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Physics Bootcamp

Samuel J. Ling

Section 41.9 AC Circuits Bootcamp

Exercises Exercises

AC Circuit Elements

5. Current and Voltage Drops Across Resistor and Inductor in Inductive Circuit.
Follow the link: Example 41.17.
7. Current in a Capacitative Circuit with Varying Frequency.
Follow the link: Example 41.22.
8. Capacitance of a Starter of a Fluorescent Light Bulb from Current and Voltage.
Follow the link: Exercise 41.4.2.1.
9. Impedance, Current, and Voltages in an RLC circuit.
Follow the link: Example 41.30.
10. Impedance and Phases of Voltages in an RLC Circuit.
Follow the link: Exercise 41.5.2.1.

Power in AC Circuits

12. Real Power, Apparent Power, and Power Factor of a Capacitative AC Circuit.
Follow the link: Example 41.34.
16. Plot Voltages Across Resitor and Capacitor Versus Frequency.
Follow the link: Example 41.36.
17. Plot Voltages Across Resitor and Inductor Versus Frequency.
Follow the link: Example 41.39.

AC Circuits Using Complex Numbers

18. Example of Series RLC Circuit in Complex Notation.
Follow the link: Example 41.42.
19. Example of Parallel RL Circuit in Complex Notation.
Follow the link: Example 41.45.
20. Example of Circuit Elements in Series and in Parallel in Complex Notation.
Follow the link: Example 41.48.
21. Complex Circuit Analysis of a Pure Inductor Circuit.
Follow the link: Exercise 41.7.6.1.
22. Complex Circuit Analysis of a Pure Capacitor Circuit.
Follow the link: Exercise 41.7.6.2.

Transformers

28. Current and Voltage in Primary and Secondary Circuits of a Transformer.
Follow the link: Example 41.56.
29. Power Adapter for a Dryer and Current Through the Dryer.
Follow the link: Exercise 41.8.2.1.
30. Primary and Secondary of a Power Adapter for Driving Speakers.
Follow the link: Exercise 41.8.2.2.
31. Energy Loss in Power Transmission Cables for Different Voltage Supply Lines.
Follow the link: Exercise 41.8.2.3.

