Question

Three capacitors of capacitance $3\mu \text{F},10\mu \text{F}$ and $15\mu \text{F}$ are connected in series to a battery of $100\text{V}$. The voltage across $10\mu \text{F}$ is

• A. $10\text{Volt}$
• B. $20\text{Volt}$
• C. $30\text{Volt}$
• D. $40\text{Volt}$

Answer 1

The correct option is B $20\text{Volt}$

In series combination, the equivalent capacitance is given by

$\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$

$\Rightarrow \frac{1}{C_{eq}}=\frac{1}{3}+\frac{1}{10}+\frac{1}{15}=\frac{1}{2}$

$\Rightarrow C_{eq}=2\mu \text{F}$

Moreover, in a series combination, the charge on each capacitor is the same, which is the same as the charge on the equivalent capacitor.

Using $Q=CV,$ we get

$Q=2\mu \text{F}\times 100\text{V}=200\mu \text{C}$

Now we know that the charge on the $10\mu \text{F}$ capacitor is $200\mu \text{C}$.

Using $V=Q/C,$ we get

$V=\frac{200\mu \text{C}}{10\mu \text{F}}=20\text{Volt}$

Alternate:

In a series combination, the charge on each capacitor is the same, which is the same as the charge on the equivalent capacitor.

$Q=C_{eq}{V}_{eq}=C_2{V}_2$

$\Rightarrow V_2=\frac{C_{eq}{V}_{eq}}{C_2}=\frac{2\times 100}{10}=20\text{V}$

Hence, option $\left(b\right)$ is correct.