Four capacitors are connected in the circuit as shown. The equivalent capacitance between P and Q will be:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 8\,\mu\mathrm{F} \)
Redraw the circuit by marking points at the same potential. The bottom wire joins P to the lower end of the middle capacitor, so that end is at the potential of P. Let X be the node where the top wire meets the middle capacitor.
The upper-left \( 6\,\mu\mathrm{F} \) lies between P and X, and the middle \( 6\,\mu\mathrm{F} \) also lies between X and P, so these two are in parallel.
\( \displaystyle C_{PX} = 6 + 6 = 12\,\mu\mathrm{F} \)
This \( 12\,\mu\mathrm{F} \) is in series with the upper-right \( 6\,\mu\mathrm{F} \) (between X and Q).
\( \displaystyle C_{\text{upper}} = \frac{12 \times 6}{12 + 6} = 4\,\mu\mathrm{F} \)
The \( 4\,\mu\mathrm{F} \) capacitor in the lower branch joins P and Q directly, so it is in parallel with this combination.
\( \displaystyle C_{PQ} = 4 + 4 = 8\,\mu\mathrm{F} \)