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Q.A cell of emf EE and internal resistance rr is connected to two resistors R1R_1 and R2R_2 in series. The current is II. When R2R_2 is removed, current becomes I/2I/2. Then rr equals

a
R1R_1
b
R2R_2
c
2R12R_1
d
R1R2/(R1+R2)R_1 R_2/(R_1 + R_2)

Correct Answer: Option A

The correct solution involves applying the fundamental concept to derive the final value step by step...

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