Ten capacitors each of 2 μF capacity were arranged in two rows such that first row contains 6 capacitors in series where as the rest were in second row. The equivalent capacitance of the network is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{5}{6}\,\mu\mathrm{F} \)
For identical capacitors in series, \( n \) capacitors of capacitance \( C \) give \( C/n \); for capacitors in parallel, the capacitances add. The two rows are each a series chain, and the rows are connected in parallel.
\( \displaystyle C_{1} = \frac{2\,\mu\mathrm{F}}{6} = \frac{1}{3}\,\mu\mathrm{F} \)
The second row has the remaining \( 10 - 6 = 4 \) capacitors in series:
\( \displaystyle C_{2} = \frac{2\,\mu\mathrm{F}}{4} = \frac{1}{2}\,\mu\mathrm{F} \)
\( \displaystyle C_{\mathrm{eq}} = C_{1} + C_{2} = \frac{1}{3}\,\mu\mathrm{F} + \frac{1}{2}\,\mu\mathrm{F} = \frac{5}{6}\,\mu\mathrm{F} \)