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(B) Let $E$ be the $EMF$ of the battery and $r$ be its internal resistance.The terminal voltage $V$ is given by $V = E - Ir$,where $I = \frac{V}{R}$.T

Vedclass · Accepted Answer

$\frac{20}{7}$

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(B) Let $E$ be the $EMF$ of the battery and $r$ be its internal resistance.The terminal voltage $V$ is given by $V = E - Ir$,where $I = \frac{V}{R}$.Thus,$E = V + \frac{V}{R}r = V(1 + \frac{r}{R})$.For the first case: $R_1 = 16 \ \Omega$,$V_1 = 12 \ V$.$E = 12(1 + \frac{r}{16}) = 12 + \frac{12r}{16} = 12 + \frac{3r}{4} \quad \dots (i)$For the second case: $R_2 = 10 \ \Omega$,$V_2 = 11 \ V$.$E = 11(1 + \frac{r}{10}) = 11 + \frac{11r}{10} \quad \dots (ii)$Equating $(i)$ and $(ii)$:$12 + \frac{3r}{4} = 11 + \frac{11r}{10}$$1 = \frac{11r}{10} - \frac{3r}{4}$$1 = \frac{22r - 15r}{20} = \frac{7r}{20}$$r = \frac{20}{7} \ \Omega$.

When a battery is connected across a resistor of $16 \Omega$,the voltage across the resistor is $12 \ V$. When the same battery is connected across a resistor of $10 \Omega$,the voltage across it is $11 \ V$. The internal resistance of the battery in ohm is

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