Two batteries A and B, each of e.m.f. 2 , V, | vedclass

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(C) The total e.m.f. of the series combination is $E_{eq} = E_A + E_B = 2 \, V + 2 \, V = 4 \, V$.The total resistance of the circuit is $R_{total} =

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$0 \, V$

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(C) The total e.m.f. of the series combination is $E_{eq} = E_A + E_B = 2 \, V + 2 \, V = 4 \, V$.The total resistance of the circuit is $R_{total} = R + r_A + r_B = 1 \, \Omega + 1.9 \, \Omega + 0.9 \, \Omega = 3.8 \, \Omega$.The current flowing through the circuit is $i = \frac{E_{eq}}{R_{total}} = \frac{4 \, V}{3.8 \, \Omega} = \frac{40}{38} \, A = \frac{20}{19} \, A$.The potential difference across the terminals of battery $A$ is given by $V_A = E_A - i r_A$.Substituting the values, $V_A = 2 \, V - (\frac{20}{19} \, A) 	imes 1.9 \, \Omega$.Since $1.9 = \frac{19}{10}$, we have $V_A = 2 - (\frac{20}{19} 	imes \frac{19}{10}) = 2 - 2 = 0 \, V$.

Two batteries $A$ and $B$, each of e.m.f. $2 \, V$, are connected in series to an external resistance $R = 1 \, \Omega$. If the internal resistance of battery $A$ is $1.9 \, \Omega$ and that of $B$ is $0.9 \, \Omega$, what is the potential difference between the terminals of battery $A$?

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