An LC circuit consists of a C=46microFarad capacitor and L=0.4Henry inductor con
ID: 1293730 • Letter: A
Question
An LC circuit consists of a C=46microFarad capacitor and L=0.4Henry inductor connected in a simple loop with zero resistance and no battery. The capacitor and inductor store a compined total energy of U=56microJoules.
(a) If all the circuits energy is stored in the capacitor at time t = 0, when is the soonest time that all the energy will be stored in the inductor?
ms
(b) How much charge is stored on one side of the capacitor at t = 0? microCoulombs. (c) How much current flows through the inductor at the time you found in (a)?
mA
(d) If the inductor is a cylindrical solenoid of length l=7centimeters and radius r=2.3centimeters, find the number of coils in the solenoid.
coils
(e) What is the maximum magnetic field inside the inductor?
mT
(f) If the maximum electric field in the capacitor is E=1.1x103N/C and the space between the plates if filled with a dielectric of constant 100,000, what is the separation between the plates?
mm
(g) What is the area of one of the plates?
cm2
Explanation / Answer
a)
frequecny
f=1/2pi*sqrt(LC) =1/2pi*sqrt(0.4*46*10-6) =37.1 Hz
T=1/f =0.02695 seconds
the time that all the energy will be stored in inductor is
t=T/4 =6.74 ms
b)
Energy stored in capacitor is
U=(1/2)(Q2/C)
=>Q=sqrt[2UC]=sqrt[2*56*10-6*46*10-6]
Q=71.8 uC
c)
Energy stored in inductor is
U=(1/2)LImax2
=>Imax=sqrt[2*U/L]=sqrt[2*56*10-6/0.4]
Imax=16.73 mA
d)
Area
A=pi*r2=pi*0.0232=1.66*10-3 m2
Inductance of a solenoid is
L=uoN2A/l
=>N=sqrt[L*l/uo*A]=sqrt[0.4*0.07/4pi*10-7*1.66*10-3]
N=3662 turns
e)
magnetic field of a solenoid is
B=uo*N*I/L =(4pi*10-7)*3662*0.01673/0.07
B=1.1 mT
f)
U=(1/2)CV2
=>V=sqrt[2U/C]=sqrt[2*56*10-6/46*10-6]
V=1.56 Volts
so
d=V/E =1.56/1100
d=1.42 mm
g)
Capacitance
C=KeoA/d
=>A=C*d/(K*eo)=(46*10-6)(1.42*10-3)/(100,000*8.85*10-12)
A=737.3 cm2
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