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In an L-C circuit, L=3.3H and C=840pF. A...

In an `L-C` circuit, `L=3.3H` and `C=840pF`. At `t=0` charge on the capacitor is `105muC` and maximum. Compute the following quantities at `t=2.0ms`.
a. The energy stored in the capacitor.
b. The total energy in the circuit,
c. The energy stored in the inductor.

Text Solution

Verified by Experts

Given, L=3.3 H, C=`840 xx 10^(-12)`F
and `q_(0) = 105 xx 10^(-6)`C
The angular frequency of L-C oscillations is,
`omega=1/(sqrt(LC))= 1/sqrt(3.3 xx 840 xx 10^(-12))`
`=1.9 xx10^(4)rads^(-1)`
Charge stored in the capacitor at time t, would be,
`q=q_(0)cosomegat`
i) At `t=2 xx 10^(-3)`s
`q=(105xx10^(-6)cos(1.9xx10^(4))(2xx10^(-3))`
`=100.3 xx 10^(-6)`C
`therefore` Energy stored in the capacitor
`U_(C) = 1/2q^(2)/C = (100.3 xx 10^(-6))^(2)/(2xx840xx10^(-12))=6.0J`
ii) Total energy in the circuit
`U=1/2q_(0)^(2)/C = (105xx10^(-6))^(2)/(2xx840xx10^(-12))=6.56` J
iii) Energy stored in inductor in the given time ltbr. =total energy in circuit -energy stored in capacitor
`=(6.56-6.0)J=0.56 J`
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