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4.The rate at which potential difference between theplates of a parallel plate capacitor with a 2 uFcapacitance be changed to produce adisplacement current of 1 A is(1) 0.5 MV/s(2) 0.3 MV/s(3) 0.2 MV/s(4) 0.1 MV/s |
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Answer» The CORRECT answer is OPTION 1) 0.5 MV/s Given that displacement current = I(d) = 2 A I(d) = ε.dΦ/dt Φ - flux = E.A dΦ/dt = change in flux E is the electric field across the capacitor and A is the AREA of the plate of capacitor. I(d) = ε.A dE/dt E = V/d = POTENTIAL difference / distance between the plates of the capacitor => I(d) = ε.A d(V/d)/dt => I(d) = (ε.A)/d × d(V)/dt => I(d) = C × dV/dt [ C = capacitance of the capacitor = (ε.A)/d ] => dV/dt = I(d)/C => dV/dt = 1/2 = 0.5 MV/s The rate of change of potential difference of the capacitor = 0.5 MV/s |
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