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An emf E=100 sin 314 t V is applied acro...

An emf `E=100 sin 314 t V` is applied across a pure capacitor of `637(mu)F`. Find
(a) the instantaneous current I
(b) the instantaneous power p
(c ) the frequency of power
(d) the maximum energy stored in the capacitor.

Text Solution

Verified by Experts

Given `E=100 sin (314 t)`
we have `omega=314 implies 2 (pi)f=100 pi implies f=50Hz`
Capacitive reactance:
`(X_c)=(1)/(omega C)=(1)/(314xx637xx10^(-6))=5 Omega`
since there is pure capacitor in the circuit, so current leads emf by `(phi)=(pi)//2`
`i-i_(0) sin (omega t + phi0=i_(0) sin [ 314t+(pi)//(2)]`
`=(E_0)/(X_C) cos (314 t)=100/5 cos (314 t)=20 cos (314 t)`
(b) Instantaneous power is given by
`p=EI=100 sin (314 t) xx20 cos (314 t)=1000 sin (628 t)`
(c ) from the expression of power above, we can calculate the frequency of power becomes maximum and minimum two times.
(d) We know that energy in the capacitor is `U=1/2 CE^(2)`. Energy stored in the cpacitor will be maximum when `E=(E_0)`.
`U_(0)=1/2 CE_(0)^(2)=1/2(637xx10^(-6))(100)^(2)=3.185 J`.
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