Question

an rlc circuit has a resistance of 180 ω, an inductance of 16 mh, and a capacitance of 32 nf. at time t = 0, the charge on the capacitor is 24 μc, and there is no current flowing. after five complete cycles, the energy stored in the capacitor is
2.9 μj.
62 mj.
160 μj.
310 mj.
1.8 mj.

Explanation:

Step1: Calculate the initial energy

The energy stored in a capacitor is given by \(U_0=\frac{Q_0^2}{2C}\).
Given \(Q_0 = 24\times10^{- 6}\text{C}\) and \(C=32\times10^{-9}\text{F}\), then \(U_0=\frac{(24\times10^{-6})^2}{2\times32\times10^{-9}}\)

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Step2: Calculate the damping factor

The damping factor for an RLC circuit is \(\alpha=\frac{R}{2L}\). Given \(R = 180\Omega\) and \(L=16\times10^{-3}\text{H}\), then \(\alpha=\frac{180}{2\times16\times10^{-3}}=\frac{180}{32\times10^{-3}} = 5625\text{s}^{-1}\)
The angular frequency \(\omega_d=\sqrt{\frac{1}{LC}-\alpha^{2}}\). Substituting \(L = 16\times10^{-3}\text{H}\) and \(C=32\times10^{-9}\text{F}\), \(\frac{1}{LC}=\frac{1}{16\times10^{-3}\times32\times10^{-9}}=\frac{1}{512\times10^{-12}}\approx1.95\times10^{9}\text{s}^{-2}\), \(\alpha^{2}=(5625)^{2}\approx3.16\times10^{7}\text{s}^{-2}\). So \(\omega_d\approx\sqrt{1.95\times10^{9}-3.16\times10^{7}}\approx 4400\text{s}^{-1}\)
The period \(T=\frac{2\pi}{\omega_d}\approx\frac{2\pi}{4400}\approx1.43\times10^{-3}\text{s}\)
After \(n = 5\) cycles, \(t=nT=5\times1.43\times10^{-3}\text{s}\approx7.15\times10^{-3}\text{s}\)
The charge as a function of time in a damped RLC circuit is \(Q(t)=Q_0e^{-\alpha t}\cos(\omega_d t+\varphi)\). Since at \(t = 0\), \(I(0) = 0\) and \(Q(0)=Q_0\), \(\varphi = 0\). The energy \(U(t)=\frac{Q^{2}(t)}{2C}=\frac{Q_0^{2}e^{-2\alpha t}\cos^{2}(\omega_d t)}{2C}\). Since \(\cos^{2}(\omega_d t)\approx1\) (we are interested in the envelope of the energy decay), \(U(t)=U_0e^{-2\alpha t}\)
Substituting \(U_0 = 9\text{mJ}\), \(\alpha=5625\text{s}^{-1}\) and \(t = 7.15\times10^{-3}\text{s}\), \(2\alpha t=2\times5625\times7.15\times10^{-3}\approx80\)
\(U(t)=9\times10^{-3}\times e^{- 80}\approx2.9\times10^{-6}\text{J}=2.9\mu\text{J}\)

Answer:

2.9 μJ.