Question

In the circuit shown in figure, the sum of charges on both capacitors at steady state will be:

• A. $5\mu \text{C}$
• B. $8\mu \text{C}$
• C. $12\mu \text{C}$
• D. $20\mu \text{C}$

Answer 1

The correct option is B $8\mu \text{C}$

At steady state, no current will flow through capacitors. Hence, current will not flow through the branches containing capacitors.

So, the current in the upper circuit at steady state.

$i_1=\frac{V_1}{R_{eq}}=\frac{10}{2+3}=2\text{A}$

$V_{BG}=V_B-V_G=i_{1}R$

$\Rightarrow V_{BG}=2\times 3=6\text{V}$

Current in the lower circuit at the steady state,

$i_2=\frac{V_2}{R_{eq}}=\frac{20}{4+6}$

$\Rightarrow i_2=2\text{A}$

Now potential difference across $\text{C}$ & $\text{F}$,

$V_{CF}=V_C-V_F=i_{2}R$

$\Rightarrow V_{CF}=2\times 4=8\text{V}$

Considering the loop $\text{BCFCB}$, the charge on both capacitors will be same, since they are in same closed loop.


Considering the potential difference across $\text{BG}$ & $\text{CF}$ and redrawing closed loop $\text{BGFCB}$ at steady state.

Using $\text{KVL}$ to the loop $\text{BGFCB}$,

$-6-\frac{q}{(6\times {10}^{-6})}+8-\frac{q}{(3\times {10}^{-6})}=0$

$\Rightarrow 2={10}^6[\frac{q}{6}+\frac{q}{3}]={10}^6\left[\frac{3q}{6}\right]$

$\therefore q=\frac{12}{3\times {10}^6}=4\times {10}^{-6}\text{C}$

$\therefore q=4\mu \text{C}$

The sum of charge on both the capacitors at steady state is,

$q+q=2q=8\mu \text{C}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option $\left(b\right)$ is correct answer.