A network of four capacitors each having capacitance 3 μF is connected across a 600 V supply as shown in the figure. The total charge stored in the network is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 2.4 mC
Reading the figure: C1, C2 and C3 form the single path A–B–C–D, so they are in series; C4 joins A directly to D, so it is in parallel with that series branch, and the whole network sits across the 600 V supply.
\( \displaystyle \frac{1}{C_{123}} = \frac{1}{3\,\mu\mathrm{F}} + \frac{1}{3\,\mu\mathrm{F}} + \frac{1}{3\,\mu\mathrm{F}} = \frac{1}{1\,\mu\mathrm{F}} \)
\( C_{123} = 1\,\mu\mathrm{F} \)
\( C_{\mathrm{eq}} = C_{123} + C_4 = 1\,\mu\mathrm{F} + 3\,\mu\mathrm{F} = 4\,\mu\mathrm{F} \)
\( Q = C_{\mathrm{eq}}V = (4\times10^{-6}\,\mathrm{F}) \times (600\,\mathrm{V}) \)
\( Q = 2.4\times10^{-3}\,\mathrm{C} = 2.4\,\mathrm{mC} \)