Question

The effective capacitance of parallel combination of two capacitors$C_1$ and$C_2$ is $10\mu F$. When these capacitors are individually connected to a voltage source of $1V$, the energy stored in the capacitor $C_2$is $4$ times of that in $C_1$. If these capacitors are connected in series, their effective capacitance will be:

• A. $1.6\mu F$
• B. $3.2\mu F$
• C. $4.2\mu F$
• D. $8.4\mu F$

Answer 1

The correct option is A

$1.6\mu F$

Step 1 : Given data

Capacitors in parallel combination

$C_1+C_2=10\mu F$

Voltage

$V=1V$

Energy

$E_2=4E_1$

Step 2: To find effective capacitance in series connection

Energy stored in the two capacitors are

$$\begin{gathered} E_1=\frac{1}{2}{C}_1{V}^2 \\ {E}_2=\frac{1}{2}{C}_2{V}^2 \end{gathered}$$

we know $E_2=4E_1$

Therefore

$$\begin{gathered} \frac{1}{2}{C}_2{V}^2=4\times C_1{V}^2 \\ {C}_2=4C_1 \end{gathered}$$

Hence we can write

$$\begin{gathered} C_1+C_2=C_1+4C_1 \\ =10\mu F \\ 5C_1=10\mu F \\ {C}_1=2\mu F \\ {C}_2=10\mu F-C_1 \\ =10\mu F-2\mu F \\ {C}_2=8\mu F \end{gathered}$$

Here the capacitors are connected in series, then

$$\begin{gathered} C_{Series}=\frac{C_1{C}_2}{C_1+C_2} \\ =\frac{2\times 8}{2+8}\mu F \\ =1.6\mu F \end{gathered}$$

Hence the effective resistance in series connection is $1.6\mu F$

Therefore the correct answer is Option A.