A 5.00F capacitor is initially charged to a potential of 16.0V . It is then conn

Question

A 5.00F capacitor is initially charged to a potential of 16.0V . It is then connected in series with a 3.75mH inductor.PartA:What is the total energy stored in this circuit?PartB:What is the maximum current in the inductor? PartC:What is the charge on the capacitor plates at the instant the current in the inductor is maximal?

Question2:A 12.0V dc battery having no appreciable internal resistance, a 150 Omega resistor, an 11.0mH inductor, and an open switch are all connected in series. PartA:After the switch is closed, what is the time constant for this circuit ?PartB:After the switch is closed, what is the maximum current that flows through it ?PartC:What is the current 73.3s after the switch is closed ?PartD:

Question3:When a certain inductor carries a current II, it stores 3.00mJ of magnetic energy.PartA:How much current (in terms of II) would it have to carry to store 9.00mJ of energy?

Explanation / Answer

44. a) Energy in an inductor = 0.5*Inductance*(Current^2)
Energy = E = 10 J
Inductance = L
Current = i = 1.5A
10 = 0.5*L*(1.5^2) = 1.125*L
L = 10/1.125 = 8.88
Therefore,
Inductance = L = 8.88
b) Cross sectional diameter = d = 4cm = 0.04 m
Density of coils = 10coils/mm = 10000coils/m = N
l = Length
mu = constant = 4*pi*10^-7
L = mu*(N^2)*A/l
8.88 = 4*pi*10^-7*10^8*pi*(0.04^2)/4*l
l = 0.017m = 1.7cm
This length is realistic for laboratory use.

48. Voltage = V = 12 volts
Resistance = R = 150 ohm
Inductance = L = 11 mH = 0.011 H
a) Time constant of the circuit = T
T = L/R = 0.011/150 = 7.33*10-5 sec
Therefore,
Time constant of the circuit = T = 7.33*10-5 seconds

b) Maximum Current that flows through it is equal to when the resistance due to the inductor becomes zero. i.e. when the inductor starts behaving like a wire.
Hence,
Maximum current = iMAX = V/R = 12/150 = 0.08 A

c) Given time elapsed = 73.3 mus = 7.33*10^-5 sec
This is equal to the time constant of the circuit!
After one time constant, the current in the circuit reaches 63.2 % of the maximum current possible(iMAX)
i.e. Current after 73.3 mus = (0.632)*iMAX = 0.632*0.08 = 0.05056 A

d) Maximum Energy stored in the inductor = W
W = 0.5*L*iMAX^2 = 0.5*0.011*0.08*0.08 = 0.0352 mJ
Hence,
Maximum energy stored in the inductor = 0.0352 mJ

54. Capacitance = C = 12 muF = 12*10^-6 F
Inductance = L = 5.25 mH = 5.25*10^-3 H
Angular frequency of charge oscillations = f
f = 1/(2*pi*sqrt(L*C))
f = 1/(2*pi*sqrt(5.25*10^-3*12*10^-6))
f = 1/2*pi*2.51*10^-4 = 634.4 Hz
Hence,
Frequency of oscillation of the charges in the circuit = f = 634.4 Hz

56. Initial charge on the capacitor = q = 175 muC = 175*10^-6 C
Capacitance = C = 15muF = 15*10^-6 F
Inductance = L = 5mH = 0.005 H
a) C = q/V
Therefore,
Voltage = q/C = 175*10^-6/15*10^-6 = 11.67 V
Max. Energy through capacitor = Max. energy through inductor
Hence,
0.5*C*V*V = 0.5*L*i*i
15*10^-6*11.67*11.67 = 0.005*i^2
i = 0.639A
Therefore,
Maximum current through the inductor = i = 0.639 A
Charge = 175 muC
b) Maximum potential across the capacitor = V = 11.67 Volts(Calculated in the previous part)
Current at this point through the inductor = 0.639 A(Calculated in the previous part)
c) Maximum energy stored in the inductor = 0.5*L*i*i = 0.5*0.005*0.639^2 = 1.02 mJ
Current in the circuit = 0.639 A(Calculated in the previous part)

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