A current of 2,A flows through a 2,Omega resi | vedclass

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(B) Let $\varepsilon$ be the electromotive force (emf) and $r$ be the internal resistance of the battery.According to Ohm's law for a complete circuit

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$\frac{1}{3}\,\Omega$

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(B) Let $\varepsilon$ be the electromotive force (emf) and $r$ be the internal resistance of the battery.According to Ohm's law for a complete circuit,the current $I$ is given by $I = \frac{\varepsilon}{R + r}$,where $R$ is the external resistance.In the first case,$I_1 = 2\,A$ and $R_1 = 2\,\Omega$:$2 = \frac{\varepsilon}{2 + r} \implies \varepsilon = 2(2 + r) = 4 + 2r$    ....$(i)$In the second case,$I_2 = 0.5\,A$ and $R_2 = 9\,\Omega$:$0.5 = \frac{\varepsilon}{9 + r} \implies \varepsilon = 0.5(9 + r) = 4.5 + 0.5r$    ....$(ii)$Equating the expressions for $\varepsilon$ from $(i)$ and $(ii)$:$4 + 2r = 4.5 + 0.5r$$2r - 0.5r = 4.5 - 4$$1.5r = 0.5$$r = \frac{0.5}{1.5} = \frac{1}{3}\,\Omega$

$A$ current of $2\,A$ flows through a $2\,\Omega$ resistor when connected across a battery. The same battery supplies a current of $0.5\,A$ when connected across a $9\,\Omega$ resistor. The internal resistance of the battery is

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