Q144 / 180 · COMEDK 2024 (Evening Shift) · Physics
The figure shows a network of five capacitors connected to a 20 V battery. Calculate the charge acquired by each 10 $\mu$F capacitor.
- A$4 \times 10^{-4} \mathrm{C}$
- B$6 \times 10^{-4} \mathrm{C}$
- C$2 \times 10^{-4} \mathrm{C}$
- Correct answerD$1 \times 10^{-4} \mathrm{C}$
Solution
Left parallel pair: $C_{LT} = 15 + 5 = 20\ \mu F$
Right parallel pair: $C_{TR} = 10 + 10 = 20\ \mu F$
$C_{LT}$ and $C_{TR}$ are in series across 20 V. Since both are equal, voltage divides equally:
$V_{TR} = \dfrac{20}{2} = 10$ V across each $10\ \mu F$ capacitor.
$Q = CV = 10 \times 10^{-6} \times 10 = 100\ \mu C = 1 \times 10^{-4}$ C
Right parallel pair: $C_{TR} = 10 + 10 = 20\ \mu F$
$C_{LT}$ and $C_{TR}$ are in series across 20 V. Since both are equal, voltage divides equally:
$V_{TR} = \dfrac{20}{2} = 10$ V across each $10\ \mu F$ capacitor.
$Q = CV = 10 \times 10^{-6} \times 10 = 100\ \mu C = 1 \times 10^{-4}$ C
Hence, the correct answer is option: 'D'.