Capacitors

JEE Main level, single correct, medium

23 of 180

Question 23 of 180

From the circuit given below, the capacitance between terminals AA and BB shown in the circuit is ______ μ\muF.
(take C1=C2=C3=1C_1=C_2=C_3=1 μ\muF and C4=2C_4=2 μ\muF.)

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Answer: (A) 22

Let the nodes in the circuit be labeled to identify the connections:
- Let the input terminal be Node A.
- Let the junction between C1C_1 and C2C_2 be Node X.
- The top wire connects Node A to the junction between C2C_2 and C3C_3. Thus, this junction is also at the potential of Node A.
- The bottom wire connects Node X to the output terminal Node B, with capacitor C4C_4 in this path.

Based on these nodes, the connections of each capacitor are as follows:
- C1C_1 is connected between Node A and Node X.
- C2C_2 is connected between Node X and Node A.
- C3C_3 is connected between Node A and Node B.
- C4C_4 is connected between Node X and Node B.

We can now simplify the circuit step-by-step:
Capacitors C1C_1 and C2C_2 are in parallel between nodes A and X. Their equivalent capacitance is:
C12=C1+C2=1 μF+1 μF=2 μFC_{12} = C_1 + C_2 = 1\,\mu\text{F} + 1\,\mu\text{F} = 2\,\mu\text{F}

This equivalent capacitor C12C_{12} is in series with C4C_4 (which is connected between X and B). The equivalent capacitance of this branch is:
CAXB=C12×C4C12+C4=2×22+2=1 μFC_{AXB} = \dfrac{C_{12} \times C_4}{C_{12} + C_4} = \dfrac{2 \times 2}{2 + 2} = 1\,\mu\text{F}

Finally, this branch is in parallel with C3C_3 (which is connected directly between A and B). The total equivalent capacitance between A and B is:
CAB=CAXB+C3=1 μF+1 μF=2 μFC_{AB} = C_{AXB} + C_3 = 1\,\mu\text{F} + 1\,\mu\text{F} = 2\,\mu\text{F}

Answer: 22

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