Question

An inductor of inductance 2.0 mH is connected across a charged capacitor of capacitance `0.5 mu F` 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 charge Q is `200 mu C`. what is the maximum value of I (in ampere)?
A. 1A
B. 1.5A
C. 2A
D. 3A

Answer 1

Correct Answer - C
For the oscillator,
` omega =1/(sqrt(LC)) = 1/(sqrt(2 xx 10^(-3) xx 5 xx 10^(-6)))`
`= `1/(10^(-4)) = 10^(4) rad//s`
Charge Q varies harmonically with time and is given by
`Q =Q_(0) cos omega t` where `Q_(0` = maximum charge at `t = 0`
`:. I = (dQ)/(dt) = -omega Q_(0) sin omega t`
`:. I_(max) = omega Q_(0) = 10^(4) xx 200 xx 10^(-6) = 2A`.