Question

When a battery sends current through a resistance $R_1$ for time t,the heat produced in the resistor is Q.When the same battery sends current through another resistance$R_2$ for time t, the heat produced in $R_2$ is again Q.The internal resistance of battery is

• A. $\frac{R_1+R_2}{2}$
• B. $\frac{R_1-R_2}{2}$
• C. $\sqrt{R_1{R}_2}$
• D. $\sqrt{\frac{R_1+R_2}{2}}$

Answer 1

The correct option is C $\sqrt{R_1{R}_2}$

As heat produced in similar time is same for two cases,

Power dissipated should be same in $R_1$ and $R_2$.

Let internal resistance of battery be $R$ and Emf be $E$.

Thus, In first case, Power dissipated in $R_1$ is $P_1=\frac{E^2}{(R+R_1{)}^2}{R}_1$

Similarly, Power dissipated in $R_2$ is $P_2=\frac{E^2}{(R+R_2{)}^2}{R}_2$

As, $P_1=P_2$

$\frac{E^2}{(R+R_1{)}^2}{R}_1=\frac{E^2}{(R+R_2{)}^2}{R}_2$

$⟹R^2(R_2-R_1)=R_1{R}_2(R_2-R_1)$

$⟹R=\sqrt{R_1{R}_2}$

Option C is correct.