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The circuit shows in Fig. is in the stea...

The circuit shows in Fig. is in the steady state with switch `S_(1)` closed. At `t = 0,S_(1)` is opened and switch `S_(2)` is closed.
a. Derive expression for the charge on capacitor `C_(2)` as a function of time.
b. Determine the first instant `t`, when the energy in the inductor becomes one-third of that in the capacitor.

Text Solution

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In the steady state, `C_(1)` and `C_(2)` are in series arrangement, their equivalent is
`C_(eq) = (C_(1)C_(2))/(C_(1) + C_(2)) = 1.2 mu F`
Charge on the capacitor `C_(2),Q_(0) = C_(eq) V = 1.2 xx 20 = 24 mu C`.
When `S_(1)` is opened, `S_(2)` is closed. Capacitor `C_(2)` starts discharging through the inductor and let at any time `t`, charge on the capacitor be `Q`. Then we know that
`Q = Q_(0) cos omegat`
`U_(E) + U_(B) = (Q_(0)^(2))/(2C_(2))`
where `omega = (1)/(sqrtLC_(2)) = (1) /(sqrt(0.2 xx 10^(-3) xx 2 xx 10^(-6))) = 50,000 rad s^(-1)`
At the times `t = t_(1),U_(B) = (1)/(3)U_(E)`, but `U_(B) + U_(E) = (Q_(0)^(2))/(2C_(2))`
Hence, `U_(E) = (3)/(4) ((1)/(2)(Q_(0)^(2))/(C_(2))) rArr (1)/(2) (Q^(2))/(C_(2)) = (3)/(4) ((1)/(2)(Q_(0)^(2))/(C_(2)))`
`Q = sqrt(3)/(2) Q_(0)` or `Q_(0) cos omegat_(1) = sqrt(3)/(2) Q_(0)`
`omegat_(1)` or `t_(1) = (pi)/(6 omega) = 1.05 xx 10^(-5) = 10.5 mu s`
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