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(B) The circuit is a Wheatstone bridge. The condition for a balanced Wheatstone bridge is $\frac{P}{Q} = \frac{R}{S}$.Here,$\frac{4}{2} = 2$ and $\fra

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$10$

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(B) The circuit is a Wheatstone bridge. The condition for a balanced Wheatstone bridge is $\frac{P}{Q} = \frac{R}{S}$.Here,$\frac{4}{2} = 2$ and $\frac{10}{5} = 2$. Since the ratios are equal,the bridge is balanced.In a balanced Wheatstone bridge,no current flows through the central branch $BD$.Let the current through the upper branch $ABC$ be $I_1$ and the current through the lower branch $ADC$ be $I_2$.Since the total current is $7 \ A$,we have $I_1 + I_2 = 7 \ A$.The potential difference across the upper branch $AC$ is $V_{AC} = I_1(4 + 2) = 6I_1$.The potential difference across the lower branch $AC$ is $V_{AC} = I_2(10 + 5) = 15I_2$.Equating the two,$6I_1 = 15I_2$,which gives $I_1 = 2.5I_2$.Substituting $I_1 = 2.5I_2$ into $I_1 + I_2 = 7$,we get $2.5I_2 + I_2 = 7$,so $3.5I_2 = 7$,which means $I_2 = 2 \ A$.Then $I_1 = 7 - 2 = 5 \ A$.The potential difference between $B$ and $C$ is $V_{BC} = I_1 	imes R_{BC} = 5 \ A 	imes 2 \ \Omega = 10 \ V$.

If a current of $7 \ A$ flows through the electric circuit,what is the potential difference between $B$ and $C$ in volts?

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