large inrush currents, heating elements, such as those in stoves, are usually made from an alloy with a very small temperature coefficient.

**Figure 3** Inrush current of a 60-W incandescent lamp

The curve in **figure 4** shows how the resistance changes with voltage for an incandescent lamp.

**Figure 4:** Incandescent Lamp Voltage-Current Characteristics Curve

A resistor with a large negative temperature coefficient, called a **thermistor,** can be used to limit inrush currents.

**Typically**, such thermistors have a resistance of over 100 Ω at room temperature, but with a current of 1 A through them, their resistance drops to less than 1 Ω after 10 to 15 s.

**These thermistors** contain semi-conductive metal oxides with a ceramic binder. Heat produced by current through the resistor breaks covalent bonds in the metal oxides, creating enough free electrons to reduce the resistance to a fraction of its value at room temperature.

Small thermistors are used to measure temperatures since a decrease in temperature of less than 20°C will more than double their resistance.

**Figure 5: **Thermistor resistance-temperature Characteristic Curve

**Varistors **depend on the nonlinear resistance characteristic of zinc oxide or silicon carbide crystals, which are formed into wafers with a clay binder.

Zinc oxide Varistors (also called metal-oxide Varistors or MOVs) are used to protect sensitive electronics from voltage surges.

Silicon carbide Varistors (commonly known by the trade name thyrite) can protect high- voltage systems. They are used as lightning arrestors on power transmission lines.

Temperature has little effect on the resistance of a varistor. Instead, a rapid increase in the number of charge carriers occurs when the potential difference across the varistor becomes greater than the threshold of the varistor. As **Figure 6** indicates, the resulting decrease in resistance is such that the current through the varistor increases greatly without appreciably increasing the voltage drop across it.

**Thus,** a varistor connected across the power input to a device can protect it from voltage surges.

**Figure 6** Typical varistor characteristic curve

A **photoresistor**, photoconductor, or light dependent resistor (LDR) contains a thin zigzag strip of cadmium sulfide or cadmium selenide.

Light falling on the strip breaks down valence bonds in the cadmium compound, creating additional charge carriers.

The resistance can range from hundreds of kilohms in the dark to less than 100 Ω in bright daylight. Photoresistors are widely used in light meters, auto-exposure circuits in cameras, and controllers for outdoor lights.

**Figure 7:** Photoresistor Characteristic Curve

Thermistors, Varistors, and photoresistors all use semiconductors engineered to become more conductive under specific conditions. The schematic symbols for these nonlinear resistors are shown in **Table 1**.

**Table 1** Schematic symbols for nonlinear resistors

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]]>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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