20. 014.76 points | Previous Answers OsUniPhys1 31.5.WA. 036 an oscillating LC c
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20. 014.76 points | Previous Answers OsUniPhys1 31.5.WA. 036 an oscillating LC circuit, L = 27.0 mH and C = 7.00 ?? At time t-0 the current is 9.50 mA, the charge on the capacitor is 3.00 ?C, and the capacitor is charg (a) What is the total energy in the circuit? 000000261 (b) What is the maximum charge on the capacitor? (c) What is the maximum current? (d) If the charge on the capacitor is given by q Q cos(at ), what is the phase angle ? (e) Suppose the data are the same, except that the capacitor is discharging at t = 0-What then is p? Additional Materials eBookExplanation / Answer
L=27mH
C=7uF
at t=0, I=9.5mA and q=3 uC
a)
total energy, U=UL+UC
=1/2*L*I^2 + q^2/2C
=1/2*27*10^-3*(9.5*10^-3)^2 + (3*10^-6)^2/(2*7*10^-6)
=1.861*10^-6 J
b)
U_max=q_max^2/2*C
1.861*10^-6=q_max^2/(2*7*10^-6)
===> q_max=5.104*10^-6 C or q_max=5.104 uC
c)
U_max=1/2*L*I_max^2
1.861*10^-6=1/2*27*10^-3*I_max^2
===> I_max=11.74*10^-3 A or 11.74 mA
d)
q=q_max*cos(w*t+phi)
if t=0
q=q_max*cos(phi)
3*10^-6=5.104*10^-6*cos(phi)
===> phi=54 degrees
phase angle,phi=54 degrees
e)
if capacitor is discharing, phase angle=-54 degrees
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