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

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

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:

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