This theorem is sometimes useful in solution of networks in which some branches may contain sources of e. m.f. It is applicable only to linear networks where current is linearly related to voltage as per Ohm’s law.
This theorem may be stated as follows :
“In any network containing more than one source of e.m.f. the current in any branch is the algebraic sum of a number of individual fictitious currents (the number being equal to the number of sources of e.m.f.), each of which is due to the separate action of each source of e.m.f., taken in order, when the remaining sources of e.m.f. are replaced by conductors, the resistances of which are equal to the internal resistances of the respective sources”.
The procedure of applying superposition theorem is as follows :
1. Replace all but one of the sources by their internal resistances. If the internal resistance of any source is small as compared to other resistances present in the network, the source is replaced by a short circuit.
2. Find the currents in different branches by using Ohm’s law.
3. Repeat the process using each of the e.m.f s. as the sole e.m.f. each time.
The total current in any branch of the circuit is the algebraic sum of currents due to each source.
When finding total current in any branch, it is necessary to take into account the directions of the currents caused by each individual source, currents flowing in the same direction being additive, currents flowing in opposite directions being subtractive.
- In Fig. 49, I 17 I2 and I represent the values of currents which are due to the simultaneous action of the two sources of e.m.f. in the network.
- In the Fig. 50 are shown the current values which would have been obtained left-hand side battery had acted alone.
- Similarly Fig. 51 represents conditions obtained when right-hand side battery acts alone.
By combining the current values of Fig. 50 and 51, the actual values of Fig. 49 can be obtained Obviously,
I 1 = I1 – I1
I2 = I2 – I2
I = I + I
Example 14. By using superposition theorem find the currents in the different branches of the network shown in Fig. 52
Solution. 11 : , 12 :, I:
First step. Refer Fig. 53.
Take e.m.f. E1 only and replace e.m.f. E2 by its zero internal resistance, the circuit is shown in Fig. 53.
= 8 + 10 x 12 / 10 / 12 = 13.45 Q
Current through 8 .Q resistance,
I1 = 20 / 13.45 = 1.487 A
Current through 10 .Q resistance
I2 = 1.487 x 10 / 12 + 10 = 0.81 A
Current through 6 .Q resistance,
I’ = 1.487 x 10 / 12 + 10 = 0.675 A
Second step. Refer Fig. 54.
E.m.f. E 1 is removed/short circuited and current due to e.m.f. E2 is found. The current is shown in the Fig. 54. · ·
Total resistance = 10 + 12 x 8 / 12 x 8 = 14 .8 Q
Current through 10 .Q resistance,
I2 = 28 / 14.8 = 1.892 A
Current through 8 n resistance,
I1 = 1.892 x 12 / 12 + 8 = 1.135 A
Current through 12 Q resistance,
I = 1.892 x 8 / 12 + 8 = 0.757 A
The total currents in different branches are :
Current through 8 n resistance,
I1 = I1‘ –I1 = 1.487 – 1.135 = 0.352 A (from L to M). (Ans.)
Current through 10 n resistance,
I2 = I2“- I2‘ = 1.892- 0.81 = 1.082 A (from N to M). (Ans.)
Current through 12 n resistance,
I= I’+ I”= 0.675 + 0.757 = 1.432 A (from M to Q). (Ans.)
Example 15. By using superposition theorem, find the current in 2 n resistance shown in Fig. 55. Internal resistances of the cells are negligible.