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(B) First,calculate the equivalent external resistance $R_{ext}$. The resistors $3 \, \Omega$ and $6 \, \Omega$ are in parallel,so their equivalent re

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$4.25 \, V, 7.5 \, V$

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(B) First,calculate the equivalent external resistance $R_{ext}$. The resistors $3 \, \Omega$ and $6 \, \Omega$ are in parallel,so their equivalent resistance is $R_p = \frac{3 	imes 6}{3 + 6} = 2 \, \Omega$.This $R_p$ is in series with the $4.5 \, \Omega$ resistor,so $R_{ext} = 2 + 4.5 = 6.5 \, \Omega$.The total resistance of the circuit is $R_{total} = R_{ext} + r_1 + r_2 = 6.5 + 0.5 + 1 = 8 \, \Omega$.The net $emf$ of the circuit is $E_{net} = E_2 - E_1 = 8 - 4 = 4 \, V$.The current in the circuit is $i = \frac{E_{net}}{R_{total}} = \frac{4}{8} = 0.5 \, A$.For cell $E_1$,the current flows into the positive terminal (charging),so the terminal voltage $V_1 = E_1 + i r_1 = 4 + (0.5 	imes 0.5) = 4 + 0.25 = 4.25 \, V$.For cell $E_2$,the current flows out of the positive terminal (discharging),so the terminal voltage $V_2 = E_2 - i r_2 = 8 - (0.5 	imes 1) = 8 - 0.5 = 7.5 \, V$.

For the given circuit,the $emf$ of the cells are $E_1 = 4 \, V$ and $E_2 = 8 \, V$,and their internal resistances are $0.5 \, \Omega$ and $1 \, \Omega$ respectively. Calculate the terminal voltage of each cell.

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