Whereas Thevenin’s theorem was used to simplify a network to a constant-voltage source and a series resistance, Norton’s theorem can be used to resolve a network into a constant-current source and a parallel resistance. The interchange of voltage sources and current sources by use of Thevenin’s and Norton’s theorems is sometimes useful in circuit analysis.

The theorem may be stated as follows :

“Any two terminal linear network containing independent voltage and current sources may be replaced by an equivalent current IN in parallel with a resistance RN where IN is the short circuit current at network terminals and RN is the equivalent resistance of network as seen from the terminals but with all voltage sources short circuited and all current sources open circuited.”

The following procedure may be adopted to determine the Norton’s equivalent circuit:

1. Calculate the short circuit current (IN) at the network tem1inals.

2. Redraw the network with each voltage source replaced by a short circuit in series with its internal resistance and each current source by an open circuit in parallel with its internal resistance.

3. Calculate the resistance (RN) of the redrawn network as seen from the network terminals.

The resistance RN is the same value as used in Thevenin’s equivalent circuit).

**Example 21.** By using Norton’s theorem find the current in the 12 Q resistance of the circuit shown in Fig. 89*.*

**Solution.**

– With 12 Q resistance removed and terminals L-M short circuited, short-circuit current,

I_{N} = 20 / 8 = 2.5 A

– With 20 V battery replaced by a short circuit, the resistance of the network as seen from terminals L and M is

R_{N} = 8 (10 + 14) / (10 + 14) = 6 Q

– The Norton’s equivalent circuit is shown in Fig. 90. The current through 12 Q resistance is I = 2.5 x 6 / 6 + 12 = 0.833 A

** **

**Example 22.** Using Norton’s theorem calculate current through a load of B Q in the circuit shown in Fig. 96.

**Solution.**

• Refer Fig. 97. It shows load impedance replaced by a short circuit.

I_{SC} = I_{N} =200 / 2 = 100 A

Norton’s resistance RN can be found by looking into the open terminals of Fig. 98. For this purpose Δ ABC has been replaced by its equivalent star. As seen RN is equal to 8 / 7Q.

Thus, Norton’s equivalent circuit consists of a 100 A source having a parallel resistance of 8/7

Q as shown in Fig. 98 (iii).

IL = 100 x (8 / 7) / 8 + (8/7) = 12.5 A