Series and Parallel AC Circuits
AC Electric Circuits
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.
Draw a phasor diagram showing the trigonometric relationship between resistance, reactance, and impedance in this series circuit:

Show mathematically how the resistance and reactance combine in series to produce a total impedance (scalar quantities, all). Then, show how to analyze this same circuit using complex numbers: regarding each of the component as having its own impedance, demonstrating mathematically how these impedances add up to comprise the total impedance (in both polar and rectangular forms).
A student is asked to calculate the phase shift for the following circuit’s output voltage, relative to the phase of the source voltage:

He recognizes this as a series circuit, and therefore realizes that a right triangle would be appropriate for representing component impedances and component voltage drops (because both impedance and voltage are quantities that add in series, and the triangle represents phasor addition):

The problem now is, which angle does the student solve for in order to find the phase shift of V_{out}? The triangle contains two angles besides the 90^{o} angle, Θ and Φ. Which one represents the output phase shift, and more importantly, why?
In this circuit, a series resistorcapacitor network creates a phaseshifted voltage for the “gate” terminal of a powercontrol device known as a TRIAC. All portions of the circuit except for the RC network are “shaded” for deemphasis:

Calculate how many degrees of phase shift the capacitor’s voltage is, compared to the total voltage across the series RC network, assuming a frequency of 60 Hz, and a 50% potentiometer setting.
Draw a phasor diagram showing the trigonometric relationship between resistance, reactance, and impedance in this series circuit:

Show mathematically how the resistance and reactance combine in series to produce a total impedance (scalar quantities, all). Then, show how to analyze this same circuit using complex numbers: regarding component as having its own impedance, demonstrating mathematically how these impedances add up to comprise the total impedance (in both polar and rectangular forms).
Determine the total current and all voltage drops in this circuit, stating your answers the way a multimeter would register them:

 L_{1} = 250 mH
 L_{2} = 60 mH
 R_{1} = 6.8 kΩ
 R_{2} = 1.2 kΩ
 V_{supply} = 13.4 V RMS
 f_{supply} = 6.5 kHz
Also, calculate the phase angle (Θ) between voltage and current in this circuit, and explain where and how you would connect an oscilloscope to measure that phase shift.
One way to vary the amount of power delivered to a resistive AC load is by varying another resistance connected in series:

A problem with this power control strategy is that power is wasted in the series resistance (I^{2}R_{series}). A different strategy for controlling power is shown here, using a series inductance rather than resistance:

Explain why the latter circuit is more powerefficient than the former, and draw a phasor diagram showing how changes in L_{series} affect Z_{total}.
A quantity sometimes used in DC circuits is conductance, symbolized by the letter G. Conductance is the reciprocal of resistance (G = ^{1}/_{R}), and it is measured in the unit of siemens.
Expressing the values of resistors in terms of conductance instead of resistance has certain benefits in parallel circuits. Whereas resistances (R) add in series and “diminish” in parallel (with a somewhat complex equation), conductances (G) add in parallel and “diminish” in series. Thus, doing the math for series circuits is easier using resistance and doing math for parallel circuits is easier using conductance:

In AC circuits, we also have reciprocal quantities to reactance (X) and impedance (Z). The reciprocal of reactance is called susceptance (B = ^{1}/_{X}), and the reciprocal of impedance is called admittance (Y = ^{1}/_{Z}). Like conductance, both these reciprocal quantities are measured in units of siemens.
Write an equation that solves for the admittance (Y) of this parallel circuit. The equation need not solve for the phase angle between voltage and current, but merely provide a scalar figure for admittance (in siemens):

Determine the total current and all component currents in this circuit, stating your answers the way a multimeter would register them:

 L_{1} = 1.2 H
 L_{2} = 650 mH
 R_{1} = 33 kΩ
 R_{2} = 27 kΩ
 V_{supply} = 19.7 V RMS
 f_{supply} = 4.5 kHz
Also, calculate the phase angle (Θ) between voltage and current in this circuit, and explain where and how you would connect an oscilloscope to measure that phase shift.
An AC electric motor operating under loaded conditions draws a current of 11 amps (RMS) from the 120 volt (RMS) 60 Hz power lines. The measured phase shift between voltage and current for this motor is 34^{o}, with voltage leading current.
Determine the equivalent parallel combination of resistance (R) and inductance (L) that is electrically equivalent to this operating motor.
Doorbell circuits connect a small lamp in parallel with the doorbell pushbutton so that there is light at the button when it is not being pressed. The lamp’s filament resistance is such that there is not enough current going through it to energize the solenoid coil when lit, which means the doorbell will ring only when the pushbutton switch shorts past the lamp:

