## This article covers Norton’s Theorem and Norton’s Equivalent Circuit Parameters, such as Norton’s Equivalent Resistance and Short-circuit Current, along with Norton’s Theorem Solved Problem with Dependent Sources.

## What is Norton’s Theorem?

Norton’s Theorem states that any two-terminal electric network comprising resistances and voltage and/or current sources may be substituted with a single current source connected in a parallel configuration with a single resistance. The current source output is the short-circuit current at the terminals of a network, whereas the parallel resistance is the resistance between the terminals of a network when all the sources are set to zero.

**Find the Norton’s Equivalent Circuit**

Suppose we are given an arbitrary circuit containing any or all of the following elements: resistors, voltage sources, and current sources (the source can be dependent as well as independent). Let us identify a pair of nodes, say nodes a and b, such that the circuit can be partitioned into two parts, as shown in Figure 1.

**Figure.1:** Circuit partitioned into two parts

Furthermore, suppose that circuit A contains no dependent source that is dependent on a variable in circuit B, and vice versa. Then, we can model circuit A by an appropriate independent current source, call it i_{sc}, which is connected in parallel with an appropriate resistance, call it R_{N}. This parallel combination of a current source and resistance is called **Norton’s equivalent of circuit A**. In other words, circuit A in Figure 1, and the circuit in the shaded box in Figure 2 have the same effect on circuit B. This result is known as Norton’s theorem.

**Figure.2:** Norton’s Equivalent Circuit

Circuit B (which is often called a load) may consist of many circuit elements, a single element ( a load resistor), or no element.

- You May Also Read: Thevenin’s Theorem with Solved Examples

**Norton’s Short Circuit Current**

To obtain Norton’s equivalent of some circuit A, determine the short-circuit current i_{sc} by placing a short circuit between nodes a and b and then calculate the resulting current (directed from the terminal a to b through the short circuit).

**Figure.3 (a):** Determination of short-circuit current

**How to Find Norton’s Equivalent Resistance **

There are several ways to find Norton’s equivalent resistance, which are given below:

**1**. To obtain the resistance R_{N}– called Norton’s equivalent resistance of circuit A:

**I.** Remove circuit B from circuit A.

**II.** Set all independent sources in circuit A to zero. (A zero voltage source is equivalent to a short circuit, and a zero current source is equivalent to an open circuit).

**III.** Determine the resistance between nodes a and b- this is R_{N}– as shown in figure 3(b).

**Figure.3 (b):** Determination of Norton’s Equivalent Resistance

**2.** Short-circuit the terminals a and b, then find the short-circuit current I_{sc}. Norton’s equivalent resistance is given by

\[{{\text{R}}_{\text{N}}}\text{ = }{{\text{V}}_{\text{oc}}}\text{/}{{\text{I}}_{\text{sc}}}\text{ = }{{\text{V}}_{\text{th}}}\text{/}{{\text{I}}_{\text{sc}}}\]

Whereas V_{oc} or V_{th} can be found as was done for the Thevenin equivalent circuit.

**3.** When the source network has a ladder structure and contains no controlled (dependent) sources, R_{N} is easily found by series-parallel reduction of the dead network. When the source network contains controlled sources, Norton’s equivalent resistance can be found using the method represented in figure 4.

**Figure.4:** Find Norton’s Equivalent Resistance

Here, the dead source network has been connected to an external test source. This test source may be any independent voltage or current source that establishes v_{o} and it at the terminals. Since the dead network contains only resistors and controlled sources, and since R_{N} equals the equivalent resistance of the dead network, the equivalent resistance theorem tells us that

\[{{\text{R}}_{\text{N}}}\text{=}{{\text{v}}_{\text{o}}}\text{/}{{\text{i}}_{\text{o}}}\]

**Norton’s Theorem Solved Example **

Find Norton’s Equivalent circuit of the following circuit.

