An LC circuit consists of a C=5OmicroFarad capacitor and L=0.7Henry inductor con
ID: 1293729 • Letter: A
Question
An LC circuit consists of a C=5OmicroFarad capacitor and L=0.7Henry inductor connected in a simple loop with zero resistance and no battery. The capacitor and inductor store a combined total energy of U=32microJoules. (a) If all the circuits energy is stored in the capacitor at time t = O, 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 = O? Micro Coulombs. Enter a number. current flows through the inductor at the time you found in (a)? mA (d) If the inductor is a cylindrical solenoid of length I=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=O.5x1O^3 N/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? Cm^2Explanation / Answer
a)
frequecny
f=1/2pi*sqrt(LC) =1/2pi*sqrt(0.7*50*10-6) =26.9 Hz
T=1/f =0.03717 seconds
the time that all the energy will be stored in inductor is
t=T/4 =9.3 ms
b)
Energy stored in capacitor is
U=(1/2)(Q2/C)
=>Q=sqrt[2UC]=sqrt[2*32*10-6*50*10-6]
Q=56.57 uC
c)
Energy stored in inductor is
U=(1/2)LImax2
=>Imax=sqrt[2*U/L]=sqrt[2*32*10-6/0.7]
Imax=9.56 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[0.7*0.07/4pi*10-7*1.66*10-3]
N=4844 turns
e)
magnetic field of a solenoid is
B=uo*N*I/L =(4pi*10-7)*4844*0.00956/0.07
B=0.831 mT
f)
U=(1/2)CV2
=>V=sqrt[2U/C]=sqrt[2*32*10-6/50*10-6]
V=1.13 Volts
so
d=V/E =1.13/500=2.26 mm
g)
Capacitance
C=KeoA/d
=>A=C*d/(K*eo)=(50*10-6)(2.26*10-3)/(100,000*8.85*10-12)
A=1278.4 cm2
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