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(D) The circuit consists of two networks connected in series.Each network forms a balanced Wheatstone bridge.For the first network,the equivalent resi

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$V_0 / 3 R$

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(D) The circuit consists of two networks connected in series.Each network forms a balanced Wheatstone bridge.For the first network,the equivalent resistance $R_1$ is calculated by simplifying the parallel combinations of resistors,which results in $R_1 = R$.For the second network,the equivalent resistance $R_2$ is calculated similarly,which results in $R_2 = 2R$.The total equivalent resistance of the circuit is $R_{AB} = R_1 + R_2 = R + 2R = 3R$.Using Ohm's law,the current $I$ is given by $I = V_0 / R_{AB} = V_0 / 3R$.

In the circuit shown,the end $A$ is at potential $V_0$ and end $B$ is grounded. The electric current $I$ indicated in the circuit is

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