AC Network Analysis
Network Analysis Techniques
Don’t just sit there! Build something!! 
Learning to mathematically analyze circuits requires much study and practice. Typically, students practice by working through lots of sample problems and checking their answers against those provided by the textbook or the instructor. While this is good, there is a much better way.
You will learn much more by actually building and analyzing real circuits, letting your test equipment provide the “answers” instead of a book or another person. For successful circuitbuilding exercises, follow these steps:
 Carefully measure and record all component values prior to circuit construction.
 Draw the schematic diagram for the circuit to be analyzed.
 Carefully build this circuit on a breadboard or other convenient medium.
 Check the accuracy of the circuit’s construction, following each wire to each connection point, and verifying these elements onebyone on the diagram.
 Mathematically analyze the circuit, solving for all voltage and current values.
 Carefully measure all voltages and currents, to verify the accuracy of your analysis.
 If there are any substantial errors (greater than a few percent), carefully check your circuit’s construction against the diagram, then carefully recalculate the values and remeasure.
For AC circuits where inductive and capacitive reactances (impedances) are a significant element in the calculations, I recommend high quality (highQ) inductors and capacitors, and powering your circuit with low frequency voltage (powerline frequency works well) to minimize parasitic effects. If you are on a restricted budget, I have found that inexpensive electronic musical keyboards serve well as “function generators” for producing a wide range of audiofrequency AC signals. Be sure to choose a keyboard “voice” that closely mimics a sine wave (the “panflute” voice is typically good), if sinusoidal waveforms are an important assumption in your calculations.
As usual, avoid very high and very low resistor values, to avoid measurement errors caused by meter “loading”. I recommend resistor values between 1 kΩ and 100 kΩ.
One way you can save time and reduce the possibility of error is to begin with a very simple circuit and incrementally add components to increase its complexity after each analysis, rather than building a whole new circuit for each practice problem. Another timesaving technique is to reuse the same components in a variety of different circuit configurations. This way, you won’t have to measure any component’s value more than once.
This phaseshifting bridge circuit is supposed to provide an output voltage with a variable phase shift from 45^{o} (lagging) to +45^{o} (leading), depending on the position of the potentiometer wiper:

Suppose, though, that the output signal is stuck at 45^{o} lagging the source voltage, no matter where the potentiometer is set. Identify a likely failure that could cause this to happen, and explain why this failure could account for the circuit’s strange behavior.
Complex number arithmetic makes possible the analysis of AC circuits using (almost) the exact same Laws that were learned for DC circuit analysis. The only bad part about this is that doing complexnumber arithmetic by hand can be very tedious. Some calculators, though, are able to add, subtract, multiply, divide, and invert complex quantities as easy as they do scalar quantities, making this method of AC circuit analysis relatively easy.
This question is really a series of practice problems in complex number arithmetic, the purpose being to give you lots of practice using the complex number facilities of your calculator (or to give you a lot of practice doing trigonometry calculations, if your calculator does not have the ability to manipulate complex numbers!).
Addition and subtraction:
(5 + j6) + (2 − j1) =  (10 − j8) + (4 − j3) =  (−3 + j0) + (9 − j12) = 
(3 + j5) − (0 − j9) =  (25 − j84) − (4 − j3) =  (−1500 + j40) + (299 − j128) = 
(25 ∠15^{o}) + (10 ∠74^{o}) =  (1000 ∠43^{o}) + (1200 ∠−20^{o}) =  (522 ∠71^{o}) − (85 ∠30^{o}) = 
Multiplication and division:
(25 ∠15^{o}) ×(12 ∠10^{o}) =  (1 ∠25^{o}) ×(500 ∠−30^{o}) =  (522 ∠71^{o}) ×(33 ∠9^{o}) = 
[(10 ∠−80^{o})/(1 ∠0^{o})] =  [(25 ∠120^{o})/(3.5 ∠−55^{o})] =  [(−66 ∠67^{o})/(8 ∠−42^{o})] = 
(3 + j5) ×(2 − j1) =  (10 − j8) ×(4 − j3) =  [((3 + j4))/((12 − j2))] = 
Reciprocation:
[1/((15 ∠60^{o}))] =  [1/((750 ∠−38^{o}))] =  [1/((10 + j3))] = 
[1/([1/(15 ∠45^{o})] + [1/(92 ∠−25^{o})])] =  [1/([1/(1200 ∠73^{o})] + [1/(574 ∠21^{o})])] =  [1/([1/(23k ∠−67^{o})] + [1/(10k ∠−81^{o})])] = 
[1/([1/(110 ∠−34^{o})] + [1/(80 ∠19^{o})] + [1/(70 ∠10)])] =  [1/([1/(89k ∠−5^{o})] + [1/(15k ∠33^{o})] + [1/(9.35k ∠45)])] =  [1/([1/(512 ∠34^{o})] + [1/(1k ∠−25^{o})] + [1/(942 ∠−20)] + [1/(2.2k ∠44^{o})])] = 
A remote speaker for an audio system is connected to the amplifier by means of a long, 2conductor cable:

