Question

A 500 uF capacitor is connected across a 1.25 H inductor At a certain time, the charge on the capacitor is zero and the current is 0.342 A. a) How much later will the capacitor charge reach its peak? b) What's the total energy in the circuit? c) What is the peak charge on the capacitor?

Answer 1

capacitance ( c) = 500 uF = 500*10^-6 F

inductance ( L) = 1.25 H,                   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 * X_L

     = 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