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In the Circuit Shown in Figure Below, E1 and E2 Are Two Cells Having Emfs 2 V and 3 V Respectively, and Negligible Internal Resistances. - Physics (Theory)

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प्रश्न

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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उत्तर

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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