6.3E: The RLC Circuit (Exercises)

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Jun 10, 2024
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- 6.3: The RLC Circuit
- 6.4: Motion Under a Central Force

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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. ${1\over20}Q''+2Q'+100Q=10\cos25t-5\sin25t$

${1\over10}Q''+2Q'+100Q=3\cos50t-6\sin50t$

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 $I_p={\omega^2RE(t)+(1/C-L\omega^2)E'(t)\over\Delta}, \nonumber$ where $\Delta=(1/C-L\omega^2)^2+R^2\omega^2. \nonumber$

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. ${1\over20}Q''+4Q'+125Q=U\cos\omega t+V\sin\omega t$

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