Question

Three capacitors of capacitances $4\mu \text{F},$ $12\mu \text{F},$ and $8\mu \text{F}$ are connected in series to a battery of $110\text{V}.$ The charge on $12\mu F$ capacitor is

• A. $520\mu \text{C}$
  
• B. $120\mu \text{C}$
• C. $240\mu \text{C}$
• D. $280\mu \text{C}$

Answer 1

The correct option is C $240\mu \text{C}$

In series combination, the equivalent capacitance is given by,

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

Substituting the values, we get

$\Rightarrow \frac{1}{C_{eq}}=\frac{1}{4}+\frac{1}{12}+\frac{1}{8}$

$\Rightarrow \frac{1}{C_{eq}}=\frac{30+10+15}{120}$

$\Rightarrow C_{eq}=\frac{24}{11}\mu \text{F}$

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,

$Q=\frac{24}{11}\mu \text{F}\times 110\text{V}=240\mu \text{C}$

Hence, option (c) is the correct answer.

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