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A bridge of capacitors has four nodes A, B, C, D with capacitances:…

Question

A bridge of capacitors has four nodes A, B, C, D with capacitances: A-C = 2 uF, A-D = 4 uF, C-B = 6 uF, D-B = 3 uF, and a bridging capacitor C-D = 5 uF. Find the equivalent capacitance between A and B. If it equals p/q uF in lowest terms, report p + q.

✓ Verified answer: 581checked by our engine — not a guess

Step-by-step solution

Apply potential V_A at A, set V_B = 0.

The inner nodes C and D float, so net charge on each inner node is zero (charge conservation).

For a node, the sum over connected capacitors of C_branch*(V_node - V_other) = 0:
Node C: 2(V_C-V_A) + 6 V_C + 5(V_C-V_D) = 0.
Node D: 4(V_D-V_A) + 3 V_D + 5(V_D-V_C) = 0.

Solve for V_C, V_D in terms of V_A.

The charge drawn from A is Q = 2(V_A-V_C) + 4(V_A-V_D).
Then C_eq = Q/V_A = 450/131 uF in lowest terms.
Therefore p+q = 450+131 = 581.

Final answer581

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