Let’s find Norton Equivalent resistance R_{N}

- Open circuit all current sources
- Short circuit all voltage sources

Now, we have the following circuit to calculate R_{N}

\[{{R}_{N}}=\text{ }5||\text{ }\left( 8+4+8 \right)\]

\[{{R}_{N}}=4\text{ }\Omega \]

In order to find short-circuit current i_{sc}, we short-circuit terminals a and b:

Using loop analysis we have;

\[{{i}_{1}}=2A\]

$20{{i}_{2}}-4{{i}_{1}}-12=0$

\[{{i}_{2}}=1A\]

So;

\[{{I}_{N}}={{i}_{sc}}={{i}_{2}}=1A\]

Hence we have followed Norton’s equivalent circuit

**Norton’s Theorem Solved Problem with Dependent Sources**

We will find I_{sc} and R_{N} for the circuit shown in the following figure.

In order to determine the short-circuit current, we’ll short-circuit terminals a and b as shown in the following figure and V_{2} = 0.

Apply KCL to node 1:

\[\begin{matrix} -0.003+\frac{{{V}_{1}}}{3000}+\frac{{{V}_{1}}}{2000}=0 & \cdots & (1) \\\end{matrix}\]

Multiplying (1) by 6000 yields:

\[\begin{align} & 5{{V}_{1}}=\text{ }18 \\ & {{V}_{1}}=\text{ }3.6\text{ }V \\\end{align}\]

Now, calculating currents:

\[i=\text{ }{{V}_{1}}/3000\text{ }=\text{ }3.6\text{ }V/3000\Omega =\text{ }0.0012\text{ }A\]

\[{{V}_{CCVS}}=\text{ }2000i\text{ }=\text{ }2.4\text{ }V\]

\[{{I}_{{{R}_{2}}}}=\text{ }{{V}_{1}}/{{R}_{2}}=1.8mA\]

\[{{I}_{{{R}_{3}}}}=\text{ }{{V}_{CCVS}}/{{R}_{3}}=2.4mA\]

Finally, I_{N} would be

\[{{I}_{N}}={{I}_{{{R}_{2}}}}+\text{ }{{I}_{{{R}_{3}}}}~=\text{ }1.8mA+\text{ }2.4mA=\text{ }4.2mA\]

In order to determine R_{N}, and deactivate the current source, a test voltage of 1 V is applied across terminals a and b as shown in the following figure. V_{t} = 1 V.

\[{{I}_{{{R}_{2}}}}=i={{V}_{t}}/({{R}_{1}}+\text{ }{{R}_{2}})\text{ }=\text{ }1\text{ }V/5\text{ }k\Omega =0.2mA\]

\[{{V}_{CCVS}}=\text{ }2000i\text{ }=\text{ }0.4\text{ }V\]

\[{{I}_{{{R}_{3}}}}=\text{ }({{V}_{t}}\text{- }{{V}_{CCVS}})/{{R}_{3}}=0.6mA\]

\[{{I}_{{{R}_{4}}}}={{V}_{t}}/{{R}_{4}}=0.25mA\]

\[{{I}_{t}}={{I}_{{{R}_{2}}}}~+{{I}_{{{R}_{3}}}}~+{{I}_{{{R}_{4}}}}=\text{ }0.2mA+\text{ }0.6mA+\text{ }0.25mA=\text{ }1.05mA\]

Noe, Norton’s equivalent resistance can be found as

\[{{R}_{N}}={{V}_{t}}/{{I}_{t}}=\text{ }952.381\Omega \]

**Summary of Norton’s Theorem Steps**

The following steps can be carried out for solving a circuit using **Norton’s Theorem**:

- The Portion of the circuit considered as the load is removed
- The short circuit current i
_{sc }is calculated at terminals - To calculate R
_{N}:

- Short circuit all independent voltage sources
- Open circuit all independent current sources

- Draw Norton’s equivalent circuit, connect the load, and calculate the load current.