Consider a LC circuit with a 26.0-mH inductor and a 32.0-$\mu$F capacitor.With a switch keeping the circuit open,the capacitor is charged with a charge of 21.0 $\mu$C.At time t = 0 s the switch is closed,allowing charge to leave the capacitor.Calculate the first time when the energy stored in the capacitor equals the energy stored in the inductor. A)7.2 * 10^-4 s B)5.9 * 10^-7 s C)4.1 * 10^-2 s D)2.9 * 10^-3 s E)1.4 * 10^-3 s

Question

Quizplus · Accepted Answer

The energy stored in the capacitor equals the energy stored in the inductor at the first quarter of the oscillation period, $T/4$, where $T$ is the period of the LC oscillation. The period $T$ is given by $T = 2\pi\sqrt{LC}$, where $L$ is the inductance and $C$ is the capacitance. Substituting $L = 26.0 \times 10^{-3}$ H and $C = 32.0 \times 10^{-6}$ F, we find $T = 2\pi\sqrt{(26.0 \times 10^{-3})(32.0 \times 10^{-6})}$. Calculating $T/4$ gives approximately $7.2 \times 10^{-4}$ s.

Consider a LC Circuit with a 26 μ\muμ F Capacitor μ\muμ CAt Time T = 0 S the Switch Is Closed,allowing

Consider a LC Circuit with a 26 $\mu$ F Capacitor $\mu$ CAt Time T = 0 S the Switch Is Closed,allowing