Three capacitors, \( C_1, C_2, C_3 \) are connected in series. This combination is connected in parallel with capacitor \( C_4 \). \( C_1 = C_2 = C_3 = 30\,\mu\mathrm{F} \) and \( C_4 = 10\,\mu\mathrm{F} \). 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) → \( 8 \times 10^{-3}\,\mathrm{C} \)
Find the equivalent capacitance of the network; the total charge drawn from the supply is then \( Q = C_{\mathrm{eq}}V \).
Three equal capacitors in series:
\( \displaystyle \frac{1}{C_s} = \frac{1}{30} + \frac{1}{30} + \frac{1}{30} = \frac{3}{30} \;\Rightarrow\; C_s = 10\,\mu\mathrm{F} \)
This branch in parallel with \( C_4 \):
\( C_{\mathrm{eq}} = C_s + C_4 = 10 + 10 = 20\,\mu\mathrm{F} \)
\( Q = C_{\mathrm{eq}}V = 20\times10^{-6} \times 400 = 8\times10^{-3}\,\mathrm{C} \)
Each branch carries \( 4\times10^{-3}\,\mathrm{C} \), and the total is the sum of the two.