Question

In the given circuit, find ratio of charges on $4\mu \text{F}$ and $2\mu \text{F}$ capacitors in steady state.

• A. $\frac{4}{3}$
  
• B. $\frac{6}{5}$
  
• C. $\frac{4}{9}$
  
• D. $\frac{7}{9}$
  

Answer 1

The correct option is C $\frac{4}{9}$


In the given circuit let current $I$ in the single loop be as shown in figure, as we know that in steady state , no current flows through the branch containing capacitor, so we do not consider any current in the branches with capacitor as shown.



Current in the single loop $AHCDEFGA$ is,

$I=\frac{20-10}{(1+1+1+2)}=2\text{A}$

Using $\text{KVL}$ to the open loop $\left(AHCB\right)$ gives,

$V_A-(2\times 1)-(2\times 1)=V_B$

$\Rightarrow V_A-V_B=4\text{V}$

Using $\text{KVL}$ to the open loop $\left(HCDE\right)$ gives,

$V_H-(2\times 1)+20=V_E$

$\Rightarrow V_H-V_E=18\text{V}$

Charges on the given $4\mu \text{F}$ and $2\mu \text{F}$capacitors as follows,

$q_{4\mu \text{F}}=C{V}_{AB}$

$\Rightarrow q_{4\mu \text{F}}=4\times 4=16\mu \text{C}$

$q_{2\mu \text{F}}=C{V}_{EH}$

$\Rightarrow q_{4\mu F}=2\times 18=36\mu \text{C}$

$\Rightarrow \frac{q_{4\mu \text{F}}}{q_{2\mu \text{F}}}=\frac{16}{36}=\frac{4}{9}$

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