Question

In the circuit shown in figure, $R_1=1\Omega ,R_2=2\Omega ,,C_1=1\mu F,C_2=2\mu F$ and $E=6V$. Calculate charge on each capacitor in steady state.

• A. $q_1=3\mu C,q_2=15\mu C$
• B. $q_1=2\mu C,q_2=12\mu C$
• C. $q_1=4\mu C,q_2=12\mu C$
• D. $q_1=2\mu C,q_2=15\mu C$

Answer 1

The correct option is B $q_1=2\mu C,q_2=12\mu C$

In steady state,

Current in the circuit$\left(I\right)=\frac{E}{R_1+R_2}$

$I=2A$

$C_2$ is directly along the battery with potential difference = $6V$

$q_2=6C_2=12\mu C$

$C_1$ is along $R_1$ and has voltage of $I{R}_1=2V$

$q_1=2C_1=2\mu C$

Option $\text{B}$ is correct answer.