Three capacitors, $C_1, C_2, C_3$ are connected in series. This combination is connected in parallel with capacitor $C_4$. $C_1= C_2 =C_3= 30 μF$ and $C_4 = 10 μF$. Now a potential 400 V is applied across the combination. The total charge across the combination of capacitors is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $8 × 10^{-3} C$
Given:
$C_1 = C_2 = C_3 = 30\ \mu F$, $C_4 = 10\ \mu F$, $V = 400\ V$
For series combination:
$\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$
$\frac{1}{C_s} = \frac{1}{30} + \frac{1}{30} + \frac{1}{30} = \frac{1}{10}$
$C_s = 10\ \mu F$
Now, $C_s$ is in parallel with $C_4$:
$C_{eq} = C_s + C_4 = 10 + 10 = 20\ \mu F$
Total charge:
$Q = C_{eq} \times V = 20 \times 10^{-6} \times 400 = 8.0 \times 10^{-3}\ C$
Therefore, the total charge across the combination is $8\ \text{mC}$.