Question

A circuit has a section $\text{AB}$ as shown in figure. If $V_A-V_B=20\text{V}$, then the potential difference across capacitors $C_1$ and $C_2$ are respectively

• A. $5\text{V}$ and $10\text{V}$
• B. $10\text{V}$ and $5\text{V}$
• C. $20\text{V}$ and $10\text{V}$
• D. $10\text{V}$ and $20\text{V}$

Answer 1

The correct option is C $20\text{V}$ and $10\text{V}$

Given, $V_A-V_B=20\text{V}$ and $E=10\text{V}$ respectively.

Let, the charge on each capacitor be $q$.

Using KVL,

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

$\Rightarrow (V_A-V_B)+E=q(\frac{1}{C_1}+\frac{1}{C_2})$

Substituting the values we get,

$20+10=q(\frac{1}{2}+\frac{1}{4})$

$\Rightarrow q=30\left(\frac{4}{3}\right)=40\mu \text{C}$

Therefore, potential difference across $C_1$:

$V_1=\frac{q}{C_1}=\frac{40}{2}=20\text{V}$

potential difference across $C_2$:

$V_2=\frac{q}{C_2}=\frac{40}{4}=10\text{V}$

Hence, option (c) is the correct answer.

Why this question? Application of KVL in a branch of a circuit.