ECE 252 Introduction to Electrical Engineering

Lesson 12. Power in AC Circuits


Introduction

You have learned the following equation for DC circuits:
P = VI

A more general equation for all circuits is this one:
p(t) = v(t) × i(t)

At any instant in time, this equation will be true. Nevertheless, engineers are usually more interested in average power than in instantaneous power.

P = I2R and P = V2/R are well known equations for power consumption in DC circuits. Similar equations apply for AC circuits:
P = Irms2R
P = Vrms2/R

P in the above two equations refers to average power.  The rms subscript is usually left off so that the equations P = I2R and P = V2/R are understood to use rms values. This will be a general rule: When I and V are used in an AC circuit, they are meant to be rms values.

Similarly, an AC circuit can consume "reactive power," given by these equations:
Q = I2X
Q = V2/X

X is the reactance, as in Z = R + jX. Unlike R, X can take on negative values. So, although P (real power) is always positive, Q (reactive power) may be positive or negative. The reactive power is not in the units of watts, because it is not real power. The units are VARs, which stands for volt-amps reactive. Real power represents the power available to do work. Reactive power can never do any work -- it's just a measure of energy storage in the circuit.

In some books, you will find Q replaced by Px for reactive power.

In a general AC circuit, the angles of the phasor voltage and current will be different; e.g.
V = 120/32° V
I = 5/95° A

It will be useful to find the difference between these two angles:
θ = θv - θi

θ is called the power factor angle.

In this specific case,
θ = 32° - 95° = -63°

Since the voltage and current angles are different, it should not be surprising that the power P is not just the product of V (the rms voltage) and I (the rms current). In fact, this is the relationship:
P = EIcosθ watts

In this equation, E is the same as V, the rms voltage. The cosθ part of the equation is referred to as the power factor. We usually work with the power factor rather than the angles themselves.

For the example above,
P = EIcosθ = (120)(5)cos(-63°) = 272 W
Power factor = .45

In manufacturers' literature, power factor is frequently expressed as a percentage.  In the above case, the power factor would be given as 45%.

The power factor may be leading or lagging. Recall that cosθ = cos(-θ). How do we differentiate between the two? When the current lags the voltage, as it does in an inductor, we get a positive value for θ, and we say that the power factor is lagging. When the current leads the voltage, as it does in a capacitor, we get a negative value for θ, and we say that the power factor is leading.  Sometimes power engineers speak of inductors absorbing magnetizing VARs and capacitors providing magnetizing VARs.

Here's a mnemonic for remembering the leading/lagging relationships for inductors and capacitors:
ELI the ICE man.
The ELI part implies current lags voltage in an inductor. The ICE part implies current leads voltage in a capacitor.

If the load is primarily inductive, Q will be positive. Most motors have lagging power factors because they are made of wire and iron, just like inductors.

L and C modify instantaneous power but do not use any average power. Their presence can, nevertheless, change the amount of average (real) power consumed in a circuit.

1 Horsepower = 745.7 watts, but most engineers use 746 W as the equivalent of 1 HP.

Motors and generators are not 100% efficient. In a motor, for example, the input electrical power is always reduced by windage, friction, and ohmic loss before it can appear as output mechanical power. If a motor has an efficiency of 89%, for example, and puts out 5 HP (mechanical), the input power (electrical) is 5 HP × 745.7 W per HP/.89 = 4189 W.

The Volt-Ampere Method

Apparent power is defined as:
S = EI

E and I are rms values of voltage and current. S is called the apparent power because it is the power that would result if the power factor angle were zero. The units are volt-amps (not watts). We now have three sets of units: watts for real power, VARs for reactive power, and volt-amps for a combination of the two.

The relationship among P, Q, and S is best understood with the power triangle, shown below.

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The relationships are just what you'd expect from a triangle:
S2 = P2 + Q2
P = Scosθ
Q = Ssinθ

This triangle clearly shows why apparent power (S) is not the same as real power. You will use this triangle repeatedly to solve power problems.

Real power in AC circuits is given by these equations:
P = EIcosθ watts
=I2R
=VR2/R
where cosθ is the power factor (pf).

Reactive power in AC circuits is given by these equations:
Q = EIsinθ VARs
= I2X
= Vx2/X
= Ptanθ = Ssinθ

X is positive for inductors and negative for capacitors.

kVA = kilo-volt-amps

By conservation of energy, total power consumed in a circuit is equal to the sum of the power consumed by each element. Also, power generated equals power consumed. Likewise, reactive power generated equals reactive power consumed.

Advice: ALWAYS DRAW A WELL-LABELED DIAGRAM SHOWING THE POWER TRIANGLE(S).

