MCQ
The ratio of potential difference that must be applied across parallel and series combination of two capacitors $C_1$ and $C_2$ with their capacitance in the ratio $1: 2$ so that energy stored in these two cases becomes same is
- A
- ✓
- C
- D
✓
Answer
Correct option: B.
(b) : Given, $\frac{C_1}{C_2}=\frac{1}{2} \Rightarrow C_2=2 C_1$
Let energy stored in series and parallel combination be $U_s$ and $U_p$ respectively.
$
\begin{aligned}
& C_p=C_1+C_2=3 C_1 \\
& U_p=\frac{1}{2} C_p V_1^2=\frac{1}{2} \times 3 C_1 V_1^2. . . . . (i) \\
& \frac{1}{C_x}=\frac{1}{C_1}+\frac{1}{C_2}=\frac{1}{C_1}+\frac{1}{2 C_1} ; C_s=\frac{2}{3} C_1
\end{aligned}
$
$\begin{aligned} & U_P=\frac{1}{2} \times C_s V_2^2=\frac{1}{2} \times \frac{2}{3} C_1 \times V_2^2 \\ & \text { As, } U_s=U_p \\ & \frac{1}{2} \times 3 C_1 V_1^2=\frac{1}{3} \times C_1 \times V_2^2 ;\left(\frac{V_1}{V_2}\right)^2=\frac{2}{9} \Rightarrow \frac{V_1}{V_2}=\sqrt{2}: 3\end{aligned}$
Let energy stored in series and parallel combination be $U_s$ and $U_p$ respectively.
$
\begin{aligned}
& C_p=C_1+C_2=3 C_1 \\
& U_p=\frac{1}{2} C_p V_1^2=\frac{1}{2} \times 3 C_1 V_1^2. . . . . (i) \\
& \frac{1}{C_x}=\frac{1}{C_1}+\frac{1}{C_2}=\frac{1}{C_1}+\frac{1}{2 C_1} ; C_s=\frac{2}{3} C_1
\end{aligned}
$
$\begin{aligned} & U_P=\frac{1}{2} \times C_s V_2^2=\frac{1}{2} \times \frac{2}{3} C_1 \times V_2^2 \\ & \text { As, } U_s=U_p \\ & \frac{1}{2} \times 3 C_1 V_1^2=\frac{1}{3} \times C_1 \times V_2^2 ;\left(\frac{V_1}{V_2}\right)^2=\frac{2}{9} \Rightarrow \frac{V_1}{V_2}=\sqrt{2}: 3\end{aligned}$
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