A `1.00 mH` inductor and a `1.00muF` capacitor are connected in series. The current in the circuit is described by `i = 20 t`, where t, is in second and `i` is in ampere. The capacitor initially has no charge. Determine
(a) the voltage across the inductor as a function of time,
(b) the voltage across the capacitor as a function of time,
(c) the time when the energy stored in the capacitor first exceeds that in the inductor.

Answer

`2 xx 10^(-2) V`; b. `10^(7)t^(2) V`; c. `2sqrt(10) xx 10^(-5) s`

Explanation

`I = 20t`

`V_("inductor") = L(dI)/(dt) = 1 xx 10^(-3) xx 20 = 2 xx 10^(-2) V`
b. `Q = int_(0)^(t) Idt rArr Q = 10t^(2)`
`V_("capacitor") = (Q)/(C ) = (10t^(2))/(10^(-6)) = 10^(7) t^(2) V` ltbr c. Energy in capacitor `= (1)/(2) xx C xx V^(2)`
`U_(C) = (1)/(2) xx 10^(-6) xx 10^(14) t^(4) = (1)/(2) xx 10^(8) t^(4)`
`U_("inductor") = (1)/(2) xx L xx I^(2) = (1)/(2) xx 1 xx 10^(-3) xx 400 t^(2)`
`= (1)/(2) xx (4 xx t^(2))/(10)`
`U_("cap") = U_("ind")`
`rArr = (1)/(2) xx 10^(8) xx t^(4) = (1)/(2) xx (4)/(10) xx t^(2)`
`rArr t = 2 sqrt(10) xx 10^(-5) s`