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Question

Physics Question on Resistance

A cell of constant emf first connected to a resistance R1R_1 and then connected to a resistance R2R_2. If power delivered in both cases is same then the internal resistance of the cell is :

A

R1R2\sqrt{R_1 R_2}

B

R1R2\sqrt{\frac{R_1}{R_2}}

C

R1R22\frac{R_1 - R_2}{2}

D

R1+R22\frac{R_1 + R_2}{2}

Answer

R1R2\sqrt{R_1 R_2}

Explanation

Solution

Current given by cell
I=ER+rI=\frac{E}{R +r}
Power delivered in first case
P1=I2R1P_{1}=I^{2} R_{1}
=(ER1+r)2R1=\left(\frac{E}{R_{1}+r}\right)^{2} R_{1}
Power delivered in second case
P2=I2R2P_{2}=I^{2} R_{2}
=(ER2+r)2R2=\left(\frac{E}{R_{2}+r}\right)^{2} R_{2}
Power delivered is same in the both the cases
(ER1+r)2R1=(ER2+r)2R2\left(\frac{E}{R_{1}+r}\right)^{2} R_{1}=\left(\frac{E}{R_{2}+r}\right)^{2} R_{2}
R1(R1+r)2=R2(R2+r)2\frac{R_{1}}{\left(R_{1}+r\right)^{2}}=\frac{R_{2}}{\left(R_{2}+r\right)^{2}}
R1(R22+r2+2R2r)R_{1}\left(R_{2}^{2}+r^{2}+2 R_{2} r\right)
=R2(R12+r3+2R1r)=R_{2}\left(R_{1}^{2}+r^{3}+2 R_{1} r\right)
R1R22+R1r2+2R1R2rR_{1} R_{2}^{2}+R_{1} r^{2}+2 R_{1} R_{2} r
=R2R12+R2r2+2R1R2r=R_{2} R_{1}^{2}+R_{2} r^{2}+2 R_{1} R_{2} r
R1R22R2R12=R2r2R1r2R_{1} R_{2}^{2}-R_{2} R_{1}^{2}=R_{2} r^{2}-R_{1} r^{2}
R1R2(R2R1)=r2(R2R1)R_{1} R_{2}\left(R_{2}-R_{1}\right)=r^{2}\left(R_{2}-R_{1}\right)
r=R1R2r=\sqrt{R_{1} R_{2}}