Fig. 1.38 (i) shows a circuit enclosed in a box with A and B terminals brought out. The network in the box may have any number of e.m.f. sources and resistors connected in any manner. But according to Norton, the complete circuit behind A and B terminals can be changed by a current source of output IN in parallel with a one resistance RN as shown in Fig. 1.38 (ii). The value of IN is found as mentioned in Norton’s theorem. The resistance RN is the exactly same as Thevenin’s resistance R0. Once Norton’s equivalent is found [See Fig. 1.38 (ii)], then-current passes through any load RL connected across terminals AB can be readily obtained.

Hence Norton’s theorem as applied to direct current circuits may be stated as under Any network having two A and B terminals can be changed by a current source of output IN in parallel with a resistance RN

(i) The output IN of the current source is exactly equal to the current that would passes through AB when A and B terminals are short-circuited.

(ii) RN is the resistance of the network measured between A and B terminals with

load (RL) removed and sources of e.m.f. replaced or changed by their internal resistances, if any. Norton’s theorem is converse of Thevenin’s theorem in that Norton equivalent uses a current producer instead of voltage generator and resistance RN (which is the same as R0 ) in parallel with the generator instead of being in series with it. Illustration. Fig. 1.39 illustrates the application of Norton’s theorem. As far as the circuit behind AB terminals is concerne [See Fig. 1.39 (i)], it can be changed by a current source of output IN in parallel with a resistance RN as shown in Fig. 1.39 (iv). The output IN of the current producer or generator is equal to the current that would passes through AB terminals when A and B terminals are short-circuited as shown in Fig. 1.39 (ii). The load R′ on the source when terminals AB are short-circuited is given by :

To determine RN, remove the load RL and change the voltage source by a short circuit because its resistance is assumed 0 [See Fig. 1.39 (iii)].

Thus the IN and RN values are known. The Norton equivalent circuit will be as shown in Fig. 1.39 (iv).

- Open the two terminals as the Thevenin theorem (i.e. remove any load) between which we need to find Norton equivalent circuit.
- (ii) make a short-circuit across the terminals under consideration. determine the short-circuit current flowing or passing in the short circuit.
**It is known as Norton’s current IN.** - (iii) Find the resistance between the two open terminals with all ideal current sources opened and all ideal voltage sources shorted (a non-ideal source is replaced by its internal resistance).
**It is called Norton’s resistance RN**. It is easy to see that RN = R0 - (iv) Connect IN and RN in parallel to develop Norton equivalent circuit between the two terminals under consideration.
- make the load resistor removed in step (i) across the terminals of the Norton equivalent circuit. The load current can now be determined by using the current-divider rule. This load current is the same as the load current in the original or first given circuit.

**Example 1.12.** **Using Norton’s theorem, determine the current in 8 Ω resistor in the network shown in** Fig. 1.40 (i).

**Solution.** We shall decrease the network to the left of AB in Fig. 1.40 (i) to Norton’s equivalent. For this reason, we need to find IN and RN.

(i) With removed load (i.e., 8 Ω) and terminals, AB short-circuited [See Fig. 1.40 (ii)], the current that passes through AB is equal to IN. Referring to Fig. 1.40 (ii),

**(ii)** With removed load (i.e., 8 Ω) and battery changed by a short (since its internal resistance is assumed zero), the resistance at AB terminals are equal to RN as shown in Fig. 1.41 (i).

The Norton’s equivalent circuit behind AB terminals is IN (= 3.24 A) in parallel with RN (= 7.4 Ω). When the load (i.e., 8 Ω) is placed across AB terminals, the circuit becomes as shown in Fig. 1.41 (ii). The current source is providing current to two resistors **7.4 Ω and 8 Ω **in parallel.

Example 1.13. **determine the Norton equivalent circuit at terminals X – Y in** Fig. 1.42.

Solution. We shall first determine the Thevenin equivalent circuit and then convert or transfer it to an equivalent current source. This will be then Norton equivalent circuit.

**Finding the Thevenin Equivalent circuit.** To find E0, refer to Fig. 1.43 (i). Since 18 V and 30 V sources are in opposition, the circuit current I is given by :

Fig. 1.44 (iii) shows the Norton equivalent circuit. analyze that the Norton equivalent resistance has the exact same value as the Thevenin equivalent resistance was. Therefore, RN is found exactly the same way.

Example 1.14. **Show that when Thevenin’s equivalent circuit of a network is transferred into Norton’s equivalent circuit, IN= E0/R0 and RN = R0. Here R0 and E0 are Thevenin resistance**

Solution. Fig. 1.45 (i) shows a network enclosed in a box with two A and B terminals brought out. Thevenin’s equivalent circuit of this network will be as shown in Fig. 1.45 (ii). To determine Norton’s equivalent circuit, we are to determine IN and RN. Referring to Fig. 1.45 (ii)

IN= Current passes through short-circuited AB in Fig. 1.45 (ii)**= E0/R0**

RN= Resistance at terminals AB in Fig. 1.45 (ii)

= R0

Fig. 1.45 (iii) shows Norton’s equivalent circuit. Hence we reached at the following two important conclusions :

(i) To convert or transfer Thevenin’s circuit into Norton’s equivalent circuit,

It is the normally practice to mount the electronic components on a metal base** known as chassis.** **For example**, in Fig. 1.46, the resistors and voltage sources are connected to the chassis. As the resistance of chassis is very small or very low, therefore, it gives a conducting path and may be considered as a piece of wire.

It is customary to refer to the chassis as ground. Fig. 1.47 shows the symbol for chassis. It will be seen that all terminals connected to chassis are shown as grounded and represent or shows the same potential. The adoption of this thing (i.e. showing points of the same potential as grounded) often make it easy for the electronic circuits. In our further discussion, we shall most use this scheme.

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