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A Wheatstone-type capacitor bridge connects node A to node B. From A…
Question
A Wheatstone-type capacitor bridge connects node A to node B. From A two branches start. Branch arm capacitors: C1 = 2 microfarad from A to node M, C2 = 4 microfarad from M to B; C3 = 4 microfarad from A to node N, C4 = 8 microfarad from N to B. A bridge capacitor C5 = 3 microfarad connects M and N. Find the equivalent capacitance between A and B, in microfarad.
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Step-by-step solution
The bridge is balanced when C1/C2 = C3/C4.
Here 2/4 = 4/8 = 1/2, so the potentials of M and N are equal and the bridge capacitor C5 carries no charge; it can be removed.
The two arms then reduce to series combinations: arm1 = C1C2/(C1+C2) = 8/6 = 4/3 uF; arm2 = C3C4/(C3+C4) = 32/12 = 8/3 uF.
These two arms are in parallel between A and B: Ceq = 4/3 + 8/3 = 4 uF.
Final answer4
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