Solution Found!
Consider the circuit shown in Fig. The circuit elements
Chapter 30, Problem 76CP(choose chapter or problem)
Consider the circuit shown in Fig. P30.76. The circuit elements are as follows: \(\mathcal{E}=32.0 \mathrm{~V}, L=0.640 \mathrm {~H}, C=2.00 ~\mu \mathrm{F} \text {, and } R=400 ~\Omega\). At time t = 0, switch S is closed. The current through the inductor is \(i_{1}\), the current through the capacitor branch is \(i_{2}\), and the charge on the capacitor is \(q_{2}\).
(a) Using Kirchhoff’s rules, verify the circuit equations
\(R\left(i_{1}+i_{2}\right)+L\left(\frac{d i_{1}}{d t}\right)=\mathcal{E}\)
\(R\left(i_{1}+i_{2}\right)+\frac{q_{2}}{C}=\mathcal{E}\)
(b) What are the initial values of \(i_{1}, i_{2} \text {, and } q_{2} ?\)
(c) Show by direct substitution that the following solutions for \(i_{1} \text { and } q_{2}\) satisfy the circuit equations from part (a). Also, show that they satisfy the initial conditions
\(i_{1}=\left(\frac{\mathcal{E}}{R}\right)\left[1-e^{-\beta t}\left\{(2 \omega R C)^{-1} \sin (\omega t)+\cos (\omega t)\right\}\right]\)
\(q_{2}=\left(\frac{\mathcal{E}}{\omega R}\right) e^{-\beta t} \sin (\omega t)\)
Where \(\beta=(2 R C)^{-1} \text { and } \omega=\left[(L C)^{-1}-(2 R C)^{-2}\right]^{1 / 2}\).
(d) Determine the time \(t_{1}\) at which \(i_{2}\) first becomes zero.
Questions & Answers
QUESTION:
Consider the circuit shown in Fig. P30.76. The circuit elements are as follows: \(\mathcal{E}=32.0 \mathrm{~V}, L=0.640 \mathrm {~H}, C=2.00 ~\mu \mathrm{F} \text {, and } R=400 ~\Omega\). At time t = 0, switch S is closed. The current through the inductor is \(i_{1}\), the current through the capacitor branch is \(i_{2}\), and the charge on the capacitor is \(q_{2}\).
(a) Using Kirchhoff’s rules, verify the circuit equations
\(R\left(i_{1}+i_{2}\right)+L\left(\frac{d i_{1}}{d t}\right)=\mathcal{E}\)
\(R\left(i_{1}+i_{2}\right)+\frac{q_{2}}{C}=\mathcal{E}\)
(b) What are the initial values of \(i_{1}, i_{2} \text {, and } q_{2} ?\)
(c) Show by direct substitution that the following solutions for \(i_{1} \text { and } q_{2}\) satisfy the circuit equations from part (a). Also, show that they satisfy the initial conditions
\(i_{1}=\left(\frac{\mathcal{E}}{R}\right)\left[1-e^{-\beta t}\left\{(2 \omega R C)^{-1} \sin (\omega t)+\cos (\omega t)\right\}\right]\)
\(q_{2}=\left(\frac{\mathcal{E}}{\omega R}\right) e^{-\beta t} \sin (\omega t)\)
Where \(\beta=(2 R C)^{-1} \text { and } \omega=\left[(L C)^{-1}-(2 R C)^{-2}\right]^{1 / 2}\).
(d) Determine the time \(t_{1}\) at which \(i_{2}\) first becomes zero.
ANSWER:
Step 1 of 5
We have to first show the equation for current and charge using Kirchhoff’s law. Then we have to state the initial condition. Then we have to show that the given expression of charge and current satisfy the derived equation. Finally we have find out what will be value of when will be zero for the first time.