6.3E: The RLC Circuit (Exercises)
- Page ID
- 18281
Q6.3.1
In Exercises 6.3.1-6.3.5 find the current in the \(RLC\) circuit, assuming that \(E(t)=0\) for \(t>0\).
1. \(R=3\) ohms; \(L=.1\) henrys; \(C=.01\) farads; \(Q_0=0\) coulombs; \(I_0=2\) amperes.
2. \(R=2\) ohms; \(L=.05\) henrys; \(C=.01\) farads’; \(Q_0=2\) coulombs; \(I_0=-2\) amperes.
3. \(R=2\) ohms; \(L=.1\) henrys; \(C=.01\) farads; \(Q_0=2\) coulombs; \(I_0=0\) amperes.
4. \(R=6\) ohms; \(L=.1\) henrys; \(C=.004\) farads’; \(Q_0=3\) coulombs; \(I_0=-10\) amperes.
5. \(R=4\) ohms; \(L=.05\) henrys; \(C=.008\) farads; \(Q_0=-1\) coulombs; \(I_0=2\) amperes.
Q6.3.2
In Exercises 6.3.6-6.3.10 find the steady state current in the circuit described by the equation.
6. \({1\over10}Q''+3Q'+100Q=5\cos10t-5\sin10t\)
7. \({1\over20}Q''+2Q'+100Q=10\cos25t-5\sin25t\)
8. \({1\over10}Q''+2Q'+100Q=3\cos50t-6\sin50t\)
9. \({1\over10}Q''+6Q'+250Q=10\cos100t+30\sin100t\)
10. \({1\over20}Q''+4Q'+125Q=15\cos30t-30\sin30t\)
Q6.3.3
11. Show that if \(E(t)=U\cos\omega t+V\sin\omega t\) where \(U\) and \(V\) are constants then the steady state current in the \(RLC\) circuit shown in Figure 6.3.1 is \[I_p={\omega^2RE(t)+(1/C-L\omega^2)E'(t)\over\Delta},\] where \[\Delta=(1/C-L\omega^2)^2+R^2\omega^2.\]
12. Find the amplitude of the steady state current \(I_p\) in the \(RLC\) circuit shown in Figure 6.3.1 if \(E(t)=U\cos\omega t+V\sin\omega t\), where \(U\) and \(V\) are constants. Then find the value \(\omega_0\) of \(\omega\) maximizes the amplitude, and find the maximum amplitude.
Q6.3.4
In Exercises 6.3.13-6.3.17 plot the amplitude of the steady state current against \(ω\). Estimate the value of \(ω\) that maximizes the amplitude of the steady state current, and estimate this maximum amplitude. HINT: You can confirm your results by doing Exercise 6.3.12.
13. \({1\over10}Q''+3Q'+100Q=U\cos\omega t+V\sin\omega t\)
14. \({1\over20}Q''+2Q'+100Q=U\cos\omega t+V\sin\omega t\)
15. \({1\over10}Q''+2Q'+100Q=U\cos\omega t+V\sin\omega t\)
16. \({1\over10}Q''+6Q'+250Q=U\cos\omega t+V\sin\omega t\)
17. \({1\over20}Q''+4Q'+125Q=U\cos\omega t+V\sin\omega t\)