Example 1

In the example below, we consider the case of an AC generator feeding a motor through a transmission line.

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One of the first things to notice is that the motor is not specified in terms of its impedance, but rather in the way it uses power. This is a 20 kW motor. This means that it uses 20 kW of real power. It has a power factor of 0.6. We could use this to calculate θ, but we will see that finding the angle is not necessary.

The task is to find the motor current, then find the voltage, power, and power factor of the generator.

We begin with a power triangle:

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Here's how we calculate the triangle values:
S = P/pf = 20×103/.6 = 33.3 kVA
Q = √S2 - P2 = 26.7 kVARS

Recalling that S = EI, we can find I, the motor current:
I = S/E = 33.3 × 103/120 = 278 A

The motor current is the same as the transmission line current, so we can make the following calculations for losses in the transmission line:
P = I2R = (278)2 × .05 = 3.86 kW
Q = I2X = (278)2 × .3 = 23.19 kVAR

Now we will use conservation of energy to determine the net power used in the motor/transmission line combination:
PT = Pmotor + Pline = 20 kW + 3.86 kW = 23.86 kW
QT = Qmotor + Qline = 26.7 kVAR + 23.19 kVAR = 48.89 kVAR

With these two values, we can now draw another power triangle for the whole circuit.

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S is found from S2 = P2 + Q2. The power factor is just P/S:
pf = P/S = 23.86/55.3 = .431 lagging

We know the power factor is lagging, because the load is inductive.

Finally, we'll find the generator voltage:
Eg = S/I = 55.3×103/278 = 199 volts

Example 2

In the above example, a 22 kVAR capacitor is placed across the motor to partially correct the pf.  Find the new values of Eg, pf, and P necessary for the generator.

Before beginning this solution, notice that the capacitor is specified in kVAR, not in microfarads. This is common practice in industrial applications. Catalogs of industrial capacitors will specify the kVAR and voltage rating of the capacitor. Recall that Q is positive for inductors and negative for capacitors. Nevertheless, the capacitor is specified as 22 kVAR and not -22 kVAR. This is because the engineer using industrial catalogs is expected to know that Q is negative for capacitors, and therefore using the minus sign is superfluous. You, the student, will also be expected to know this. In the diagram below, the actual (negative) value is used. Warning: This may not always be the case with other diagrams.

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From Example 1, we know that for the motor,
Q = +26.7 kVAR

Therefore the total Q for the motor and capacitor together is,
Q = 26.7 - 22 = 4.7 kVAR

This results in the following power triangle:

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Notice that the power of the motor did not change as a result of adding the capacitor. Now, we'll calculate the current in the line:
I = S/E = 20.54×103/120 = 171 amps

Before going on, let's note that the line current (171 A) is less than the motor current (278 A). This occurs because the capacitor has corrected the power factor to be more nearly unity.

For the transmission line:
P = I2R = 1.47 kW
Q = I2X = 8.79 kVAR

Now we apply conservation of energy to get the totals:
PT = 20 kW + 1.47 kW = 21.5 kW
QT = 4.7 + 8.79 = 13.5 kVAR

This allows us to draw the composite power triangle.

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pf = P/S = 21.5/25.4 = 0.85 lagging

Eg = S/I = 25.4×103/171 = 149 volts

Notice that addition of the capacitor has allowed us to reduce the generator voltage. This is a direct result of improving the power factor. The electric power utility now needs to provide lower voltage and lower current without changing the amount of power delivered to the customer.

Procedure for Solving Power Problems

  1. Draw a power triangle for each device.
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    The devices on the right have negative values of Q.
  2. Find the total real power by adding all the Ps.
  3. Find the total reactive power by adding all the Qs.
  4. Make a new power triangle using the total P and the total Q.
  5. The total current that flows can be found from S = EI.
  6. The resulting power factor is P/S, but note whether the power factor is lagging (positive Q) or leading (negative Q).

Before going on, you should complete Tutorial 12 on AC power.


Electric Utility Rate Structures

(The author acknowledges the assistance of Howard Bush Of Kentucky Utilities and Bob White of Louisville Gas and Electric in the preparation of this section.)

Electric utility rates can be very complex, especially for industrial customers, but even residential users may be subject to some of these complications.

Energy Charge

The energy charge is just the rate for the number of kWh of energy that are used. A typical value for residential users is 10˘/kWh. For example, if you used 650 kWh in a billing period (usually a month), you would pay 10˘ × 650 = $65 for your electricity use. The energy charge for commercial and (especially) industrial users will probably be much lower.

Users of electricity may also be charged in "tiers." That is, they may be charged one price for the first X kWh, and another price for energy use above X. This new price may be lower (if the utility wants to encourage further energy use for market reasons) or higher (if the utility wants to dicourage further energy use for environmental reasons).

