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(A) The total number of batteries is $n = 4$. Each battery has an $EMF$ $\varepsilon = 1.5 \ V$ and internal resistance $r = 0.1 \ \Omega$.When batter

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$3 \ V, 0.4 \ \Omega$

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(A) The total number of batteries is $n = 4$. Each battery has an $EMF$ $\varepsilon = 1.5 \ V$ and internal resistance $r = 0.1 \ \Omega$.When batteries are connected in series,if one battery is connected in reverse,its $EMF$ opposes the others.Resultant $EMF$ $\varepsilon' = (\varepsilon + \varepsilon + \varepsilon) - \varepsilon = 2\varepsilon$.$\varepsilon' = 2 	imes 1.5 \ V = 3 \ V$.Since internal resistances are always in series regardless of the polarity of the $EMF$,the resultant internal resistance $r'$ is the sum of all individual internal resistances.$r' = r + r + r + r = 4r$.$r' = 4 	imes 0.1 \ \Omega = 0.4 \ \Omega$.Therefore,the resultant $EMF$ is $3 \ V$ and the resultant internal resistance is $0.4 \ \Omega$.

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