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 charge on $15\mu \text{F}$ is

• A. $50\mu \text{C}$
• B. $100\mu \text{C}$
• C. $200\mu \text{C}$
• D. $280\mu \text{C}$

Answer 1

The correct option is C $200\mu \text{C}$

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.

Now using $Q=CV$, we get

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

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

Key Concept : Charge on a capacitor in a series arrangement in a circuit.