Nortons Theorem

Electrical Circuit Analysis > Network Theorems

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What is Norton's Theorem?

Norton's theorem states that a linear two-terminal circuit can be replaced by an equivalent circuit consisting of a current source $I_N$ in parallel with a resistor $R_N$, where $I_N$ is the short-circuit current through the terminals and $R_N$ is the input or equivalent resistance at the terminals when the independent sources are turned off.

Thus, the circuit in [Fig. 1(a)] can be replaced by the one in [Fig. 1(b)].

Fig. 1: (a) Original circuit, (b) Norton equivalent circuit.

Fig. 2:Finding Norton current IN.

How to find Norton's current ($I_N$) and Norton's Resistance $R_N$?

Here we are mainly concerned with how to get $R_N$ and $I_N$. We find $R_N$ in the same way we find $R_{Th}$ in Thevenin's theorem. In fact, from what we know about source transformation, the Thevenin and Norton resistances are equal; that is,

$$\bbox[10px,border:1px solid grey] {R_N = R_{Th}}$$

To find the Norton current $I_N$, we determine the short-circuit current flowing from terminal a to b in both circuits in [Fig. 1]. Since the two circuits are equivalent in [Fig. 1(a)] and [(b)]. Thus, $I_N = i_{sc}$ as shown in [Fig. 2]. Dependent and independent sources are treated the same way as in Thevenin's theorem. If the two terminals a to b with open circuit are closed, then a current limited by $R_{Th}$ will start flowing between the terminals. Hence from $V_{Th}$ and $R_{Th}$ we can find norton's current $I_N$. Thus,

$$\bbox[10px,border:1px solid grey]{I_N = { V_{Th} \over R_{Th}}} \tag{1}$$

Who developed Norton's Theorem?

In 1926, about 43 years after Thevenin published his theorem, E. L. Norton, an American engineer at Bell Telephone Laboratories, proposed a similar theorem.

Thevenin-Norton transformation

Since $V_{Th}$, $I_N$, and $R_{Th}$ are related according to Eq. (1), to determine the Thevenin or Norton equivalent circuit requires that we find:

The open-circuit voltage $v_{oc}$ across terminals a and b.
The short-circuit current $i_{sc}$ at terminals a and b.
The equivalent or input resistance Rin at terminals a and b when all independent sources are turned off.

We can calculate any two of the three using the method that takes the least effort and use them to get the third using Ohm's law. Example 1. will illustrate this. Also, since

$$ \begin{aligned} V_{Th} &= v_{oc} & (2)\\ I_N &= i_{sc} & (3)\\ R_{Th} &= {v_{oc} \over i_{sc}} = R_N & (4) \end{aligned}$$

the open-circuit and short-circuit tests are sufficient to find any Thevenin or Norton equivalent.

Example 1: Find the Norton equivalent circuit of the circuit in [Fig. 3].

Fig. 3: For Example 1.

Solution:
We find $R_N$ in the same way we find $R_{Th}$ in the Thevenin equivalent circuit. Set the independent sources equal to zero. This leads to the circuit in [Fig. 4(a)], from which we find $R_N$. Thus,

$$\begin{split} R_N &= 5\, ||\, (8 + 4 + 8) = 5\, ||\, 20\\ &= {20\, || \,5 \over 25} = 4 Ω \end{split}$$

To find $I_N$, we short-circuit terminals a and b, as shown in [Fig. 4(b)]. We ignore the 5-Ω resistor because it has been short-circuited. Applying mesh analysis, we obtain

$$i_1 = 2 A, \, 20i_2 - 4i_1 - 12 = 0$$

From these equations, we obtain

$$i_2 = 1 A = i_{sc} = I_N$$

Alternatively, we may determine $I_N$ from $V_{Th}/R_{Th}$. We obtain $V_{Th}$ as the open-circuit voltage across terminals a and b in [Fig. 4(c)]. Using mesh analysis, we obtain

$$i_3 = 2 A$$

$$25i_4 - 4i3 - 12 = 0 \Rightarrow i_4 = 0.8 A$$

and

$$v_{oc} = V_{Th} = 5i_4 = 4 V$$

Fig. 4: For Example 1; finding: (a) $R_N$, (b) $I_N$, (c) $V_{Th}$

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