In an oscillating LC circuit, L = 27.0 mH and C = 7.00 In an oscillating LC circ

Question

In an oscillating LC circuit, L = 27.0 mH and C = 7.00

In an oscillating LC circuit, L = 27.0 mH and C = 7.00 mu F. At time t = 0 the current is 9.40 mA, the charge on the capacitor is 3.10 mu C, and the capacitor is charging. (a) What is the total energy in the circuit? J (b) What is the maximum charge on the capacitor? C (c) What is the maximum current? A (d) If the charge on the capacitor is given by q = Q cos(?t + phi), what is the phase angle phi? (e) Suppose the data are the same, except that the capacitor is discharging at t = 0. What then is phi?

Explanation / Answer

a)

Total Energy in the circuit

E=(1/2)[Q2/C +LI2]=(1/2)[(3.1*10-6)2/(7*10-6)+(27*10-3)(9.4*10-3)2]

E=1.88*10-6 J

b)

E=(1/2)Qmax2/C

=>Qmax=sqrt[2*E*C]=sqrt[2*1.88*10-6*7*10-6]

Qmax=5.13*10-6 C or 5.13 uC

c)

W=1/sqrt[LC)=1/sqrt[7*10-6*27*10-3]=2300.2 rad/s

Maximum current

Imax=WQmax=2300.2*5.13*10-6

Imax=0.0118 A or 11.8 mA

d)

q=QCos(Wt+phi)

Cos(Wt+phi)=q/Q =3.1/5.13

Wt+phi=Cos-1(0.6043)=52.82o

During charging at t=0

phi=-52.82o

e)

During discharging at t=0

phi=52.82o

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