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Question

A 5 V battery and a 2 V battery are connected to the resistances as shown. The current in the 10 Ω resistor is :


A
0.03 A, P1 to P2
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B
0.03 A, P2 to P1
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C
0.27 A, P1 to P2
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D
0.27 A, P2 to P1
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Solution

The correct option is B 0.03 A, P2 to P1
Let's assume potential at P1 to be zero, and potential at P2 to be x.


Now let's take currents in the branches and apply KCL

i1 + i2 + i3 = 0

x52 + x010 + x+21 = 0

x = 516 V

So, current through 10 Ω resistor will be

i2 = 5/16010 = 0.03 A

So, 0.03 A current will flow from P2 to P1.

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