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

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(C) Let the electromotive force of the battery be $\varepsilon$ and its internal resistance be $r$.According to Ohm's law for a circuit,the current $I

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$1/3\,\Omega$

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(C) Let the electromotive force of the battery be $\varepsilon$ and its internal resistance be $r$.According to Ohm's law for a circuit,the current $I$ is given by $I = \frac{\varepsilon}{R + r}$.For the first case: $2 = \frac{\varepsilon}{2 + r} \implies \varepsilon = 2(2 + r) = 4 + 2r$  (Equation $1$).For the second case: $0.5 = \frac{\varepsilon}{9 + r} \implies \varepsilon = 0.5(9 + r) = 4.5 + 0.5r$ (Equation $2$).Equating the two expressions for $\varepsilon$:$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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