Question

A capacitor of $4\mu F$ is connected as shown in the circuit. The internal resistance of the battery is $0.5\Omega$. The amount of charge on the capacitor plates will be

• A. $0$
• B. $4\mu C$
• C. $16\mu C$
• D. $8\mu C$

Answer 1

The correct option is D

$8\mu C$

Step 1: Given data

EMF, $E=2.5V$

Resistance in loop $1:R_1=10\Omega$

Resistance in loop $2:R_2=2\Omega$

Internal resistance, $r=0.5\Omega$

Capacitance, $C=4\mu F$

Step 2: Formula used

Charge on a capacitor is given as $Q=CV$, where $C$ is the capacitance and $V$ is the voltage difference across the capacitor.

Step 3: Determine the current in loop 2

The current $I$ in loop 2 can be determined as,

$$\begin{gathered} I=\frac{E}{R_2+r} \\ =\frac{2.5}{2+0.5} \\ =1A \end{gathered}$$

(We know that, $I=\frac{Volatge}{Resis\tan ce}$. In this loop, the voltage is $E$ and the resistance is $R_2+r$)

Step 4: Determine the potential difference across AB

$$\begin{gathered} V_{AB}=E-Ir \\ =2.5-1(0.5) \\ =2V \end{gathered}$$

(The voltage is the sum of the voltage due to the current and the voltage provided in the loop.)

Step 5: Determine the charge on the capacitor

$$\begin{gathered} Q=CV \\ =4\times 2 \\ =8\mu C \end{gathered}$$

Hence, option D is the correct option.