Suppose that such a doorbell circuit suddenly stops working one day, and the home owner assumes the power source has quit since the bell will not ring when the button is pressed and the lamp never lights. Although a dead power source is certainly possible, it is not the only possibility. Identify another possible failure in this circuit which would result in no doorbell action (no sound) and no light at the lamp.
Calculate the total impedance offered by these two inductors to a sinusoidal signal with a frequency of 60 Hz:

Show your work using two different problemsolving strategies:
 Calculating total inductance (L_{total}) first, then total impedance (Z_{total}).
 Calculating individual impedances first (Z_{L1} and Z_{L2}), then total impedance (Z_{total}).
Do these two strategies yield the same total impedance value? Why or why not?
Calculate the total impedance offered by these two inductors to a sinusoidal signal with a frequency of 120 Hz:

Show your work using three different problemsolving strategies:
 Calculating total inductance (L_{total}) first, then total impedance (Z_{total}).
 Calculating individual admittances first (Y_{L1} and Y_{L2}), then total admittance (Y_{total}), then total impedance (Z_{total}).
 Using complex numbers: calculating individual impedances first (Z_{L1} and Z_{L2}), then total impedance (Z_{total}).
Do these two strategies yield the same total impedance value? Why or why not?
Is this circuit’s overall behavior capacitive or inductive? In other words, from the perspective of the AC voltage source, does it “appear” as though a capacitor is being powered, or an inductor?

Now, suppose we take these same components and reconnect them in parallel rather than series. Does this change the circuit’s overall “appearance” to the source? Does the source now “see” an equivalent capacitor or an equivalent inductor? Explain your answer.

Calculate the total impedance offered by these two capacitors to a sinusoidal signal with a frequency of 3 kHz:

Show your work using two different problemsolving strategies:
 Calculating total capacitance (C_{total}) first, then total impedance (Z_{total}).
 Calculating individual impedances first (Z_{C1} and Z_{C2}), then total impedance (Z_{total}).
Do these two strategies yield the same total impedance value? Why or why not?
A student measures voltage drops in an AC circuit using three voltmeters and arrives at the following measurements:

Upon viewing these measurements, the student becomes very perplexed. Aren’t voltage drops supposed to add in series, just as in DC circuits? Why, then, is the total voltage in this circuit only 10.8 volts and not 15.74 volts? How is it possible for the total voltage in an AC circuit to be substantially less than the simple sum of the components’ voltage drops?
Another student, trying to be helpful, suggests that the answer to this question might have something to do with RMS versus peak measurements. A third student disagrees, proposing instead that at least one of the meters is badly out of calibration and thus not reading correctly.
When you are asked for your thoughts on this problem, you realize that neither of the answers proposed thus far are correct. Explain the real reason for the “discrepancy” in voltage measurements, and also explain how you could experimentally disprove the other answers (RMS vs. peak, and bad calibration).
A quantity sometimes used in DC circuits is conductance, symbolized by the letter G. Conductance is the reciprocal of resistance (G = ^{1}/_{R}), and it is measured in the unit of siemens.
Expressing the values of resistors in terms of conductance instead of resistance has certain benefits in parallel circuits. Whereas resistances (R) add in series and “diminish” in parallel (with a somewhat complex equation), conductances (G) add in parallel and “diminish” in series. Thus, doing the math for series circuits is easier using resistance and doing math for parallel circuits is easier using conductance:

In AC circuits, we also have reciprocal quantities to reactance (X) and impedance (Z). The reciprocal of reactance is called susceptance (B = ^{1}/_{X}), and the reciprocal of impedance is called admittance (Y = ^{1}/_{Z}). Like conductance, both these reciprocal quantities are measured in units of siemens.
Write an equation that solves for the admittance (Y) of this parallel circuit. The equation need not solve for the phase angle between voltage and current, but merely provide a scalar figure for admittance (in siemens):

