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Physics Bootcamp

Samuel J. Ling

Section 40.6 EM-Oscillations Bootcamp

Exercises Exercises

LC Circuits

1. Frequency of Oscillations of an LC-circuit and Voltage at an Instant.
Follow the link: Example 40.4.
2. Inductance, Energy, and Current of an LC circuit from Period of Oscillations.
Follow the link: Example 40.5.

Damped RLC Circuits

4. Practice with Determining Underdamped, Overdamped, or Critically Damped Circuits.
Follow the link: Example 40.14.
7. Oscillation Versus Natural Frequency in a Damped Electrical Circuit.
Follow the link: Exercise 40.2.5.2.

Driven RLC Circuits

8. Practicing Resonance Frequency of Current in RLC Circuits.
Follow the link: Example 40.26.
9. Current Amplitude in a Driven RLC Circuit.
Follow the link: Example 40.28.
10. Average Power of the Source in a Driven RLC Circuit.
Follow the link: Example 40.29.
11. Current and Voltages in a Driven Series RLC Circuit.
Follow the link: Example 40.30.

Resonance of Electrical Circuits

12. Current and Voltages in a Series RLC Circuit.
Follow the link: Example 40.41.
16. Average Powerer Delivered to a Circuit at Different Frequencies.
Follow the link: Exercise 40.5.1.
17. Power Dissipating in Resistor of a Driven RLC Circuit.
Follow the link: Exercise 40.5.2.

