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In the following circuit, the equivalent capacitance between terminal A and terminal B is:
Bridge ratio is 2/2 = 2/2 = 1 -> Central 2 μF capacitor is inactive -> Two parallel branches of (2 in series) -> 1 + 1 = 2 μF!
The given circuit consists of five capacitors each having capacitance C = 2 μF arranged in a Wheatstone bridge configuration. The condition for balance in a capacitance bridge is:
Because the ratio of adjacent arms is equal, the electric potential at the upper junction and lower junction is identical. No charge flows into or through the central 2 μF capacitor, so it can be completely removed from the circuit without affecting the equivalent capacitance.
With the central capacitor removed, the circuit simplifies to two parallel branches, each containing two 2 μF capacitors in series:
The top and bottom branches are connected in parallel across terminals A and B. Their capacitances add directly:
Diagnose every choice: Know exactly why the right option works and why distractors fail.
Incorrect; would occur if all 4 outer capacitors were assumed to be in series.
Incorrect; would occur if branches were in parallel without series reduction.
Standard balanced bridge result: C_eq equals the value of a single arm capacitor (2 μF).
Represents only a single branch capacitance, forgetting the other parallel branch.
Students unfamiliar with the Wheatstone bridge spend 2+ minutes attempting Kirchhoff loop equations or star-delta transformations on the central capacitor.
"In a symmetrical balanced bridge with identical components C in all 4 arms, the equivalent capacitance is always identically equal to C."
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