Calculate the total impedance offered by these three resistors to a sinusoidal signal with a frequency of 10 kHz:
 R_{1} = 3.3 kΩ
 R_{2} = 10 kΩ
 R_{3} = 5 kΩ

State your answer in the form of a scalar number (not complex), but calculate it using two different strategies:
 Calculate total resistance (R_{total}) first, then total impedance (Z_{total}).
 Calculate individual admittances first (Y_{R1}, Y_{R2}, and Y_{R3}), then total impedance (Z_{total}).
Calculate the total impedance offered by these three capacitors to a sinusoidal signal with a frequency of 4 kHz:
 C_{1} = 0.1 μF
 C_{2} = 0.047 μF
 C_{3} = 0.033 μF

State your answer in the form of a scalar number (not complex), but calculate it using two different strategies:
 Calculate total capacitance (C_{total}) first, then total impedance (Z_{total}).
 Calculate individual admittances first (Y_{C1}, Y_{C2}, and Y_{C3}), then total impedance (Z_{total}).
Calculate the total impedance offered by these two capacitors to a sinusoidal signal with a frequency of 900 Hz:

Show your work using three different problemsolving strategies:
 Calculating total capacitance (C_{total}) first, then total impedance (Z_{total}).
 Calculating individual admittances first (Y_{C1} and Y_{C2}), then total admittance (Y_{total}), then total impedance (Z_{total}).
 Using complex numbers: calculating individual impedances first (Z_{C1} and Z_{C2}), then total impedance (Z_{total}).
Do these two strategies yield the same total impedance value? Why or why not?
Due to the effects of a changing electric field on the dielectric of a capacitor, some energy is dissipated in capacitors subjected to AC. Generally, this is not very much, but it is there. This dissipative behavior is typically modeled as a seriesconnected resistance:

Calculate the magnitude and phase shift of the current through this capacitor, taking into consideration its equivalent series resistance (ESR):

Compare this against the magnitude and phase shift of the current for an ideal 0.22 μF capacitor.
A technician needs to know the value of a capacitor, but does not have a capacitance meter nearby. In lieu of this, the technician sets up the following circuit to measure capacitance:

You happen to walk by this technician’s workbench and ask, “How does this measurement setup work?” The technician responds, “You connect a resistor of known value (R) in series with the capacitor of unknown value (C_{x}), then adjust the generator frequency until the oscilloscope shows the two voltage drops to be equal, and then you calculate C_{x}.”
Explain how this system works, in your own words. Also, write the formula you would use to calculate the value of C_{x} given f and R.
If the dielectric substance between a capacitor’s plates is not a perfect insulator, there will be a path for direct current (DC) from one plate to the other. This is typically called leakage resistance, and it is modeled as a shunt resistance to an ideal capacitance:

Calculate the magnitude and phase shift of the current drawn by this real capacitor, if powered by a sinusoidal voltage source of 30 volts RMS at 400 Hz:

Compare this against the magnitude and phase shift of the current for an ideal capacitor (no leakage).
Voltage divider circuits may be constructed from reactive components just as easily as they may be constructed from resistors. Take this capacitive voltage divider, for instance:

Calculate the magnitude and phase shift of V_{out}. Also, describe what advantages a capacitive voltage divider might have over a resistive voltage divider.
Determine the total current and all voltage drops in this circuit, stating your answers the way a multimeter would register them:

 C_{1} = 125 pF
 C_{2} = 71 pF
 R_{1} = 6.8 kΩ
 R_{2} = 1.2 kΩ
 V_{supply} = 20 V RMS
 f_{supply} = 950 kHz
Also, calculate the phase angle (Θ) between voltage and current in this circuit, and explain where and how you would connect an oscilloscope to measure that phase shift.
The input impedance of an electrical test instrument is a very important parameter in some applications, because of how the instrument may load the circuit being tested. Oscilloscopes are no different from voltmeters in this regard:

Typical input impedance for an oscilloscope is 1 MΩ of resistance, in parallel with a small amount of capacitance. At low frequencies, the reactance of this capacitance is so high that it may be safely ignored. At high frequencies, though, it may become a substantial load to the circuit under test:

Calculate how many ohms of impedance this oscilloscope input (equivalent circuit shown in the above schematic) will impose on a circuit with a signal frequency of 150 kHz.
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