Electromagnetic Induction and Alternating Current 4 Question 20

####20. An inductor of inductance 2.0mH is connected across a charged capacitor of capacitance 5.0μF and the resulting LC circuit 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 Q is 200μC.

(1998,8M)

(a) When Q=100μC, what is the value of |dI/dt| ?

(b) When Q=200μC, what is the value of I ?

(c) Find the maximum value of I.

(d) When I is equal to one-half its maximum value, what is the value of |Q| ?

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Answer:

Correct Answer: 20.  (a) 104A/s (b) zero (c) 2.0A (d) 1.732×104C

Solution:

  1. This is a problem of LC oscillations. Charge stored in the capacitor oscillates simple harmonically as

Q=Q0sin(ωt±φ)

Here, Q0= maximum value of Q=200μC=2×104C

ω=1LC=1(2×103)(5.0×106)=104s1

Let at

t=0,Q=Q0, then 

Q(t)=Q0cosωtI(t)=dQdt=Q0ωsinωt and 

dI(t)dt=Q0ω2cos(ωt)

(a) Q=100μC or Q02 at cosωt=12 or ωt=π3

At cos(ωt)=12, from Eq. (iii) :

dIdt=(2.0×104C)(104s1)212dIdt=104A/s

(b) Q=200μC or Q0 when cos(ωt)=1 i.e. ωt=0,2π

At this time I(t)=Q0ωsinωt

 or I(t)=0(sin0=sin2π=0)

(c) I(t)=Q0ωsinωt

Maximum value of I is Q0ω.

Imax=Q0ω=(2.0×104)(104)Imax=2.0A

(d) From energy conservation,

12LImax2=12LI2+12Q2C or Q=LC(Imax2I2)I=Imax2=1.0AQ=(2.0×103)(5.0×106)(2212)Q=3×104C or Q=1.732×104C



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