Question:
The equivalent capacitance between \(A\) and \(B\) in the circuit given below, is :
The equivalent capacitance between \(A\) and \(B\) in the circuit given below, is :
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Identify nodes: the lower wire joins the right end of the top 2 \(\mu\)F capacitor, so the two 5 \(\mu\)F capacitors and the 2 \(\mu\)F capacitor are in parallel; likewise the 4 \(\mu\)F and bottom 2 \(\mu\)F. Then combine three groups in series.
Updated On: Nov 11, 2026
- \(2.4\,\mu\mathrm{F}\)
- \(4.9\,\mu\mathrm{F}\)
- \(3.6\,\mu\mathrm{F}\)
- \(5.4\,\mu\mathrm{F}\)
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The Correct Option is A
Solution and Explanation
Reading the circuit: from A a 6 \(\mu\)F capacitor leads to node N1 (top wire). From N1, the top 2 \(\mu\)F capacitor goes to node N2, and each of the two 5 \(\mu\)F capacitors goes from the top wire (N1) to the lower wire. The lower wire is joined by a plain wire to N2 (the right end of the top 2 \(\mu\)F capacitor). So between N1 and N2 there are three capacitors in parallel: \(2+5+5=12\ \mu\mathrm{F}\).
From N2 to B: the 4 \(\mu\)F capacitor and the bottom 2 \(\mu\)F capacitor are both connected between N2 and B, so they are in parallel: \(4+2=6\ \mu\mathrm{F}\).
The three groups 6, 12 and 6 \(\mu\)F are in series:
\(\frac{1}{C}=\frac16+\frac1{12}+\frac16=\frac{2+1+2}{12}=\frac{5}{12}\), so \(C=\frac{12}{5}=2.4\ \mu\mathrm{F}\).
Answer: (1) \(2.4\,\mu\mathrm{F}\)
From N2 to B: the 4 \(\mu\)F capacitor and the bottom 2 \(\mu\)F capacitor are both connected between N2 and B, so they are in parallel: \(4+2=6\ \mu\mathrm{F}\).
The three groups 6, 12 and 6 \(\mu\)F are in series:
\(\frac{1}{C}=\frac16+\frac1{12}+\frac16=\frac{2+1+2}{12}=\frac{5}{12}\), so \(C=\frac{12}{5}=2.4\ \mu\mathrm{F}\).
Answer: (1) \(2.4\,\mu\mathrm{F}\)
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