A capacitor with a capacitance of 4.10 nF is in an LC circuit with a 9.50 mH ind

Question

A capacitor with a capacitance of 4.10 nF is in an LC circuit with a 9.50 mH inductor. At time t = 0 the voltage across the capacitor is a maximum.
At what time will the voltage across the capacitor next be a maximum (ignoring the sign)?
What is the next time after t = 0 that the charge on one of the plates of the capacitor will have exactly the same charge it has at time t=0 Do NOT ignore sign?
What is the first time after t = 0 that current in the inductor will have its maximum value (ignore direction of the current)?

Explanation / Answer

it will be an LC resonant circuit with w = 1/LC = 160 krad/s

=> T = 2/w = 39.2s
at t=0 cap voltage is max => Vc = Vmax cos (wt)

next immediate '-'vemax occurs at wt= ; i.e at half time period = T/2 = 19.6 s

next immediate '+'vemax occurs at wt= 2; i.e at full time period = T = 39.2s

current in inductor = current in capacitor = c dVc /dt = -cwVmax sin(wt)

hence at wt = /2 i.e at 1/4 rth time period current max occurs =T/4 = 9.8s

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