Question

31.2 damped oscillations in an rlc circuit
an rlc circuit has a resistance of 240 ω, an inductance of 12.0 mh, and a capacitance of 40.0 nf. at time t = 0, the charge on the capacitor is 25.0 μc, and there is no current flowing. after three complete cycles, what is the energy stored in the capacitor?
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Explanation:

Step1: Calculate the damping factor

The damping factor \(\beta=\frac{R}{2L}\). Given \(R = 240\space\Omega\) and \(L=12.0\times10^{- 3}\space H\), then \(\beta=\frac{240}{2\times12.0\times10^{-3}} = 10^{4}\space s^{-1}\).

Step2: Calculate the natural frequency

The natural frequency \(\omega_0=\frac{1}{\sqrt{LC}}\). Given \(L = 12.0\times10^{-3}\space H\) and \(C=40.0\times10^{-9}\space F\), then \(\omega_0=\frac{1}{\sqrt{12.0\times10^{-3}\times40.0\times10^{-9}}}\approx4.56\times10^{4}\space rad/s\).

Step3: Calculate the angular frequency of damped oscillation

The angular frequency of damped oscillation \(\omega_d=\sqrt{\omega_0^{2}-\beta^{2}}\). Substituting the values of \(\omega_0\) and \(\beta\), \(\omega_d=\sqrt{(4.56\times10^{4})^{2}-(10^{4})^{2}}\approx4.45\times10^{4}\space rad/s\).

Step4: Calculate the time for three - cycle

The time period of damped oscillation \(T=\frac{2\pi}{\omega_d}\). The time for three - cycle \(t = 3T=\frac{6\pi}{\omega_d}\). Substituting \(\omega_d\approx4.45\times10^{4}\space rad/s\), \(t=\frac{6\pi}{4.45\times10^{4}}\approx4.27\times10^{-4}\space s\).

Step5: Calculate the charge at time \(t\)

The charge in a damped RLC circuit is given by \(q = q_0e^{-\beta t}\cos(\omega_dt+\varphi)\). At \(t = 0\), \(i = 0\) and \(q = q_0\), so \(\varphi = 0\). Then \(q = q_0e^{-\beta t}\). Substituting \(q_0=25.0\times10^{-6}\space C\), \(\beta = 10^{4}\space s^{-1}\) and \(t = 4.27\times10^{-4}\space s\), \(q=(25.0\times10^{-6})e^{-10^{4}\times4.27\times10^{-4}}\approx25.0\times10^{-6}\times e^{- 4.27}\approx25.0\times10^{-6}\times0.0138 = 3.45\times10^{-7}\space C\).

Step6: Calculate the energy stored in the capacitor

The energy stored in the capacitor \(U=\frac{q^{2}}{2C}\). Substituting \(q = 3.45\times10^{-7}\space C\) and \(C = 40.0\times10^{-9}\space F\), \(U=\frac{(3.45\times10^{-7})^{2}}{2\times40.0\times10^{-9}}=\frac{1.19\times10^{-13}}{8\times10^{-8}}=1.49\times10^{-6}\space J\).

Answer:

\(1.49\times 10^{-6}\space J\)