Question

If at $t=0$, switch $\text{S}$ is closed, then at steady state find potential drop across each capacitor as shown in figure.

• A. $\frac{4}{3}\text{V},\frac{20}{3}\text{V},\frac{5}{3}\text{V}$
  
• B. $\frac{20}{3}\text{V},\frac{4}{3}\text{V},\frac{4}{3}\text{V}$
  
• C. $\frac{10}{3}\text{V},\frac{20}{3}\text{V},10\text{V}$
  
• D. $\frac{10}{3}\text{V},\frac{4}{3}\text{V},\frac{5}{3}\text{V}$
  

Answer 1

The correct option is B $\frac{20}{3}\text{V},\frac{4}{3}\text{V},\frac{4}{3}\text{V}$


Let us solve this problem by assigining potential to known points let us take potential at junction $\text{P}$ be $V_0$ as shown in figure.


Using Kirchhoff`s junction law, we can find $V_0$:

$(V_0-8)\times 1+(V_0-0)\times 2+(V_0-0)\times 3=0$

$\Rightarrow V_0-8+2V_0+3V_0=0$

$\Rightarrow 6V_0=8$

$\therefore V_0=\frac{4}{3}\text{V}$

Using the value of $V_0$ we can get, potential drop across each capacitor,

Potential drop across $C_1$ is, $V_1=8-V_0=8-\frac{4}{3}=\frac{20}{3}\text{V}$

Similarly, let potential drop across $C_2$ and $C_3$ be $V_2$ and $V_3$ respectively.

$V_2=(V_0-0)=\frac{4}{3}\text{V}$

$V_3=(V_0-0)=\frac{4}{3}\text{V}$

Hence, option (b) is the correct answer.

Why this question: Assigning potential to known points, we can get potential of an unknown point.