Question

A capacitor C is charged to an initial potential of 50.0 V, with an initial charge of Qo. It is in a circuit with a switch and an inductor with inductance L = 3.61×10-2 H and resistance RL -Th Rr. At t=0, the switch is closed, and the curve below shows the potential V across the capacitor as a function of time t. 40 38 20 18 1B P-28 58 0.1 8.2 .3 84 8.5 .6 8.7 8.8 89 1 1. t, in illiseconds) Calculate the energy in the circuit after a time of 14 periods. Note that the curve passes through a grid intersection point. 1.98x 10 Submit Angwer Incorrect. Tries 5/12 Previous Tries Calculate the time required for 87% of the initial energy to be dissipated 1.00x 10-3s Submil Answer Incorrect. Tries 3/12 Previous Tries

Answer 1

Part (a):  The number of periods completed between time t_o = 0 ms and t₁ = 0.9 ms are 10. Therefore the time period is,

  

Thus the angular frequency is,

   $\small \omega =\frac{2\pi }{T}=\frac{2\pi }{0.00009}=69813.17\: Hz$

Peak voltage at t_o is 70 V and at t₁ is 40 V. The voltage in a damped oscillating circuit can be written as,

   $\small V=V_{o}e^{\frac{-Rt}{2L}}cos\omega t$

Here for peak value cos( $\small \omega t$ ) =1. So the time constant is,

$\small \Rightarrow \frac{R}{2L}=\frac{-1}{t}ln\left ( \frac{V}{V_{o}} \right )$

   $\small \Rightarrow \frac{R}{2L}=\frac{-1}{0.9\times 10^{-3}}ln\left ( \frac{40}{50} \right )=247.94\: s^{-1}$

Now the resonant frequency is,

   $\small \omega _{o}=\sqrt{\omega^{2}+\left ( \frac{R}{2L} \right )^{2} }=\sqrt{(69813.17)^{2}+(247.94)^{2}}=69814.94\: Hz$

Now the capacitance can be calculated as,

   $\small \omega _{o}=\frac{1}{\sqrt{LC}}$

   $\small \Rightarrow C=\frac{1}{L\omega _{o}^{2}}=\frac{1}{3.61\times 10^{-2}\times (69814.24)^{2}}=5.68\times 10^{-9}\: F$

Initial energy in the capacitor can be given as,

   $\small U_{o}=\frac{1}{2}CV^{2}=14.2\times 10^{-6}\: J$

Now the energy after 17 periods can be calculated as,

   $\small U=U_{o}e^{\frac{-Rt}{L}}=U_{o}e^{\frac{-2R(17T)}{2L}}=14.2\times 10^{-6}\left ( e^{-2(247.94)(14\times 0.00009)} \right )$

$\small \Rightarrow U=7.60\times 10^{-6}\: J$

Part (b): When 87% of initial energy is dissipated, that is,

$\small U_{87}=(1-0.87)U_{o}=0.13U_{o}$

Then the time is,

   $\small U_{87}=U_{o}e^{\frac{-Rt_{92}}{L}}$

   $\small \Rightarrow 0.13=e^{\frac{-2Rt_{92}}{2L}}=e^{-2(247.87)(t_{87})}$

   $\small \small \ s=4.11\: ms$