This system may be schematically modeled as an AC voltage source connected to a load resistor:

Suppose we decided to use the 2conductor cable for more than just conveying an audio (AC) signal  we want to use it to carry DC power as well to energize a small lamp. However, if we were to simply connect the DC power source in parallel with the amplifier output at one end, and the lamp in parallel with the speaker at the other, bad things would happen:

If we were to connect the components together as shown above, the DC power source will likely damage the amplifier by being directly connected to it, the speaker will definitely be damaged by the application of significant DC voltage to its coil, and the light bulb will waste audio power by acting as a second (nonaudible) load. Suffice to say, this is a bad idea.
Using inductors and capacitors as “filtering” components, though, we can make this system work:

Apply the Superposition Theorem to this circuit to demonstrate that the audio and DC signals will not interfere with each other as they would if directly connected. Assume that the capacitors are of such large value that they present negligible impedance to the audio signal (Z_{C} ≈ 0 Ω) and that the inductors are sufficiently large that they present infinite impedance to the audio signal (Z_{L} ≈ ∞).
Calculate the impedance value necessary to balance this AC bridge, expressing your answer in both polar and rectangular forms:

Also, describe what sort of device might be appropriate to serve as a “null detector” to indicate when bridge balance has been achieved, and where this device would be connected to in the bridge circuit.
An AC bridge circuit commonly used to make precision measurements of inductors is the MaxwellWien bridge. It uses a combination of standard resistors and capacitors to “balance out” the inductor of unknown value in the opposite arm of the bridge:

Suppose this bridge circuit balances when C_{s} is adjusted to 120 nF and R_{s} is adjusted to 14.25 kΩ. If the source frequency is 400 Hz, and the two fixedvalue resistors are 1 kΩ each, calculate the inductance (L_{x}) and resistance (R_{x}) of the inductor being tested.
Electrical engineers often represent impedances in rectangular form for the sake of algebraic manipulation: to be able to construct and manipulate equations involving impedance, in terms of the components’ fundamental values (resistors in ohms, capacitors in farads, and inductors in henrys).
For example, the impedance of a seriesconnected resistor (R) and inductor (L) would be represented as follows, with angular velocity (ω) being equal to 2 πf:

Using the same algebraic notation, represent each of the following complex quantities:
 Impedance of a single capacitor (C) =
 Impedance of a series resistorcapacitor (R, C) network =
 Admittance of a parallel inductorresistor (L, R) network =
 Admittance of a parallel resistorcapacitor (R, C) network =
Mathematical analysis of the MaxwellWien bridge is as follows:












