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Question

When a battery sends current through the resistances $R_1$ and $R_2R$, the internal resistance of battery is :

A
R1R22
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B
R1R2
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C
R1R22
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D
R1+R22
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Solution

The correct option is D R1R2
Here current through the resistance R1 is
I1=εr+R1
and current through the resistance R2 is
I2=εr+R2.
Also the heat produced due to passage of current through both the resistances is same.
Hence,
Q=I21R1=I22R2
Using the values of I1 and I2 in above equation we get
(εr+R1)2R1=(εr+R2)2R2
where, ε is the emf and r is the internal resistance of battery.
ε2R1(r+R1)2=ε2R2(r+R2)2
r2(R1R2)=R1R2(R1R2)
r2=R1R2
r=R1R2

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