Question

Two capacitors when connected in series have a capacitance of $3\mathrm{\mu F}$, and when connected in parallel have a capacitance of $16\mathrm{\mu F}$. Their individual capacities are :

• A. 1μF, 2 μF
• B. 6 μF, 2 μF
• C. 12 μF, 4 μF
• D. 3 μF, 6 μF

Answer 1

The correct option is C

12 μF, 4 μF

Step 1: Given data

Resultant series capacitance ${\mathrm{C}}_{\mathrm{ser}}=3\text{μF}$

Resultant parallel capacitance ${\mathrm{C}}_{\mathrm{par}}=16\text{μF}$

Step 2: Calculate the individual capacity

$C_{1,}{C}_2$ are individual capacitance.

When the capacitors are connected in series, $\frac{1}{{\mathrm{C}}_1}+\frac{1}{{\mathrm{C}}_2}=\frac{1}{3\text{μF}}$ …($1$)

When the capacitors are connected in parallel, ${\mathrm{C}}_1+{\mathrm{C}}_2=16\mathrm{\mu F}$ …($2$)

After solving equations ($1$) and ($2$) get the values of individual capacitance are,

$$\begin{gathered} \frac{1}{{\mathrm{C}}_1}+\frac{1}{{\mathrm{C}}_2}=\frac{1}{3\text{μF}} \\ \frac{{\mathrm{C}}_1+{\mathrm{C}}_2}{{\mathrm{C}}_1{\mathrm{C}}_2}=\frac{1}{3\text{μF}} \\ \frac{16}{{\mathrm{C}}_1{\mathrm{C}}_2}=\frac{1}{3\text{μF}}\left[\because {\mathrm{C}}_1+{\mathrm{C}}_2=16\mathrm{\mu F}\right] \\ {\mathrm{C}}_1{\mathrm{C}}_2=3\left(16\right) \\ {\mathrm{C}}_1{\mathrm{C}}_2=48...\left(3\right) \end{gathered}$$

From equations (2) and (3), we get

$$\begin{gathered} {\mathrm{C}}_1+{\mathrm{C}}_2=16\mathrm{\mu F} \\ {\mathrm{C}}_1+\frac{48}{{\mathrm{C}}_1}=16\left[\because {\mathrm{C}}_2=\frac{48}{{\mathrm{C}}_1}\right] \\ {{\mathrm{C}}_1}^2-16{\mathrm{C}}_1+48=0 \end{gathered}$$

or,

$$\begin{gathered} {{\mathrm{C}}_1}^2-12{\mathrm{C}}_1-4{\mathrm{C}}_2+48=0 \\ {\mathrm{C}}_1({\mathrm{C}}_1-12)-4({\mathrm{C}}_1-12)=0 \\ ({\mathrm{C}}_1-12)({\mathrm{C}}_1-4)=0 \\ {\mathrm{C}}_1=12,4\mathrm{\mu F} \end{gathered}$$

Now, from equation (3),

For ${\mathrm{C}}_1=12,{\mathrm{C}}_2=\frac{48}{12}=4\mathrm{\mu F}$

For ${\mathrm{C}}_1=4,{\mathrm{C}}_2=\frac{48}{4}=12\mathrm{\mu F}$

Hence, option (C) is correct.