Note that neither of the two equations solving for unknown quantities (R_{x} = [(R^{2})/(R_{s})] and L_{x} = R^{2} C_{s}) contain the variable ω. What does this indicate about the MaxwellWien bridge?
Suppose we have two equivalent LR networks, one series and one parallel, such that they have the exact same total impedance (Z_{total}):

We may write an equation for the impedance of each network in rectangular form, like this:


Since we are told these two networks are equivalent to one another, with equal impedances, these two expressions in rectangular form must also be equal to each other:

Algebraically reduce this equation to its simplest form, showing how R_{s}, R_{p}, X_{s}, and X_{p} relate.
Challenge question: combine the result of that simplification with the equations solving for scalar impedance of series and parallel networks (Z_{s}^{2} = R_{s}^{2} + X_{s}^{2} for series and Z_{p}^{2} = [1/([1/(R_{p}^{2})] + [1/(X_{p}^{2})])] for parallel) to prove the following transformative equations, highly useful for “translating” a series network into a parallel network and visaversa:


This interesting bridge circuit is a variable phaseshifter. It works best when the excitation frequency is such that X_{C} = R in each arm of the bridge:

Supposing that X_{C} does equal R in each arm of the bridge, and that the potentiometer resistance is sufficiently high to limit current through it to a negligible level (in other words, R_{pot} >> R). Calculate the phase shift of V_{out} with respect to the excitation source voltage when:
 The potentiometer wiper is fully left:
 The potentiometer wiper is fully right:
 The potentiometer wiper is perfectly centered:
This phaseshifting bridge circuit is supposed to provide an output voltage with a variable phase shift from 45^{o} (lagging) to +45^{o} (leading), depending on the position of the potentiometer wiper:

Suppose, though, that there is a solder “bridge” between the terminals of resistor R_{1} on the circuit board. What effect will this fault have on the output of the circuit? Be as complete as you can in your answer.
This phaseshifting bridge circuit is supposed to provide an output voltage with a variable phase shift from 45^{o} (lagging) to +45^{o} (leading), depending on the position of the potentiometer wiper:

Suppose, though, that the output signal is stuck at +45^{o} leading the source voltage, no matter where the potentiometer is set. Identify a likely failure that could cause this to happen, and explain why this failure could account for the circuit’s strange behavior.
This phaseshifting bridge circuit is supposed to provide an output voltage with a variable phase shift from 45^{o} (lagging) to +45^{o} (leading), depending on the position of the potentiometer wiper:

Suppose, though, that the output signal registers as it should with the potentiometer wiper fully to the right, but diminishes greatly in amplitude as the wiper is moved to the left, until there is practically zero output voltage at the fullleft position. Identify a likely failure that could cause this to happen, and explain why this failure could account for the circuit’s strange behavior.
Complex quantities may be expressed in either rectangular or polar form. Mathematically, it does not matter which form of expression you use in your calculations.
However, one of these forms relates better to realworld measurements than the other. Which of these mathematical forms (rectangular or polar) relates more naturally to measurements of voltage or current, taken with meters or other electrical instruments? For instance, which form of AC voltage expression, polar or rectangular, best correlates to the total voltage measurement in the following circuit?

You should know that the line voltage of a threephase, Yconnected, balanced system is always greater than the phase voltage by a factor of √3.

Apply Kirchhoff’s Voltage Law (KVL) to the upper “loop” in this Yconnected alternator schematic to prove how 120 V ∠ 0^{o} and 120 V ∠ 120^{o} makes 208 V. Show the “polarity” marks for each of the voltages as part of your answer.
Calculate all voltage drops in this circuit, expressing your answers in complex (polar) form:

The load resistor’s color code is as follows:
Brown, Black, Black, Brown, Violet
Assume the resistor’s error is 0%. That is, its resistance value is precisely equal to what the “digit” and “multiplier” color bands declare. The signal generator’s output is 25 volts RMS, at a frequency of 2 kHz.
Challenge question: what practical function does this circuit perform?
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