Question

In the adjoining figure, a section of a complicated circuit is shown, in which $E=10\text{V},$ $C_1=2\mu \text{F}$, $C_2=3\mu \text{F}$ and $(V_A-V_B)=10\text{V}.$ The potential on $C_1$ will be

• A. $0\text{V}$
• B. $4\text{V}$
• C. $12\text{V}$
• D. $16\text{V}$

Answer 1

The correct option is C $12\text{V}$

Given, $V_A-V_B=10\text{V};E=10\text{V}{C}_1=2\mu \text{F};C_2=3\mu \text{F}$


Using, KVL to the given open loop, we get,

$V_A-V_1+E-V_2=V_B$

$V_A-V_B=V_1+V_2-E$

$V_A-V_B=\frac{q}{C_1}+\frac{q}{C_2}-E$

$10=\frac{q}{2\times {10}^{-6}}+\frac{q}{3\times {10}^{-6}}-10$

$20=\frac{q}{{10}^{-6}}[\frac{1}{2}+\frac{1}{3}]=\frac{5q}{6\times {10}^{-6}}$

$\therefore q=24\times {10}^{-6}\text{C}$

$V_1=\frac{q}{C_1}=\frac{24\times {10}^{-6}}{2\times {10}^{-6}}=12\text{V}$

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