A network of four capacitors of 5 µF is connected to a 500 V supply as shown in figure. Find the equivalent capacitance of the network.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 6.67 µF
Between the two points where the 500 V is applied there are two branches in parallel. The upper branch is the left, top and right capacitors joined one after the other, which is a series combination of three; the lower branch is the single bottom capacitor.
Upper branch (three capacitors of \( 5\,\mu\mathrm{F} \) in series)
\( \displaystyle \frac{1}{C_{1}} = \frac{1}{5} + \frac{1}{5} + \frac{1}{5} = \frac{3}{5} \Rightarrow C_{1} = \frac{5}{3}\,\mu\mathrm{F} \)
Whole network (upper branch in parallel with the bottom capacitor)
\( \displaystyle C_{\mathrm{eq}} = \frac{5}{3} + 5 = \frac{20}{3} \approx 6.67\,\mu\mathrm{F} \)
Note: NTA picked this question from an NCERT (2026-27) example and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Capacitors in Parallel – …
