Question

A cell of E.M.F. E and internal resistance r supplies currents for the same time t through
external resistances R1 and R2 respectively. If the heat produced in both cases is the
same, then the internal resistance is
(A) 1/r= 1/R1+1/R2
(B) r = (R1+R2)/2
(C) r = √R1R2
(D) R1 + R2​

Answer 1

Answer:

(C) $r=\sqrt{R_1R_2}$

Solution:

Heat produced in a resistor of resistance R through which current I for time t is given by:

$\boxed{H=I^2Rt}---(2)$

Since :

$I=\frac{V}{R}\\\\I=\frac{E-Ir}{R}\\\\\boxed{I=\frac{E}{r+R}}---(1)$

From (1) and (2)

∴$\boxed{H=(\frac{E}{r+R})^2Rt}$

Since H₁=H₂;

⇒ $(\frac{E}{r+R_1})^2R_1t =(\frac{E}{r+R_2})^2R_2t$

⇒$\frac{R_1}{(r+R_1)^2} =\frac{R_2}{(r+R_2)^2}$

On solving the above, we get $\boxed{r=\sqrt{R_1R_2}}$

Hope this answer helped you :)