AC Power
AC Electric Circuits
If a sinusoidal voltage is applied to an impedance with a phase angle of 90^{o}, the resulting voltage and current waveforms will look like this:

Given that power is the product of voltage and current (p = i e), plot the waveform for power in this circuit. Also, explain how the mnemonic phrase ËLI the ICE man” applies to these waveforms.
If a sinusoidal voltage is applied to an impedance with a phase angle of 90^{o}, the resulting voltage and current waveforms will look like this:

Given that power is the product of voltage and current (p = i e), plot the waveform for power in this circuit. Also, explain how the mnemonic phrase ËLI the ICE man” applies to these waveforms.
A generator is coupled to a bicycle mechanism, so that a person can generate their own electricity:

The person pedaling this bicycle/generator notices that it becomes more difficult to pedal when the generator is connected to a load such as a light bulb, or when it is charging a battery. When the generator is opencircuited, however, it is very easy to spin. Explain why this is, in terms of work and energy transfer.
If the power waveform is plotted for an AC circuit with a 90 degree phase shift between voltage and current, it will look something like this:

What is the significance of the power value oscillating equally between positive (above the zero line) and negative (below the zero line)? How does this differ from a scenario where there is zero phase shift between voltage and current?
If this circuit is built and operated, it will be found that the resistor becomes much hotter than the inductor, even though both components drop the exact same amount of voltage and carry the exact same amount of current:

Explain why there is such a remarkable difference in heat output between these two components, given their identical voltage drops and currents.
A student is pondering the behavior of a simple series RC circuit:

It is clear by now that the 4 kΩ capacitive reactance does not directly add to the 3 kΩ resistance to make 7 kΩ total. Instead, the addition of impedances is vectorial:



It is also clear to this student that the component voltage drops form a vectorial sum as well, so that 4 volts dropped across the capacitor in series with 3 volts dropped across the resistor really does add up to 5 volts total source voltage:


What surprises the student, though, is power. In calculating power for each component, the student arrives at 4 mW for the capacitor (4 volts times 1 milliamp) and 3 mW for the resistor (3 volts times 1 milliamp), but only 5 mW for the total circuit power (5 volts times 1 milliamp). In DC circuits, component power dissipations always added, no matter how strangely their voltages and currents might be related. The student honestly expected the total power to be 7 mW, but that doesn’t make sense with 5 volts total voltage and 1 mA total current.
Then it occurs to the student that power might add vectorially just like impedances and voltage drops. In fact, this seems to be the only way the numbers make any sense:

However, after plotting this triangle the student is once again beset with doubt. According to the Law of Energy Conservation, total power in must equal total power out. If the source is inputting 5 mW of power total to this circuit, there should be no possible way that the resistor is dissipating 3 mW and the capacitor is dissipating 4 mW. That would constitute more energy leaving the circuit than what is entering!
What is wrong with this student’s power triangle diagram? How may we make sense of the figures obtained by multiplying voltage by current for each component, and for the total circuit?
In this circuit, three common AC loads are modeled as resistances, combined with reactive components in two out of the three cases. Calculate the amount of current registered by each ammeter, and also the amount of power dissipated by each of the loads:

If someone were to read each of the ammeters’ indications and multiply the respective currents by the figure of 120 volts, would the resulting power figures (P = I E) agree with the actual power dissipations? Explain why or why not, for each load.
The three different types of power in AC circuits are as follows:
 •
 S = apparent power, measured in VoltAmps (VA)
 •
 P = true power, measured in Watts (W)
 •
 Q = reactive power, measured in VoltAmps reactive (VAR)
Explain the names of each of these power types. Why are they called äpparent,” “true,” and “reactive”?
Power calculation in DC circuits is simple. There are three formulae that may be used to calculate power:

Calculating power in AC circuits is much more complex, because there are three different types of power: apparent power (S), true power (P), and reactive power (Q). Write equations for calculating each of these types of power in an AC circuit:
In this circuit, three common AC loads are represented as resistances, combined with reactive components in two out of the three cases. Calculate the amount of true power (P), apparent power (S), reactive power (Q), and power factor (PF) for each of the loads:

Also, draw power triangle diagrams for each circuit, showing how the true, apparent, and reactive powers trigonometrically relate.
A common analogy used to describe the different types of power in AC circuits is a mug of beer that also contains foam:

Explain this analogy, relating the quantities of beer and foam to the different types of power in an AC circuit, and also why this analogy is often employed to describe the “desirability” of each power type in a circuit.
If an electrical device is modeled by fixed values of resistance, inductance, and/or capacitance, it is not difficult to calculate its power factor:


In real life, though, things are not so simple. An electric motor will not come labeled with an idealcomponent model expressed in terms of R and L. In fact, that would be impossible, as the resistance R in the circuit model represents the sum total of mechanical work being done by the motor in addition to the energy losses. These variables change depending on how heavily loaded the motor is, meaning that the motor’s power factor will also change with mechanical loading.
However, it may be very important to calculate power factor for electrical loads such as multithousand horsepower electric motors. How is this possible to do when we do not know the equivalent circuit configuration or values for such a load? In other words, how do we determine the power factor of a real electrical device as it operates?

