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Question

An inductor of inductance 2.0 mH is connect across a charge capacitor of capacittance 5.0 uF , and the resulting LC circuit is set oscillating at natural frequency. Let Q denote the instantaneous charge on the capacitor, and I the current in the circuit. It is found the maximum value of Q is 200 uC .When Q=100uC the value of dIdt is found to be 10x A/s. Find the value of x is then

A
x=-4
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B
x=-5
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C
x=4
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D
x=5
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Solution

The correct option is A x=-4
In an Le system he charge oscillation between I and Q
Applying KVL gives,
LdIdt+9c=0 where L=2 mH
c=5μF
also, maximum charge Q0=200 μc
Thus, total energy of system, μ=Q02c=200Z22×5=4000 J
when 9=100 uc, we get from (1),
dIdtdfrac+9Lc=+1002×103×5=+104 A/s
as dIdt=10x in gives x=4


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