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A capacitor of capacitance 12.0 muF is j...

A capacitor of capacitance `12.0 muF` is joined to an `AC` source of frequency `200 Hz`. The rms current in the circuit is `2.00 A`. (a) Find the rms voltage across the capacitor. (b) Find the average energy stored in the electric field between the plates of the capacitor.

Text Solution

Verified by Experts

The impedance of the capacitor `= 1/(omegaC)`
`= 1/((2pi xx 200 s^(-1)) (12 muF)) = 66.3 Omega`.
The rms voltage across the capacitor
`= i_(rms) Z = 2.0 A xx 66.3 Omega = 133 V`.
(b) The energy stored in the electric field `= 1/2 CV^2`.
Hence the average energy stored `= 1/2 C vec V^2`.
But `vec V^2 = (V_(rms)^2`.
Thus, the average energy stored
`= 1/2 xx (12 muF) xx (133 V)^2 = 0.106 J`.
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