In 1926, about 43 years after Thevenin published his theorem, E.L. Norton, an American engineer at Bell Telephone, proposed a similar theorem.
Make sure to read what is electric circuit first.
These circuit analysis theorems are classified as:
Norton’s Theorem Theory
Before learning about Norton’s theorem, let us read the brief explanation below :
Norton’s theorem states that a linear twoterminal 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 shortcircuit 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 Figure.(1a) can be replaced by the one in Figure.(1b).
Figure 1 
We find R_{N} in the same way we find R_{Th}. In fact, from what we know about source transformation, the Thevenin and Norton resistances are equal; that is,
(1) 
To find the Norton current I_{N}, we determine the shortcircuit current flowing from the terminal a to b in both circuits in Figure.(1). It is evident that the shortcircuit current in Figure.(1b) is I_{N}.
This must be the same shortcircuit current from the terminal a to b in Figure.(1a), since the two circuits are equivalent. Thus,
(2) 
shown in Figure.(2). Dependent and independent sources are treated the same way as in Thevenin’s theorem.
Figure 2 
Observe the close relationship between Norton’s and Thevenin’s theorems: R_{N} = R_{Th} as in Equation.(1), and
(3) 
This is essentially source transformation. For this reason, source transformation is often called TheveninNorton transformation.
Since V_{Th}, I_{N}, and R_{Th} are related according to Equation.(3), to determine the Thevenin or Norton equivalent circuit requires that we find :

The opencircuit voltage v_{oc} across terminals a and b.

The shortcircuit current i_{sc} at terminals a and b.

The equivalent or input resistance R_{in} 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. Also, since
(4a) 
(4b) 
(4c) 
the opencircuit and shortcircuit test are sufficient to find any Thevenin or Norton equivalent, of a circuit which contains at least one independent source.
Norton’s Theorem Examples
For better understanding let us review the examples below :
1. Find the Norton equivalent circuit of the circuit in Figure.(3) at terminals ab.
Figure 3 
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 Figure.(4a), from which we find R_{N}. Thus,
To find I_{N}, we shortcircuit terminals a and b, as shown in Figure.(4b). We ignore the 5 Ω resistor because it has been shortcircuited. Applying mesh analysis, we obtain
From these equations, we obtain
Figure 4 
Alternatively, we may determine I_{N} from V_{Th/}R_{Th}. We obtain V_{Th} as the opencircuit voltage across terminals a and b in Figure.(4c). Using mesh analysis, we obtain
and
as obtained previously. This also serves to confirm Equation.(4c) that R_{Th} = v_{oc}/i_{sc} = 4/1 = 4 Ω. Thus, the Norton equivalent circuit is as shown in Figure.(5).
Figure 5 
2. Using Norton’s theorem, find R_{N} and of the circuit in Figure.(6) at terminals ab.
Figure 6 
Solution :
To find R_{N}, we set the independent voltage source equal to zero and connect a voltage source of v_{o} = 1 V (or any unspecified voltage v_{o}) to the terminals. We obtain the circuit in Figure.(7a).
We ignore the 4 Ω resistor because it is shortcircuited. Also due to the short circuit, the 5 Ω resistor, the voltage source, and the dependent current source are all in parallel.
Hence i_{x} = 0. At node a, i_{o} = 1v/5Ω = 0.2 A, and
To find I_{N}, we shortcircuited terminals a and b and find the current i_{sc}, as indicated in Figure.(7b).
Note from this figure that the 4 Ω resistor, the 10 V voltage source, the 5 Ω resistor, and the dependent current source are all in parallel.
Hence,
At node a, KCL gives
Thus
Figure 7 
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