**Norton’s theorem** is used for simplifying a network in terms of currents instead of voltages. This theorem can be used with either single-source or multiple-source circuits. In certain cases, analyzing the division of currents may be easier than voltage analysis.

Norton’s theorem simplifies a resistive network and represents it with a **Norton equivalent current source ( I_{N})** in parallel with an equivalent

**Norton resistance (**as shown in

*R*),_{N}**Figure 1**.

The basis of Norton’s theorem is the use of a current source to supply a total load current that is divided among parallel branches.

**Figure 1** Norton equivalent circuit.

**Norton’s Equivalent Circuit**

As with similar theorems, certain steps have to be performed to arrive at an equivalent circuit. The following steps are used to convert a resistive network into its Norton equivalent circuit:

- Calculate the Norton equivalent current source. This is equal to the current that would flow between terminals
*A*and*B*if the load resistor was removed and replaced with a short circuit. - Calculate the Norton equivalent resistance. This is equal to the resistance between terminals
*A*and*B*when the voltage source is removed and replaced with a short circuit.

**Norton’s Equivalent Circuit Example 1**

**Problem:** Derive the Norton equivalent circuit for the simple resistive network circuit shown in **Figure 2**. Calculate the load current and voltage for the 6-Ω load resistor *R _{L}*. Repeat the calculations for a load resistor of 3 Ω.

**Figure 2** Circuit for Example 1.

**Solution:**

** Step 1.** Short-circuit the load terminals

*A*–

*B*as shown in

**Figure 3**. Calculate the resulting current flow. This is the value of the Norton equivalent current source

*I*

_{N}. Note that the short circuit across terminals

*A*–

*B*short-circuits both

*R*and the parallel

_{L }*R*

_{2}. Then the only resistance in the circuit is

*R*

_{1}in series with the voltage source.

**Figure 3** Circuit for step 1.

**Step 2**. The Norton equivalent resistance (R_{N}) is equal to the resistance between terminals A and B with the load removed and the voltage source replaced with a short circuit, as shown in **Figure 4**. The resistance seen looking back into the circuit from terminals A-B is then R_{1} in parallel with R_{2}.

**Figure 4** Circuit for step 2.

**Step 3.** The resultant Norton equivalent circuit is shown in **Figure 5**. It consists of a 6-A current source (I_{N}) in parallel with a 3-Ω resistance (R_{N}).

**Figure 5** Norton equivalent circuit.

**Step 4.** To calculate IRL and ERL, connect the load resistor to the equivalent Norton circuit, as shown in **Figure 6**. The current source still delivers 6 A, but now that current is divided between the two branches of R_{N} and R_{L}. The load current can be calculated using the current divider rule, and the load voltage can be calculated using Ohm’s law.

**Figure 6** Circuit for step 4.

** Step 5.** When the value of

*R*is changed, the values of

_{L}*I*

_{N}and

*R*

_{N}remain the same. Therefore the load current and voltage for a 3-Ω load can be determined as follows without recalculating the entire circuit.

**Norton’s Equivalent Circuit Example 2**

**Problem:** Derive the Norton equivalent circuit for resistive network shown in **Figure 7**.

**Figure 7** Circuit for Example 2.

**Solution:**

** Step 1.** Short-circuit the load, as shown in

**Figure 8**, to determine the Norton equivalent current. Note that the short circuit current

*I*

_{N}in this example is a branch current, not the main line current. Resistors

*R*

_{2}and

*R*

_{3}are now in parallel so the total resistance seen by the voltage source is

**Figure 8** Circuit for step 1.

The total current is then

The Norton equivalent current is then found by using the current divider rule:

**Step 2**. Short-circuit the voltage source, as shown in **Figure 9** to find the Norton equivalent resistance of the circuit. R_{1} and R_{3} are now in parallel, and R_{2} is in series with this parallel combination.

**Figure 9** Circuit for step 2.

**Step 3.** The Norton equivalent circuit can then be drawn as shown in **Figure 10**.

**Figure 10** Norton equivalent circuit.

The Norton equivalent circuit may also be determined directly from the Thevenin equivalent circuit, and vice versa. **Figure 11** shows the relationship between the two circuits. The following formulas can be used to convert from one equivalent circuit to the other:

From Thevenin to Norton

From Norton to Thevenin

**Figure 11** Norton–Thevenin conversions.