Miscellaneous

18. Oscillations, Energy, and Charge on Capacitor of an LC-Circuit.
A \(500\text{-pF}\) capacitor is charged fully by a \(12\text{-V}\) battery, and then connected to a \(125\text{-mH}\) inductor at \(t=0\text{.}\) Assume the resistance in the circuit to be negligible. Determine (a) the frequency of oscillation, (b) the current as a function of time, (c) the maximum energy stored in the capacitor, (d) the maximum energy stored in the magnetic field of the inductor, and (e) the charge on the capacitor when the current in the inductor is half the maximum.
Solution 1. a
The frequency of oscillation of an LC-circuit is given by
\begin{equation*} f = \dfrac{1}{2\pi\sqrt{125\:\textrm{mH}\times 500\:\textrm{pF}}} = \dfrac{10^{6}}{2\pi\sqrt{12.5\times 5}} = 20,132\:\textrm{Hz}. \end{equation*}
Solution 2. b
\(I(t) = -\omega\: C\: V_0\: \sin(\omega t) = 760\:\mu\textrm{A}\:\sin(126500\: t)\text{.}\)
Solution 3. c
\(U = \frac{1}{2}\: C V_0^2 = 36\:\textrm{nJ}\text{.}\)
Solution 4. d
Same as (c) based on the energy conservation in purely LC-circuit. [No resistance in the circuit assumed.]
Solution 5. e
The energy in the inductor at this instant will be
\begin{equation*} U_L = \dfrac{1}{2} L I^2 = \dfrac{1}{2}\times 0.125\:\textrm{H} \times (0.5\times 760\:\mu\textrm{A})^2 = 10.8\:\textrm{nJ}. \end{equation*}
Therefore, the energy in the capacitor at this instant will be
\begin{equation*} U_C = U - U_L = 36-10.8 = 25.2\:\textrm{nJ}. \end{equation*}
This will give the following for the charge on the capacitor plates at this instant
\begin{equation*} Q = \sqrt{ 2 C U_C} = \sqrt{2\times 500\:\textrm{pF} \times 36\:\textrm{nJ} } = 6 \:\textrm{nC}. \end{equation*}
19. \(Q\) Factor of a Lightly Damped Oscillator.
In a damped oscillating circuit the energy is dissipated in the resistor. The Q-factor is a measure of the persistence of the oscillator against the dissipative loss. (a) Prove that for a lightly damped circuit the energy \(E\) in the circuit decreases according to the following equation.
\begin{equation*} \frac{dE}{dt} = -2\:\beta\: E,\ \textrm{where}\ \beta = \frac{R}{2L}. \end{equation*}
(b) Using the definition of the Q-factor as energy divided by the loss over next cycle, prove that Q-factor of a lightly damped oscillator as defined in this problem is
\begin{equation*} Q \equiv \frac{E_{\textrm{begin}}}{\Delta E_{\textrm{one cycle}}}= \frac{1}{2\pi R}\: \sqrt{\frac{L}{C}}. \end{equation*}
Solution.
(a) Start with \(E = \dfrac{1}{2} m v^2 + \dfrac{1}{2} k x^2\) and use the approximation \(\beta \ll \omega_0\) to simplify terms.
Time for one cycle is
\begin{equation*} T = \dfrac{1}{f} = 2\pi\sqrt{LC}. \end{equation*}
Therefore, change over one cycle will be
\begin{equation*} \Delta E \approx -2\beta E_{\textrm{ave}}\times T = \dfrac{R}{L}\:2\pi\sqrt{LC}\: E_{\textrm{ave}}. \end{equation*}
For a good oscillator, there will be very little loss in any one cylce and we can replace \(E_{\textrm{ave}}\) by \(E_{\textrm{begin}}\text{.}\) This gives the following for \(Q\text{.}\)
\begin{equation*} Q = \dfrac{E_{\textrm{begin}}}{\Delta E} = \frac{1}{2\pi R}\: \sqrt{\frac{L}{C}}. \end{equation*}
20. Width of Power Resonance Curve of an RLC Circuit.
A series RLC-circuit is driven by a sinusoidal EMF. Find the width of the power resonance curve at half the height.
Solution.
The power resonance curve is a plot of power versus frequency. We will do the calculations in \(\omega\) rather than \(f\) to simplify writing. The average power \(P\) as a function of \(\omega\) for a series \(RLC\) circuit is given by
\begin{equation*} P(\omega) = \dfrac{V_0^2 R}{2\left[R^2 + \left(\omega L - 1/\omega C \right)^2 \right]}.\ \ \ \ (1) \end{equation*}
The maximum of this function occurs at \(\omega = 1/\sqrt{LC}\text{.}\) Setting this for \(\omega\) in this equation gives the value at maximum.
\begin{equation*} P_{\textrm{max}} = \dfrac{V_0^2}{2R}. \ \ \ \ (2) \end{equation*}
Let us illustrate the power curve by a figure of \(P\) vs \(\omega\text{.}\) I have used \(R = 5\:\Omega\text{,}\) \(L = 3\) mH, \(C = 1.6\:\mu\)F, and \(V_0 = 10\) V. The values have been picked for illustrative purpose only.
Figure 40.49.
To find the width of the curve we set \(P(\omega)\) in (1) to \(\dfrac{1}{2} P_{\textrm{max}}\text{.}\) This give the following equation for \(\omega\text{.}\)
\begin{equation*} \left(\omega L - 1/\omega C \right)^2 = R^2. \end{equation*}
This equation is an polynomial equation of fourth power in \(\omega\text{.}\) Keeping only the positive solutions for \(\omega\) we get
\begin{equation*} \omega _1 = \dfrac{1}{LC}\left[-RC + \sqrt{R^2C^2 + 4 LC} \right],\ \ \omega _2 = \dfrac{1}{LC}\left[RC + \sqrt{R^2C^2 + 4 LC} \right]. \end{equation*}
