Question

Find the potential drop across $4\mu F$ capacitor and $6\Omega$ resistor in figure.

• A. $0V,0V$
• B. $0V,3V$
• C. $0V,2V$
• D. $2V,0V$

Answer 1

The correct option is D

$2V,0V$

Step 1: Find the current in the circuit

We know a capacitor draws no current.

Therefore, no current flows through the branch containing the capacitor.

The total resistance of the circuit, excluding the branch containing the capacitor,

$$\begin{gathered} R=2+4 \\ \Rightarrow R=6\Omega \end{gathered}$$

Current,

$$\begin{gathered} I=\frac{V}{R} \\ \Rightarrow I=\frac{3}{6} \\ \Rightarrow I=\frac{1}{2} \end{gathered}$$

Step 2: Finding the potential drop across $4\mu F$ capacitor

The potential drop across $4\Omega$ resistor,

$$\begin{gathered} V_{4\Omega }=I\times 4 \\ \Rightarrow V_{4\Omega }=\frac{1}{2}\times 4\left(\because I=\frac{1}{2}\right) \\ \Rightarrow V_{4\Omega }=2V \end{gathered}$$

The potential drop across $4\Omega$ resistor is $2\hspace{0.17em}V$

Since no current flows through the branch containing the capacitor,

The whole potential drop occurs across the $4\mu F$ capacitor.

Since it is a parallel combination,

The potential drop across the $4\mu F$ capacitor = The potential drop across the $4\Omega$ resistor

Therefore, the potential drop across the $4\mu F$ capacitor is $2\hspace{0.17em}V$

Step 3: Find the potential drop across $6\Omega$ resistor

Since no current flows through the branch containing the capacitor,

The potential drop across the $6\Omega$ resistor, $V_{6\Omega }=0\times 6=0V$

Therefore, the potential drop across the $6\Omega$ resistor is $0\hspace{0.17em}V$

So, the potential drop across the $4\mu F$ capacitor is $2\hspace{0.17em}V$ and the potential drop across the $6\Omega$ resistor is $0\hspace{0.17em}V$.

hence, option D is the correct answer.