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Question

# A 5 V battery with internal resistance 2 Ω and a 2 V battery with internal resistance 1 Ω are connected to a resistor as shown in the figure. The current in the resistor is

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

## The correct option is C 0.03 A,P2 to P1Let potential at P1 is 0 V and potetial at P2 is V0. Now apply KCL at P2. V0−52+V0−010+V0−(−2)1=0 V0=516 So, current through 10 Ω resistor is, V=IR (Ohm's law) ⇒I=ΔVR=51610=132 A or I=0.03125≈0.03 A (From P2 to P1)

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