Miscellaneous

33. RC Circuit as a Low Pass Filter.
Follow the link: Example 41.23.
34. Wien Bridge Circuit.
RC oscillators commonly use a Wein bridge circuit shown in Figure 41.57. A galvanometer G is connected between the points marked A and B to detect any current flow between A and B. The Wein bridge circuit is said to be balanced when no current flows between A and B. Prove that the following relations must hold true when the circuit is balanced.
\begin{align*} \amp \frac{R_1}{R_2} + \frac{C_1}{C_2} = \frac{R_3}{R_4},\\ \amp \omega = \frac{1}{\sqrt{R_1 R_2 C_1 C_2}}. \end{align*}
Figure 41.57. Wien Bridge.
Hint.
Replace passive elements by their complex impedances.
Solution.
Let us replace the resistors and capacitors by their complex impedances. Suppose current \(I_1\) flows in the upper branch and \(I_2\) in the lower branch - note that the current through G is zero at the balance. Now, we write the Kirchhoff’s loop rules for the loop on the left of the galvanometer G and the loop on the right of G for the condition that there be no current through G. This
\begin{align*} \amp I_1 \left( R_1 - \dfrac{i}{\omega C_1} \right) = I_2 R_2,\\ \amp I_1 \left[\dfrac{R_2\times \frac{-i}{\omega C_2}}{R_2 - \frac{i}{\omega C_2}} \right] = I_2 R_2. \end{align*}
Dividing out \(I_1\) and \(I_2\) gives the following complex equation.
\begin{equation*} \dfrac{(\omega R_1 C_1 -i)(\omega R_2 C_2 - i)}{-i\omega C_1 R_2} = \dfrac{R_3}{R_4}. \end{equation*}
Separating out the real and imaginary parts of this equation gives
\begin{align*} \amp \omega^2 R_1 C_1 R_2 C_2 -1 = 0,\\ \amp \omega(R_1 C_1 + R_2 C_2) = \dfrac{R_3}{R_4}\:R_2 \omega C_2. \end{align*}
These equations yield the results we seek.
35. Crossover Circuit.
A cross-over circuit is used deliver power to two different parts of the circuit depending on the frequency of the driving signal.
For instance one would want more power be delivered to woofer at low frequency and to tweeter at high frequency. For a given \(L\text{,}\) \(C\text{,}\) \(R_1\) and \(R_2\) in the circuit in Figure 41.58, find the cross-over frequency such that below that frequency more power goes to \(R_2\) and above that frequency more power goes to \(R_1\text{.}\)
Figure 41.58. Crossover Circuit.
Hint.
Use complex analysis.
Answer.
See solution.
Solution.
Let \(Z_1\) and \(Z_2\) be the complex impedances of the upper and lower branches respectively.
\begin{align*} \amp Z_1 = R_1 - \dfrac{i}{\omega C},\ \ |Z_1| = \sqrt{R_1^2 + \dfrac{1}{\omega^2 C^2}},\\ \amp Z_2 = R_2 - i \omega L,\ \ |Z_2| = \sqrt{R_2^2 + \omega^2 L^2}. \end{align*}
The amplitudes of the currents in the two branches will be
\begin{align*} \amp I_{10} = \dfrac{V_0}{|Z_1|},\\ \amp I_{20} = \dfrac{V_0}{|Z_2|}. \end{align*}
Equating the power in the two resitors will give us a condition for the cross-over frequency \(\omega\text{.}\)
\begin{equation*} \dfrac{1}{2}I_{10}^2\:R_1 = \dfrac{1}{2}I_{20}^2\:R_2. \end{equation*}
Therefore,
\begin{equation*} \dfrac{R_1}{|Z_1|^2} = \dfrac{R_2}{|Z_2|^2}. \end{equation*}
This gives a quartic equation in \(\omega\text{,}\) which student is encouraged to solve. I found the following answer for the positive real \(\omega\text{.}\)
\begin{align*} \omega = \amp \dfrac{1}{\sqrt{2}\;LC}\left[\left(R_1R_2 - R_2^2\right)C^2 + \sqrt{\left(R_2^2-R_1R_2\right)^2C^4 + \dfrac{4 R_2L^2C^2}{R_1}} \right]. \end{align*}
36.
An AC generator is connected to a device whose internal circuits are not known.
We only know current and voltage outside the device as shown in the Figure. Based on the information given what can you infer about the electrical nature of the device and its power usage.
Figure 41.59.
Solution.
From the given current in the circuit and the voltage of the source we can determine the amplitude and phase of the net impedance of the device. Write \(Z = |Z|\angle\phi\text{.}\) Then we have
\begin{align*} \amp |Z| = \dfrac{V_0}{I_0} = \dfrac{170\:\textrm{V}}{0.5\:\textrm{A}} = 340\:\Omega,\\ \amp \phi = -\dfrac{\pi}{4}. \end{align*}
Since the phase of the impedance is negative, the device is a net capacitative device. The power into the device will be
\begin{equation*} P_{\textrm{ave}} = \dfrac{1}{2}I_0 V_0 \cos\phi = 30\:\textrm{W}. \end{equation*}
37. Impedance Matching in a DC Circuit.
In many instances we are interested in connecting two circuits such that maximum power is delivered from one to the other. Consider circuit shown in the Figure.
For the purpose of this problem we think of DC power supply V and the resistor \(R_1\) as Circuit \# 1 and resistor \(R_2\) as Circuit \#2. For fixed values of \(V\) and \(R_1\text{,}\) find the value of \(R_2\) so that power delivered to \(R_2\) is greatest. Note circuits 1 and 2 are open and when they are connected they form a closed circuit.
Figure 41.60.
Solution.
Let \(I\) be the current through the resistor \(R_2\text{.}\) From the loop equation we obtain
\begin{equation*} I = \dfrac{V}{R_1 + R_2}. \end{equation*}
The power \(P\) dissipated in \(R_2\) will be
\begin{equation*} P = I^2R = \left(\dfrac{V}{R_1 + R_2}\right)^2\: R_2. \end{equation*}
We now treat this \(P\) as a function of \(R_2\) to find the value of \(R_2\) for which \(P\) will be a maximum.
\begin{equation*} \dfrac{dP}{dR_2} = 0,\ \ \Longrightarrow\ \ \dfrac{V^2}{(R_1 + R_2)^2} - \dfrac{2V^2 R_2}{(R_1 + R_2)^3} = 0. \end{equation*}
This gives the condition \(R_2 = R_1\text{.}\)
38. Impedance Matching in an AC Circuit.
To study the impedance matching in an AC circuit consider circuit shown in Figure 41.61. For the purpose of this problem we think of the AC power supply \(V_0\; \cos(\omega t)\) and circuit elements whose equivalent impedance is \(Z_1\) as Circuit no. 1 and circuit elements whose impedance is \(Z_2\) as Circuit no. 2.
For fixed values of \(V_0\text{,}\) \(\omega\text{,}\) and \(Z_1\text{,}\) find the value of \(Z_2\) so that the average power delivered to \(Z_2\) is greatest. Note circuits 1 and 2 are open and when they are connected they form a closed circuit.
Figure 41.61.
Hint.
This problem will formally be identical to the last problem if done using complex number representation.
39. Matching Impedance of an Ausio Speaker.
An output impedance of an audio amplifier has an impedance of 500 \(\Omega\) and has a mismatch with a low impedance 8 \(\Omega\) loudspeaker. You are asked to insert an appropriate transformer to match the impedances. What turns ratio will you use and why? Use the following simplified circuit. Hint: Assume \(N_p\) and \(N_s\) and derive current in primary, \(I_p = V_p/[(N_p/N_s)^2 R_{\textrm{speaker}}]\text{.}\)
Figure 41.62.
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