(1) An unusual type of capacitor is made using four identical plates P1, P2, P3 and P4, each of area A arranged in air with the same separation d as shown in the figure. A thin wire outside the system of plates connects the plates P2 and P4. Wires soldered to the plates P1 and P3 serve as terminals T1 and T2 of the system. What is the effective capacitance between the terminals T1 and T2?
(a) (2ε0A)/3d
(b) (3ε0A)/2d
(c) (3ε0A)/d
The arrangement contains 3 identlcal capacitors C1, C2 and C3 each of capacitance (ε0A)/d arranged as shown in the adjoining figure. Plate P2 is common for C1 and C2. Likewise, plates P3 is common for C2 and C3. The capacitors C2 and C3 are connected in parallel (to produce an effective value 2ε0A/d) and this parallel combination is connected in series with the capacitor C1 of value ε0A/d . The effective capacitance C between the terminals T1 and T2 is therfore given by
C = [(2ε0A/d) ×(ε0A/d)] / [(2ε0A/d) +(ε0A/d)].
Therefore, C = (2ε0A)/3d
(a) 3 μF
(b) 2 μF
(c) 1.5 μF
(d) 1 μF
(e) 0.5 μF
This is a very simple question once you identify the circuit to be a balanced Wheatstone’s bridge. The 3 μF capacitor is connected between equipotential points and you can ignore it. Without the diagonal branch, there are four 2 μF capacitors only. The capacitors in the upper pair are in series and produce a value of 1 μF. The capacitors in the lower pair also produce a value of 1 μF. Since they are in parallel the effective capacitance between the points A and B is 2 μF.
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