Question

A cell of E.M.F. E and internal resistance r supplies currents for the same time t through external resistance$R_1$ and $R_2$ respectively. If the heat produced in both cases is the same then the internal resistance is

• A. 1/r = 1/$R_1$ + 1/$R_2$
• B. r = ($R_1$ + $R_2$)/2
• C. r = ($R_1$$R_2$)
• D. r = $R_1$+$R_2$

Answer 1

The correct option is C r = ($R_1$$R_2$)

If $R_1$ is there

heat released $=l^2{R}_{1}t$ $\to \left(i\right)$

$i=\frac{\epsilon }{R_1+r}$ $\to \left(2\right)$

Put (2) in (1)

Heat released $=\frac{{\epsilon }^2}{{(R}_1+r{)}^2}{R}_{1}t$ $\to \left(3\right)$

If $R_2$ is there

heat released $=i^2{R}_{2}t\to \left(5\right)$

$i=\frac{\epsilon }{R_2+r}\to \left(6\right)=\frac{{\epsilon }^2}{(R_2+r{)}^2}{R}_{2}t\to \left(4\right)$

Given (4) and (3) are same

$\frac{{\epsilon }^2{R}_{1}t}{{(R}_1+r{)}^2}=\frac{{\epsilon }^2{R}_{2}t}{(R_2+r)}\frac{R_1}{{(R}_1+r{)}^2}=\frac{R_2}{{(R}_2+r{)}^2}\to R_1(R_2^2+r^2+2R_{2}r)\text{=}{R}_2(R_1^2+r^2+2R_{1}r)\to R_1{R}_2^2+R_1{r}^2+2R_1{R}_{2}r=R_2{R}_1^2+R_2{r}^2+2R_1{R}_{2}r\to R_1{R}_2^2+R_1{r}^2=R_2{R}_1^2+R_2{R}^2\to R_1{R}_2^2-R_2{R}_1^2=r^2(R_2-R_1)\to R_1{R}_2(R_2-R_1)=r^2(R_2-R_1)\to r^2=R_1{R}_2$