A 84.0 nF capacitor is charged to 12.0 V , then disconnected from the power supp
ID: 1452358 • Letter: A
Question
A 84.0 nF capacitor is charged to 12.0 V , then disconnected from the power supply and connected in series with a coil that has L = 4.30×102 H and negligible resistance. A.) At an instant when the charge on the capacitor is 0.450 C , what is the magnitude of the current in the inductor? Express your answer with the appropriate units.
B.) At an instant when the charge on the capacitor is 0.650 C, what is the magnitude of the rate of change of this current?
If you would show your work so i can understand how to do similar problems it would be appreciated!
i got A to be 7.48 mA and that was incorrect if that helps
Explanation / Answer
given
C = 84 nF
Vmax = 12 volts
L = 4.3*10^-2 H
A) when capacitor is fully charged,
Qmax = C*Vmax
= 84*10^-9*12
= 1.008*10^-6 C
Apply conservation of energy
Qmax^2/(2*C) = Q^2/(2*C) + L*I^2/2
L*I^2 = Qmax^2/c - Q^2/c
I = sqrt( (Qmax^2 - Q^2)/(L*C))
= sqrt( (1.008^2 - 0.45^2)10^-12/(0.043*84*10^-9))
= 0.015 A or 15 mA
B) Apply KVL in the ckt
VC - VL = 0
VL = VC
L*dI/dt = Q/C
dI/dt = Q/(c*L)
= 0.65*10^-6/(84*10^-9*0.043)
= 180 A/s
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