An inductor of inductance 2.0 mH is connected across a charged capacitor of capacitance 5.0 μ F and the resulting L-C circuit is set oscillating at its natural frequency. Let Q denote the instantaneou

Question

An inductor of inductance 2.0 mH is connected across a charged capacitor of capacitance 5.0 μ F 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 200 μ C . When Q = 100 μ C, what is the value of dI / dt ?

TestHub · Accepted Answer

This is a problem of L - C oscillations.
Charge stored in the capacitor oscillates simple harmonically as

$Q = Q_{0} \sin (ω t \pm ϕ)$

Here, $Q_{0} =$ maximum value of $Q = 200 μ C = 2 \times 10^{- 4} C$

$ω = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{(2 \times 10^{- 3}) (5.0 \times 1 0^{- 6})}} = 10^{4} s^{-1}$

At $t = 0, Q = Q_{0}$ then

$Q (t) = Q_{0} \cos ω t$ ... (i)

$I (t) = \frac{dQ}{dt} = - Q_{0} ω sin ω t$ and ... (ii)

$\frac{dI (t)}{dt} = - Q_{0} ω^{2} cos (ω t)$ ... (iii)

$Q = 100 μ C$ or $\frac{Q_{0}}{2}$

$cos ω t = \frac{1}{2}$ or $ω t = \frac{π}{3}$

$|\frac{dI}{dt}| = (2.0 \times 1 0^{- 4} C) {(1 0^{4} s^{-1})}^{2} (\frac{1}{2})$

$|\frac{dI}{dt}| = 10000 A / s$

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