show steps please! An LC circuit consists of a C=46microFarad capacitor and L=O.
ID: 1293695 • Letter: S
Question
show steps please!
An LC circuit consists of a C=46microFarad capacitor and L=O.3Henry inductor connected in a simple loop with zero resistance and no battery. The capacitor and inductor store a compined total energy of U=6OmicroJoules. (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? 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 I= 8 centimeters and radius r=2.9centimeters, 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.6x1O^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(0.3*46*10-6) =42.84 Hz
T=1/f =0.02334 seconds
the time that all the energy will be stored in inductor is
t=T/4 =5.84 ms
b)
Energy stored in capacitor is
U=(1/2)(Q2/C)
=>Q=sqrt[2UC]=sqrt[2*60*10-6*46*10-6]
Q=74.3 uC
c)
Energy stored in inductor is
U=(1/2)LImax2
=>Imax=sqrt[2*U/L]=sqrt[2*60*10-6/0.3]
Imax=20 mA
d)
Area
A=pi*r2=pi*0.0292=2.64*10-3 m2
Inductance of a solenoid is
L=uoN2A/l
=>N=sqrt[L*l/uo*A]=sqrt[0.3*0.08/4pi*10-7*2.64*10-3]
N=2689 turns
e)
magnetic field of a solenoid is
B=uo*N*I/L =(4pi*10-7)*2689*0.02/0.08
B=0.845 mT
f)
U=(1/2)CV2
=>V=sqrt[2U/C]=sqrt[2*60*10-6/46*10-6]
V=1.615 Volts
so
d=V/E =1.615/600
d=2.69 mm
g)
Capacitance
C=KeoA/d
=>A=C*d/(K*eo)=(46*10-6)(2.69*10-3)/(100,000*8.85*10-12)
A=1399.2 cm2
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