Question

In the given circuit diagram when the current reaches steady state in the circuit, the charge on the capacitor of capacitance C will be:

• A. $CE$
• B. $CE\left(\frac{{\displaystyle r_1}}{{\displaystyle r_2+r}}\right)$
• C. $CE\left(\frac{{\displaystyle r_2}}{{\displaystyle r_2+r}}\right)$
• D. $CE\left(\frac{{\displaystyle r_1}}{{\displaystyle r_1+r}}\right)$

Answer 1

The correct option is C

$CE\left(\frac{{\displaystyle r_2}}{{\displaystyle r_2+r}}\right)$

Step 1: At the steady-state, the current through the capacitor is zero

Since the capacitor behaves as an open circuit, then the current through $r_1$ is $0$, therefore

Current through the circuit is given by $I=\frac{E}{r+r_2}$,-------------(1)

Step 2: Calculate the voltage across the capacitor $V_C$ and remove resistance $r_1$ since no current flows through it

The voltage across the capacitor is obtained by

$$\begin{gathered} V_C=I·r_2----(From1) \\ {V}_C=\left(\frac{E}{r+r_2}\right)r_2 \\ {V}_C=\frac{E{r}_2}{r+r_2}-----\left(2\right) \end{gathered}$$

Step 3: Use the formula $Q=CV$ to find the charge stored in the capacitor

$\begin{array}{rcl}Q& =& C{V}_C-----(From2)\\ Q& =& C·\frac{E{r}_2}{r+r_2}\\ & =& CE\left(\frac{r_2}{r+r_2}\right)\end{array}$

Hence, the correct option is (C).