A 500 uF capacitor is connecte
A 500 uF capacitor is connected across a 1.25 H inductor At acertain time, the charge on the capacitor is zero and the currentis 0.342 A. a) How much later will the capacitor charge reach itspeak? b) What’s the total energy in the circuit? c) What is thepeak charge on the capacitor?
Answer:
capacitance ( c) = 500 uF = 500*10-6 F
inductance ( L) = 1.25H, I ( current ) = 0.342A
a) frequency of oscillations in the circuit:
f = ( 1/ 2*pie* ( LC) 1/2 )
time period of oscillations, T = 1/f
T = 2*pie* ( LC) 1/2
the capacitor will get maximum charge at t = T / 4
t = 2*pie * ( LC) 1/2 / 4
t = 2*3.14* ( 1.25*500*10-6)1/2 / 4
t = 0.0392 sec
b) total energy of the circuit = ( 1/2) * L * I2
= ( 1/2) * 1.25* ( 0.342)2
= 0.0731 J
c) peak charge , Q = CV
V = I * XL
= I *wL ( w = 2*pie*f = 1/ ( LC)1/2 = ( 1/ (1.25*500*10-6 )1/2 ) = 40 rad/s)
= 0.342*40*1.25
V = 17.1 volt
Q = 500*10-6 * 17.1
Q = 8.55*10-3 C
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