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Question
In the circuit shown in the figure below, E1 and E2 are two cells having emfs 2 V and 3 V respectively, and negligible internal resistance. Applying Kirchhoff’s laws of electrical networks, find the values of currents l1 and I2.
Solution
The distribution of current in the circuit is as shown in figure
Applying Kirchoff's laws (Loop law) to loop ABEFA
-2 (I1 + I2) - I1 × 1 + 2 = 0
2 - I1 - 2(I1 + I2) = 0
⇒ 2 - 3I1 - 2I2 = 0 .....(i)
Applying to loop BCDEB
-3 + 6I2 + 2(I1 + I2) = 0
⇒ 3 - 6I2 - 2I1 - 2I2 = 0
⇒ 3 - 8I2 - 2I1 = 0 ....(ii)
Solving equations (i) and (ii), we can write
I1 = `1/2`A , I2 = `1/4` A
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