A battery of emf E and internal resistance r is connected to a resistor of resistance r₁and Q joules of heat is produced in a certain time t. When the same battery is connected to another resistor of resistance r₂, the same quantity of heat is produced in the same time t, the value of r is

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$\frac{{r_{1}}^{2}}{r_{2}}$

$\frac{{r_{2}}^{2}}{r_{1}}$

$\frac{1}{2} (r_{1} + r_{2})$

$\sqrt{r_{1} r_{2}}$

answer is D.

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Detailed Solution

In the first case, the current in the circuit is

$I_{1} = \frac{E}{r_{1} + r}$

$∴ Q_{1} = I^{2} r_{1} t = {(\frac{E}{r_{1} + r})}^{2} \times r_{1} t$ (i)

In the second case,

$Q_{1} = {(\frac{E}{r_{2} + r})}^{2} \times r_{2} t$ (ii)

Equating (i) and (ii) we get

$\frac{r_{1}}{{(r_{1} + r)}^{2}} = \frac{r_{2}}{{(r_{2} + r)}^{2}} o r r_{1} {(r_{2} + r)}^{2} = r_{2} {(r_{1} + r)}^{2} o r r_{1} ({r_{2}}^{2} + 2 r r_{2} + r^{2}) = r_{2} ({r_{1}}^{2} + 2 r r_{1} + r^{2}) o r r^{2} (r_{1} - r_{2}) = r_{1} r_{2} (r_{1} - r_{2}) o r r = \sqrt{r_{1} r_{2}}$