Answer

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**Hint**Find the total current of the given circuit from the rearranged ohm’s law. Again substitute the value of the resistance and the calculated current in it to find the potential drop across the two branches with the capacitor and the $6\,\Omega $ resistor.

Useful formula:

The ohm’s law is given by

$V = IR$

Where $V$ is the potential difference, $I$ is the current and $R$ is the resistance developed in the circuit.

**Complete step by step answer**

From the given circuit diagram, it is understood that it possesses three branches with the capacitor at the middle branch. Let us use the formula of the ohm’s law,

$V = IR$

Let us find the total current through the circuit, by rearranging the above formula, we get

$I = \dfrac{V}{R}$

Let us consider that the capacitor is fully charged. Since it is fully charged, it will not draw any current from the total current. Hence the current through the branch of the capacitor is zero. Let us substitute this in the above formula, we get

$I = \dfrac{3}{{2 + 4}}$

By performing basic arithmetic operations, we get

$I = \dfrac{1}{2}\,A$

The potential drop across the $6\,\Omega $ resistor is found using the ohm’s law.

$V = IR = 0 \times 6$

$V = 0$

Hence the potential drop across the capacitor circuit is zero.

The potential drop across the $4\,\Omega $ resistor is calculated using the ohm’s law.

$V = \dfrac{1}{2} \times 4$

By simplification of the above step, we get

$V = 2\,V$

Hence the potential drop across the $4\,\Omega $ resistor is calculated as $2$ volt. Since the voltage drop across the parallel circuit is the same, the potential drop in the branch of $4\,\Omega $ resistor must be equal to that of the potential drop across the $4\,\mu F$ capacitor.

**Thus the option (D) is correct.**

**Note:**Remember that in the parallel circuit, the total current through the circuit will be equal to the sum of the current through the resistors. The voltage drop across each branch will be equal to the voltage drop across other branches and also the total voltage drop.

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