Of course, there do exist special meters to measure true power (wattmeters) and reactive power (“var” meters), as well as power factor directly. Unfortunately, these instruments may not be readily available for our use. What we need is a way to measure power factor using nothing more than standard electrical/electronic test equipment such as multimeters and oscilloscopes. How may we do this?
Hint: remember that the angle Θ of the SQP “power triangle” is the same as the angle in a circuit’s ZXR impedance triangle, and also the same as the phase shift angle between total voltage and total current.
Suppose that a singlephase AC electric motor is performing mechanical work at a rate of 45 horsepower. This equates to 33.57 kW of power, given the equivalence of watts to horsepower (1 HP ≈ 746 W).
Calculate the amount of line current necessary to power this motor if the line voltage is 460 volts, assuming 100% motor efficiency and a power factor of 1.
Now recalculate the necessary line current for this motor if its power factor drops to 0.65. Assume the same efficiency (100%) and the same amount of mechanical power (45 HP).
What do these calculations indicate about the importance of maintaining a high power factor value in an AC circuit?
An oscilloscope is connected to a lowcurrent AC motor circuit to measure both voltage and current, and plot them against one another as a Lissajous figure:

The following Lissajous figure is obtained from this measurement:

From this figure, calculate the phase angle (Θ) and the power factor for this motor circuit.
A very highpower AC electric motor needs to have its power factor measured. You and an electrician are asked to perform this measurement using an oscilloscope. The electrician understands what must be done to measure voltage and current in this dangerous circuit, and you understand how to interpret the oscilloscope’s image to calculate power factor.
It would be impractical to directly measure voltage and current, seeing as how the voltage is 4160 volts AC and the current is in excess of 200 amps. Fortunately, PT (“potential transformer”) and CT (“current transformer”) units are already installed in the motor circuit to facilitate measurements:

After the electrician helps you safely connect to the PT and CT units, you obtain a Lissajous figure that looks like this:

Calculate the power factor of the AC motor from this oscilloscope display.
A large electrical load is outfitted with a wattmeter to measure its true power. If the load voltage is 7.2 kV and the load current is 24 amps, calculate the load’s apparent power (S). Calculate the power factor and also the phase angle between voltage and current in the circuit if the wattmeter registers 155 kW at those same voltage and current values.
Draw a “power triangle” for this circuit, graphically showing the relationships between apparent power, true power, and phase angle.
The power factor of this circuit is as low as it can possibly be, 0:

Calculate the apparent, true, and reactive power for this circuit:
 •
 S =
 •
 P =
 •
 Q =
Now, suppose a capacitor is added in parallel with the inductor:

Recalculate the apparent, true, and reactive power for this circuit with the capacitor connected:
 •
 S =
 •
 P =
 •
 Q =
The “power triangle” is a very useful model for understanding the mathematical relationship between apparent power (S), true power (P), and reactive power (Q):

Explain what happens to the triangle if power factor correction components are added to a circuit. What side(s) change length on the triangle, and what happens to the angle Θ?
When a capacitor is to be connected in parallel with an inductive AC load to correct for lagging power factor, it is important to be able to calculate the reactive power of the capacitor (Q_{C}). Write at least one equation for calculating the reactive power of a capacitor (in VARs) given the capacitor’s reactance (X_{C}) at the line frequency.
An inductive AC load draws 13.4 amps of current at a voltage of 208 volts. The phase shift between line voltage and line current is measured with an oscilloscope, and determined to be 23^{o}. Calculate the following:
 •
 Apparent power (S) =
 •
 True power (P) =
 •
 Reactive power (Q) =
 •
 Power factor =
An electrician suggests to you that the lagging power factor may be corrected by connecting a capacitor in parallel with this load. If the capacitor is sized just right, it will exactly offset the reactive power of the inductive load, resulting in zero total reactive power and a power factor of unity (1). Calculate the size of the necessary capacitor in Farads, assuming a line frequency of 60 Hz.
A dualtrace oscilloscope is used to measure the phase shift between voltage and current for an inductive AC load:

Calculate the following, given a load voltage of 110 volts, a load current of 3.2 amps, and a frequency of 60 Hz:
 •
 Apparent power (S) =
 •
 True power (P) =
 •
 Reactive power (Q) =
 •
 Θ =
 •
 Power factor =
 •
 Necessary parallel C size to correct power factor to unity =
If an AC circuit has a lagging power factor, the way to correct for that power factor is to add a capacitor to the circuit to create leading reactive power. This leading reactive power will cancel the lagging reactive power of the load, ideally negating one another so that there is no reactive power demands placed on the source (zero net reactive power in the circuit):

Define a stepbystep procedure for calculating the size of compensating capacitor needed (in Farads) in order to correct any lagging power factor to a value of unity. Make your procedure general enough that it may apply to any scenario.
Most methods of power factor correction involve the connection of a parallel capacitance to an inductive load:

It is technically possible to correct for lagging power factor by connecting a capacitor in series with an inductive load as well, but this is rarely done:

Explain why series capacitance is not considered a practical solution for power factor correction in most applications.
In AC power systems, a common way of thinking about reactive power among engineers is in terms of production and consumption. Inductive loads, it is said, consume reactive power. Conversely, capacitive loads produce reactive power.
Explain how the models of “production” and “consumption” relate to reactive power in capacitors and inductors, respectively. Being that neither type of component actually dissipates or generates electrical energy, how can these terms be appropriate in describing their behavior?
A very large 3phase alternator at a hydroelectric dam has the following continuous fullpower ratings:
 •
 600 MW power output
 •
 15 kV line voltage
 •
 23.686 kA line current
Calculate the continuous fullload apparent power for this alternator (in MVA), its continuous fullload reactive power (in MVAR), and its power factor (in percent).
Large power distribution centers are often equipped with capacitors to correct for lagging (inductive) power factor of many industrial loads. There is never any one value for capacitance that will precisely correct for the power factor, though, because load conditions constantly change. At first it may seem that a variable capacitor would be the answer (adjustable to compensate for any value of lagging power factor), but variable capacitors with the ratings necessary for power line compensation would be prohibitively large and expensive.
One solution to this problem of variable capacitance uses a set of electromechanical relays with fixedvalue capacitors:

Explain how a circuit such as this provides a stepvariable capacitance, and determine the range of capacitance it can provide.