Q145 / 160 · TS EAMCET 2024 (09 May Shift 1) · Physics
Three capacitors of capacitances $10 \mu \mathrm{~F}, 5 \mu \mathrm{~F}$ and $20 \mu \mathrm{~F}$ are connected in series with a 14 V dc supply. The charge on $5 \mu \mathrm{~F}$ capacitor is
- A$20 \mu \mathrm{C}$
- Correct answerB$40 \mu \mathrm{C}$
- C$70 \mu \mathrm{C}$
- D$2.8 \mu \mathrm{C}$
Solution
Charge given by DC supply, $\mathrm{Q}=\mathrm{C}_{\text {eq }} \mathrm{V}$
$\frac{1}{C_{e q}}=\frac{1}{10}+\frac{1}{5}+\frac{1}{20}$
or, $\mathrm{C}_{\mathrm{eq}}=\frac{20}{7}$
$\therefore Q=\frac{20}{7} \times 14=40 \mu \mathrm{c}$
As capacitors are connected in series so charge on $5 \mu \mathrm{~F}$ capacitor $=40 \mu \mathrm{c}$.
Hence, the correct answer is option: 'B'.