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(B) The total electromotive force (emf) of the circuit is $E_{eq} = E_1 - E_2 = 8 \ V - 2 \ V = 6 \ V$.The total resistance of the circuit is $R_{eq}

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$6$

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(B) The total electromotive force (emf) of the circuit is $E_{eq} = E_1 - E_2 = 8 \ V - 2 \ V = 6 \ V$.The total resistance of the circuit is $R_{eq} = r_1 + r_2 = 2 \ \Omega + 4 \ \Omega = 6 \ \Omega$.The current flowing in the circuit is $I = \frac{E_{eq}}{R_{eq}} = \frac{6 \ V}{6 \ \Omega} = 1 \ A$.Since the cell $E_2$ is being charged (current enters its positive terminal),its terminal potential difference $V$ is given by $V = E_2 + Ir_2$.Substituting the values,we get $V = 2 \ V + (1 \ A 	imes 4 \ \Omega) = 2 \ V + 4 \ V = 6 \ V$.

Two cells are connected in opposition as shown. Cell $E_1$ has an electromotive force (emf) of $8 \ V$ and an internal resistance of $2 \ \Omega$; cell $E_2$ has an emf of $2 \ V$ and an internal resistance of $4 \ \Omega$. The terminal potential difference of cell $E_2$ is: (in $V$)

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