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An inductor of inductance 2mH is connected cross a charged capacitor of capacitance 5 microfarads and the resulting LC circiut is set oscillating at its natural frequency.Let Q denote the instantaneous charge on the capacitor and I the current in the circuit.it is found that the maximum value of charge is 200 micro coulumbs. 1)When Q=100 micro coulumbs what is the value of di/dt? 2)when Q=200 micro coulumbs what is the value of I? 3)Find the maximum value of I 4)when I equals half its maximin value,what is the value of Q?


An inductor of inductance 2mH is connected cross a charged capacitor of capacitance 5 microfarads and the resulting LC circiut is set oscillating at its natural frequency.Let Q denote the instantaneous charge on the capacitor and I the current in the circuit.it is found that the maximum value of charge is 200 micro coulumbs.


1)When Q=100 micro coulumbs what is the value of di/dt?


2)when Q=200 micro coulumbs what is the value of I?


3)Find the maximum value of I


4)when I equals half its maximin value,what is the value of Q?


Grade:upto college level

1 Answers

ROSHAN MUJEEB
askIITians Faculty 829 Points
one year ago

Solution :
This is a problemL−CL-Coscillations.
Charge stored in the capacitor oscillates simple harmonically as
Q=Q0sin(ωt±ϕ)Q=Q0sin(ωt±ϕ)
HereQ0=Q0=maximum value ofQ=200μC=2×10−4CQ=200μC=2×10-4C
ω=1LC−−−√=1(2×10−3)(5.0×10−6)−−−−−−−−−−−−−−−−−−√=104s−1ω=1LC=1(2×10-3)(5.0×10-6)=104s-1
Let att=0,Q=Q0t=0,Q=Q0then
Q(t)=Q0cosωtQ(t)=Q0cosωt........i
I(t)=dQdt=−Q0ωsinωtI(t)=dQdt=-Q0ωsinωtand ........ii
dI(t)dt=−Q0ω2cosωtdI(t)dt=-Q0ω2cosωt............iii
a.Q=100μCQ=100μC
orQ02Q02atcosωt=12cosωt=12
orωt=π3ωt=π3
Atcosωt=12,cosωt=12,from eqn iii:
∣∣∣dIdt∣∣∣=(2.0×10−4C)(104s−1)2(12)|dIdt|=(2.0×10-4C)(104s-1)2(12)
∣∣∣dIdt∣∣∣=104A/s|dIdt|=104A/s
b.Q=200μCQ=200μCorQ0Q0whencosωt=1,cosωt=1,i.e.cosωt=0,2πcosωt=0,2π........
At the timeI(t)=−Q0ωsinωtI(t)=-Q0ωsinωt
orI(t)=0(sin02=sin2π=0)I(t)=0(sin02=sin2π=0)
c.I(t)=−Q0ωsinωtI(t)=-Q0ωsinωt
∴∴Maximum value ofIIisQ0ωQ0ω
Imax=Q0ω=(2.0×10−4)(104)Imax=Q0ω=(2.0×10-4)(104)
Imax=2.0AImax=2.0A
d. From energy conservation
12LI2max=12LI2+12Q2C12LImax2=12LI2+12Q2C
orQ=LC(I2max−I2)−−−−−−−−−−−−√Q=LC(Imax2-I2)
I=Imax2=1.0AI=Imax2=1.0A
∴Q=(2.0×10−3)(5.0×10−−6)(22−12)−−−−−−−−−−−−−−−−−−−−−−−−−−−−√∴Q=(2.0×10-3)(5.0×10--6)(22-12)
Q=3–√×10−4CQ=3×10-4C
orQ=1.732×10−4C

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