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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.
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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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