Question

The circuit shown in the figure consists of a battery of emf $ϵ=10V$; a capacitor of capacitance $C=1.0\mu F$ and three resistor of values $R_1=2\Omega ,R_2=2\Omega$ and $R_3=1\Omega$. Initially the capacitor is completely uncharged and the switch $S$ is open. The switch $S$ is closed at $t=0$.

• A. The current through resistor $R_3$ at the moment the switch closed is zero
• B. The current through resistor $R_3$ a long time after the switch closed is $5A$
• C. The ratio of current through $R_1$ and $R_2$ is always constant
• D. The maximum charge on the capacitor during the operation is $5\mu C$

Answer 1

The correct option is D The maximum charge on the capacitor during the operation is $5\mu C$

When the switch S is just closed , the capacitor will start charging through resistors $R_1,R_2$ and no current will pass through resistor $R_3$.
The current through $R_1$ is $i_1=i_{10}{e}^{-t/C{R}_1}$ and current through $R_2$ is $i_2=i_{20}{e}^{-t/C{R}_2}$
As $R_1=R_2=2,i_{10}=i_{20}=10/2=5$ and $C{R}_1=C{R}_2=2$
Thus the ratio $\left(i_1/i_2\right)$ is always constant.
After switch closed long time the charging of capacitor will stop and the circuit will treat as shown in figure below.
$R_{eq}=\frac{2\times 2}{2+2}=1\Omega$
Current through $R_3$ is $I=\frac{10}{1+1}=5A$
The capacitor charged maximum when potential across $C$ is $V_C=V_{R_3}=I{R}_3=5V$
thus, max charge on $C$ is $Q=C{V}_C=5\mu C$