Question: 34. In the following circuit, the equivalent capacitance between terminal A and terminal B is :
(1) 2 \mu \mathrm{~F}
(2) 1 \mu \mathrm{~F}
(3) 0.5 \mu \mathrm{~F}
(4) 4 \mu \mathrm{~F}
Answer: Option (1)
Explanation:
All capacitors in the circuit have equal capacitance of 2 \mu \mathrm{F}.
The circuit is symmetric about the vertical line passing through the central capacitor.
Due to symmetry, the potentials at the midpoints of the top and bottom branches are equal.
Hence, there is no potential difference across the central vertical capacitor.
Therefore, the central capacitor does not contribute to the equivalent capacitance and can be ignored.
The remaining circuit consists of two parallel branches between A and B.
Each branch has two capacitors of 2 \mu \mathrm{F} connected in series.
The equivalent capacitance of two equal capacitors in series is
C_s = \frac{C}{2} C_s = \frac{2}{2} = 1 \mu \mathrm{F}Now, the two branches are in parallel, so their capacitances add.
C_{\text{eq}} = 1 + 1 C_{\text{eq}} = 2 \mu \mathrm{F}Hence, the equivalent capacitance between terminals A and B is 2 \mu \mathrm{F}.