The difference of these solutions gives the width \(\Delta\omega\) of the resonance curve.
\begin{equation*} \Delta \omega = \omega _2 - \omega_1 = \dfrac{R}{L}. \end{equation*}
21. Adjusting Resistance to Sharpen the Current Resonance Curve.
A series RLC circuit with \(R = 4\, \text{k}\Omega\text{,}\) \(C = 0.3\, \text{pF}\text{,}\) and \(H = 2\, \text{pH}\) is driven at the frequency of the current resonance. How should the resistance in the circuit be adjusted so as to sharpen the width of the current resonance to half the original width?
Solution.
The amplitude of the current in teh series RLC circuit is given by
\begin{equation*} I_0 = \dfrac{I_{\textrm{max}}\: R}{\sqrt{R^2 + \left( \omega L - 1/\omega C \right)^2}}. \end{equation*}
This expression gives the peak at \(\omega - 1/\sqrt{LC}\) and the width at half-height at
\begin{equation*} \Delta \omega = \sqrt{3}\dfrac{R}{L}.\ \ \ \ (1) \end{equation*}
Let \(R\) be adjusted from \(R_1 = 4\:\textrm{k}\Omega\) to \(R_{2}\) keeping everything else same so that \(\Delta \omega\) becomes half as much. This requirement on (1) gives
\begin{equation*} R_2 = \frac{1}{2} R = 2\:\textrm{k}\Omega. \end{equation*}
22. Tuning an AM-Radio by Varying Resistance and Capacitance.
A series RLC circuit for tuning into an AM radio station. The circuit has a variable resistor of resistance between \(2\, \Omega\) and \(2000\, \Omega\text{,}\) a variable capacitor whose capacitance could be varied between \(0.1\, \text{pF}\) and \(100\, \text{pF}\text{,}\) and a fixed inductance of \(4\, \text{pH}\text{.}\) What settings of the resistor and capacitor would you need so that the circuit would resonate at (i.e. tunes into) a \(450\, \text{kHz}\) signal?
Solution.
To tune into a \(4\, \text{kHz}\) signal you would the resonance frequency of the circuit equal this frequency. This gives the following condition
\begin{equation*} \dfrac{1}{2\pi\sqrt{LC}} = 450\times 10^3\:\textrm{Hz}. \end{equation*}
With \(L = 4\) pH we need
\begin{equation*} C = \dfrac{1}{4\pi^2\times 4\times 10^{-12}\:\textrm{H}\times (450\times 10^3\:\textrm{Hz})^2} = 31.3\:\textrm{mF}. \end{equation*}
The tuning condition itself does not put restriction on \(R\text{.}\) But, we would need to watch out for the width of the peak so as to stay away from other frequencies being broadcast. If we knew the width of the peak to look out for, we would use choose \(R\) according to the width \(R/L\) of the peak desired.
23. Resonance of the Induced EMF.
(a) In the series RLC circuit connected to a sinusoidally varying EMF source, find the amplitude of the induced EMF as a function of time. (b) What is the resonance frequency of the induced EMF?
Solution 1. a
The induced EMF is the EMF associated with the self-inductance in the circuit. This means we are looking at
\begin{equation*} \mathcal{E}_L = -L\dfrac{dI}{dt}. \end{equation*}
By taking the derivative of \(I\) in a series RLC-circuit we obtain
\begin{equation*} \mathcal{E}_L = \omega\: L\: I_0\sin(\omega t + \phi+ \frac{\pi}{2}). \end{equation*}
The amplitude is
\begin{equation*} \mathcal{E}_{0L} = \omega\: L\: I_0 = \dfrac{\omega L V_0}{\sqrt{R^2 + (\omega L - 1/\omega C)^2}}. \end{equation*}
Solution 2. b
Setting the derivative of \(\mathcal{E}_{0L}\) with respect to \(\omega\) zero gives the following for the resonance frequency.
\begin{equation*} \omega_R = \dfrac{\sqrt{2}}{\sqrt{2LC - R^2 C^2}}. \end{equation*}
24. Voltage Across Capacitor in RLC Circuit.
(a) In the series \(RLC\) circuit connected to a sinusoidally varying EMF source, find the amplitude of the voltage \(V_C\) across the capacitor as a function of time.
(b) What is the resonance frequency of \(V_C\text{?}\)
(c) Using results of the problem Exercise 40.6.23, find the frequency at which the magnitude of \(V_C\) would equal the magnitude of the induced EMF. What is the significance of this frequency?
Solution 1. a
In the series \(RLC\) circuit the voltage across the capacitor was shown in the textbook to be
\begin{equation*} V_C = V_{0C} \cos(\omega t + \phi), \textrm{with}\ V_{0C} = \dfrac{V_0/\omega C}{\sqrt{R^2 + (\omega L - 1/\omega C)^2}}, \end{equation*}
where \(V_0\) the amplitude of the driving EMF and \(\phi\) is the phase lag.
Solution 2. b
By setting the derivative of \(V_{0C}\) to zero you can verify that the resonance of \(V_C\) will occur at
\begin{equation*} \omega_R = \sqrt{\dfrac{1}{LC} - \dfrac{R^2}{2L^2}}. \end{equation*}
Solution 3. c
The condition gives
\begin{equation*} \dfrac{\omega L V_0}{\sqrt{R^2 + (\omega L - 1/\omega C)^2}} = \dfrac{V_0/\omega C}{\sqrt{R^2 + (\omega L - 1/\omega C)^2}}, \end{equation*}
which gives
\begin{equation*} \omega = \dfrac{1}{\sqrt{LC}}. \end{equation*}
This is the condition for the resonance of power.
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