JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Capacitance

The equivalent capacitance between points A and B in below shown figure will be ______

μ F

Answer: 6

Solution

In the given circuit, two capacitors are short-circuited (their terminals are directly connected by wires with zero resistance), so they can be removed from the circuit.

After removing the short-circuited capacitors, we are left with two capacitors: one of 24 μF and one of 8 μF in series between points A and B.

For capacitors in series:

C_{e q} = \frac{C _{1} \times C _{2}}{C _{1} + C _{2}} = \frac{24 \times 8}{24 + 8} = \frac{192}{32} = 6 μ F

JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Integer answer

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