Seven capacitors each of capacitance $2 \mu F$ are to be connected to obtain equivalent capacitance of…
- 3 in parallel and 4 in series
- 2 in parallel and 5 in series
- 5 in parallel and 2 in series
- 4 in parallel and 3 in series
Solution
For B: \(\mathrm{C}_{\mathrm{p}}=2+2+2=6 \mu \mathrm{F}\) and \(\mathrm{C}_{\mathrm{s}}=(1 / 2+1 / 2+1 / 2+1 / 2)^{-1}=1 / 2 \mu \mathrm{F}\) so \(\mathrm{C}_{\mathrm{eq}}=\frac{\mathrm{C}_{\mathrm{p}} \mathrm{C}_{\mathrm{s}}}{\mathrm{C}_{\mathrm{p}}+\mathrm{C}_{\mathrm{s}}}=\frac{6 \times 1 / 2}{6+1 / 2}=3 \times \frac{2}{13}=\frac{6}{13} \mu \mathrm{F}\).
For C: \(\mathrm{C}_{\mathrm{p}}=2+2+2+2=8 \mu \mathrm{F}\) and \(\mathrm{C}_{\mathrm{s}}=(1 / 2+1 / 2+1 / 2)^{-1}=2 / 3 \mu \mathrm{F}\) so \(\mathrm{C}_{\mathrm{eq}}=\frac{\mathrm{C}_{\mathrm{p}} \mathrm{C}_{\mathrm{s}}}{\mathrm{C}_{\mathrm{p}}+\mathrm{C}_{\mathrm{s}}}=\frac{8 \times 2 / 3}{8+2 / 3}=\frac{16}{3} \times \frac{3}{26}=\frac{8}{13} \mu \mathrm{F}\).
For D: \(C_p=2+2+2+2+2=10 \mu \mathrm{F}\) and \(C_s=(1 / 2+1 / 2)^{-1}=1 \mu \mathrm{F}\) so \(\mathrm{C}_{\mathrm{eq}}=\frac{\mathrm{C}_{\mathrm{p}} \mathrm{C}_{\mathrm{s}}}{\mathrm{C}_{\mathrm{p}}+\mathrm{C}_{\mathrm{s}}}=\frac{10 \times 1}{10+1}=\frac{10}{11} \mu \mathrm{F}\).
Asked in: MHT CET 2020 (13 Oct Shift 2)