Question

An L-C circuit containing an 82.0-mH inductor and a 1.60-nF capacitor oscillates with a maximum current of 0.800 A . Assuming the capacitor had its maximum charge at time t= 0, calculate the energy stored in the inductor after 2.40 ms of oscillation.

Answer 1

Answer:

19.5 mJ

Explanation:

Assuming perfect components without resistance or losses, the circuit should oscillate indefinetly.

The circuit will have a natural pulsation of

$w = \frac{1}{\sqrt{L * C}} = \frac{1}{\sqrt{82e-3 * 1.6e-9}} = 87304 rad/s$

$f = \frac{w}{2\pi} = \frac{87304}{2\pi} = 13895 Hz$

$T = \frac{1}{f} = \frac{1}{13895} = 7.2 \mu s$

So, by the time t = 2.4 ms, 333.33 cycles would have passed

$\frac{2.4 ms}{7.2 \mu s} = \frac{2400 \mu s}{7.2 \mu s} = 333.33$

Therefore it would be at one third after the beginning of the cycle. The circuit would be in an equivalent state as t = (7.2 us)/3 = 2.4us

At t=0 the capacitor is fully charged, so the voltage is maximum and the current is 0. The current will increase towards a maximum of 800 mA at t=T/4, then decreas to 0 at t=T/2, decrease to -800 mA at 3T/4 and go back to 0 at t=T following a sine wave.

The equation of this sine wave would be

$I(t) = I0 * sin(w * t)$

$I(T/3) = 0.8 * sin(w * T/3)$

Since w = 2π/T

w * T = 2π

$I(T/3) = 0.8 * sin(2 \pi /3) = 0.8 * 0.866 = 0.69 A$

The current stored in an inductor is

$E = \frac{1}{2} * L * I^2$

$E = \frac{1}{2} * 82e-3 * 0.69^2 = 0.0195 J = 19.5 mJ$