How To Nortonize a Given Circuit?

This procedure is based on the first statement of the theorem given above.
1. Remove the resistance (if any) across the two given tenninals and put a short-circuit across
them.
2. Compute the short-circuit current Isc-
3. Remove all voltage sources but retain their internal resistances, if any. Similarly, remove
all current sources and replace ..'1~n> Jpen-circuits i.e. by infinite resistance.
4. Next, find the resistance RI (also called RN)of the network as looked into from the given
terminals. It is exactly the same as R'h
5. The current source (Isc)joined in parallel across Rj between the two terminals gives Norton's
equivalent circuit. .

(b) Norton Equivalent Circuit
A short placed across tenninals A and B will short out R2 as well. Hence, Isc =EIR1. The
Norton equivalent resistance is exactly the same as Thevenin resistance except that it is connected in
parallel with the current source as shown in
 Using Norton's theorem, find the constant-current equivalent of the circuit

Solution. When tenninals A and B are short-circuited as shown, total resistance
of the circuit, as seen by the battery, consists of a 10 Q resistance in series with a parallel
combination of 10 Q and 15 Q resistances.This current is divided into two parts at point C of .
Current through A B is Isc =6.25 x 10/25=2.5 A
Since the battery has no internal resistance, the input resistance of the network when viewed
fromA and B consists of a 15.0 resistance in series with the parallel combination of 10.0 and 10.0.
Hence, RI =15 + (10/2) =20 .a
Hence, the equivalent constant-current source is as shown in .
. Apply Norton's theorem to calculate currentflowing through 5 - Q resistor of

Solution. (i) Remove 5 - .a resistor and put a short across terminals A and B as shown in
. As seen, 10 -.a resistor also becomes short-circuited.
(ii) Let us now find Isc' The battery sees a parallel combination of 4.0 and 8.0 in series with
a 4 .a resistance. Total resistance seen by the battery =4 + 4 II8 =20/3.0. Hence, 1= 20 + 20/3 =
3 A. This current divides at point C of Fig. 2.205 (b). Current going along path CAB gives Isc- Its
value =3 x 4/12 = 1 A
The negative sign shows that the actual direction of flow of 12is opposite to that shown in Fig.

As Applicable to Voltage Sources
This Theorem is a combination of Thevenin's and Norton's theorems. It is used for finding the
common voltage across any network which contains a number of parallel voltage sources as shown
in . Then common voltage VABwhich appears across the output terminals A and B is
affected by the voltage sources EI' E2 and E3' The value of the voltage is given by
E)IR) + EzlRz +E31R3 _ I) +/z +/3 _ ~
VAB = --
11R) + 11Rz+ 11R3 G) + Gz + G3 1:G
This voltage represents the Thevenin's voltage Vth. The resistance Rthcan be found, as usual, by
replacing each voltage source by a short circuit. If there is a load resistance RL across the terminals
A and B, then load current IL is given by
IL = VtJ(Rth + RJ
 a branch does not contain any voltage source, the same procedure
is used except that the value of the voltage for that branch is equated to zero as illustrated in . Hence, ISh =IN =12 = - 5/4 A i.e. current flows from point B to A.
After the terminals A and B are open-circuited and the three batteries are replaced by shortcircuits
(since their internal resistances are zero), the internal resistance of the circuit, as viewed
from these terminals' is
Rj = RN=2+4114=4Q
The Norton's equivalent circuit consists of a constant current source of 5/4 A in parallel with a
resistance of 4 Q as shown in . When 6 Q resistance is connected across the equivalent
circuit, current through it can be found by the current-divider rule (Art).
Current through 6 Q resistor = % x 1~ =0.5 from B to A
General instructions For Finding Norton Equivalent Circuit As Applicable to Current Sources
This theorem is applicable to a mixture of parallel voltage and current sources that are reduced
to a single flam equivalent source which is either a constant current or a constant voltage source.
This theorem can be stated as follows :
Any number of constant current sources which are directly connected in parallel can be converted
into a single current source whose current is the algebraic sum of the individual source currents and
whose total internal resistances equals the combined individual source resistances in parallel

Procedure for fmding Norton equivalent circuit of a given network has already been given in
Art. That procedure applies to circuits which contain resistors and independent voltage or current
sources. Similar procedures for circuits which contain both dependent and independent sources or
only dependent sources are given below:
(a) Circuits Containing Both Dependent and Independent Sources
(i) Find the open-circuit voltage Vo<with all the sources activated or 'alive'.
(ii) Find short-circuit current ishby short-circuiting the terminals a and b but with all sources
activated.
(iii) RN=Vo!ish
(b) Circuits Containing Dependent Sources Only
(i) ish= o.
(ii) Connect I A source to the terminals a and b calculate v00.
(iii) RN = v0011.