A series circuit contains an 80-μF capacitor,a 0.020-H inductor,and a switch.The resistance of the circuit is negligible.Initially,the switch is open,the capacitor voltage is 50 V,and the current in the inductor is zero.At time t = 0 s,the switch is closed.How long after closing the switch does the magnetic energy in the inductor first reach its maximum value?
A)2.0 ms
B)4.0 ms
C)6.0 ms
D)8.0 ms
E)10 ms

Question

Quizplus · Accepted Answer

The magnetic energy in the inductor reaches its maximum value when the capacitor is fully discharged and all its energy has been transferred to the inductor. This occurs at a quarter of the oscillation period of the LC circuit. The oscillation period $T$ is given by $T = 2\pi\sqrt{LC}$, where $L$ is the inductance and $C$ is the capacitance. Substituting the given values, $L = 0.020 \, \text{H}$ and $C = 80 \times 10^{-6} \, \text{F}$, we find $T = 2\pi\sqrt{(0.020)(80 \times 10^{-6})} = 2\pi\sqrt{1.6 \times 10^{-3}} \approx 2\pi\sqrt{1.6} \times 10^{-1.5}$. A quarter of this period, $T/4$, is when the magnetic energy is at its maximum, which is approximately $2.0 \, \text{ms}$.

A Series Circuit Contains an 80-μF Capacitor,a 0