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Question: Answered & Verified by Expert
An inductor of inductance 2.0mH is connected across a charged capacitor of capacitance 5.0μF and the resulting L-C 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. When Q=100μC, what is the value of dI / dt?
PhysicsAlternating CurrentJEE Main
Options:
  • A 10000
  • B 1000
  • C 100000
  • D 100
Solution:
2883 Upvotes Verified Answer
The correct answer is: 10000
This is a problem of L - C oscillations.
Charge stored in the capacitor oscillates simple harmonically as

Q= Q 0 sin( ωt±ϕ )
 
Here, Q 0 = maximum value of Q=200μC=2× 10 4 C
 
ω = 1 LC = 1 2 × 10 - 3 5.0 × 1 0 - 6 = 10 4  s -1
 
At   t=0,Q= Q 0 then
 
Q( t )= Q 0 cosωt                               ... (i)
 
I t = dQ dt = - Q 0 ω sin ω t           and             ... (ii)
 
dI t dt = - Q 0 ω 2 cos ω t                         ... (iii)
 
Q=100μC or Q 0 2

cos ω t = 1 2    or   ω t = π 3

dI dt = 2.0 × 1 0 - 4 C 1 0 4 s -1 2 1 2
 
dI dt = 1 0000  A / s

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