Question

In the given circuit, the potential difference between $\text{A}$ and $\text{B}$ in the steady state is

• A. $25\text{V}$
• B. $50\text{V}$
• C. $75\text{V}$
• D. $35\text{V}$

Answer 1

The correct option is C $75\text{V}$

We can redraw the given circuit as follows,


At steady state, current through the capacitor is zero, hence we can redraw the given circuit as follows.

From the figure,

Let, $C_1=6\mu \text{F}{C}_2=2\mu \text{F}$

As $C_1,C_2$ are in series $q$ is same, $q=C_1{V}_1=C_2{V}_2$

$\frac{V_1}{V_2}=\frac{C_2}{C_1}=\frac{2}{6}=\frac{1}{3}...\left(1\right)$

$V_1+V_2=\mathrm{100...}\left(2\right)$

Solving $\left(1\right)\&\left(2\right)$
$\therefore V_2=\left(\frac{3}{1+3}\right)100=\frac{3}{4}\times 100$

$\begin{array}{c}{V}_2=75V\end{array}$