Base Charge

The base charge may also be called a "service charge," a "connection charge," or an "administrative charge." This is a fixed charge that is independent of the amount of electricity used. For a residential user, it may be about $9/month. Commercial and industrial customers pay a lot more. Ocasionally, the base charge is just a minimum charge; for example, if the base charge is $12, but you use $130 of electricity, your total bill may be just the $130; the $12 may only kick in if you use less that $12 worth of electricity over the month.

Time of Day Rate

The charge for energy, peak power use, or both may vary depending upon the time of day. The utility may want to discourage energy use during peak demand periods, which may typically be from noon to 6 PM. The peak-time-of-day rate may be considerably more than the normal energy charge or demand charge (see below). Conversely, the utility may want to encourage energy use in off-peak hours, say, between midnight and 5 AM. For this period of time, the energy charge or demand charge may be considerably lower.

Seasonal Rate

In most markets, electrical energy use peaks in the summer, when air conditioning is in heavy use, and is at a minimum in fall and spring when little electricity is used for air conditioning or heat. The energy charge (and/or demand charge) may be seasonally adjusted to reflect this change in demand.

Demand Charge

Commercial and (especially) industrial customers are subject to a "demand charge." Whatever the industry, power use varies throughout the month. Heavy users of electricity, such as foundries, require lots of electrical infrastructure from their electrical utility provider: transformers, switch gear, and conductors, to name a few. The utility must provide the necessary equipment for the maximum power the user requires, even if that power is only required for a short period of time. Therefore, the utility assesses a "demand charge." The utility does this by averaging the power use over a period of time, typically 15 to 30 minutes. The demand charge is assessed based on the customer's highest power use in any one of these blocks.

Power Factor

Industrial and (frequently) commercial customers must also be concerned about power factor. Most industries make heavy use of motors. Motors are made of iron and wire and have a significant inductive component. These motors produce a "lagging power factor." This means that the current lags the voltage. Since the current and voltage are out of phase, a motor will use more apparent power (Vrms × Irms) in kilovolt-amps (kVA) than real power in kilowatts (kW). The electrical utility must size its distribution equipment for the apparent power, not the real power. Therefore, the utility will penalize the customer by charging a higher demand charge if the power factor is different from one (the perfect power factor where apparent power = real power). This is frequently done in one of two ways:

Therefore, it is in the best interest of the customer to get the power factor as close to unity as possible. This is typically done by adding capacitor banks to offset the effect of the inductive motors.

Residential Users

It's simple for residential users of electric power. They pay about 10˘ per kWh. Note that this is an energy charge, not a power charge. In a few markets, residential users may be charged at a different rate dependent on the time of day or season of year. Typically, there is also a base charge.

Commercial Users

Commercial users, such as banks and shops, may pay a lower energy charge (perhaps 4˘/kWh) than residential users, but they may also pay a significant base charge plus a maximum load charge (demand charge). They may also be required to keep the power factor above some specified minimum. Some markets have a demand charge based on apparent power in kVA rather than real power in kW.

Industrial Users

Industrial customers pay the lowest energy charge of all, perhaps 3˘/kWh or less. Instead they may pay in many other ways:

Industrial User Example:
The Ferrous Iron Works pays a base charge of $325/month for electric service. The energy charge is 3˘/kWh, and the customer uses 150,000 kWh of energy. The demand charge is $4.50/kVA, and the customer's peak apparent power use is 340 kVA.

Base Charge: $325
Energy Charge: 150,000 kWh × 3˘/kWh = $4,500
Demand Charge: 340 kVA × $4.50 = $1,530

Total bill for the month = $325 + $4,500 + $1,530 = $6,355

This calculation does not consider differences in price based on time of day, day of week, season of year, or tiered pricing.

Other Users

The electric utility may also have rate structures for other specific users, such as farms and outdoor lighting.


Maximum Power Transfer in Complex Circuits

We learned earlier that we could maximize the power consumed in a resistive load if we set the load resistance equal to the Thevenin resistance of the circuit the load was connected to, (i.e. RL = RTH).  For complex circuits, the result is similar:

ZL = ZTH*

The asterisk (*) indicates complex conjugate.  To find the complex conjugate of a complex number, change the sign of the imaginary part (if in rectangular form) or change the sign of the angle (if in polar form).  Below are two examples.

(5 + j6)* = (5 - j6)
20/-35° * = 20/35°

Consider the circuit below.

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For ZL to consume the most power, it must have an impedance of (8 - j7)* = 8 + j7 Ω.

Now let's calculate the power consumed in ZL.  Note that the total impedance seen by the voltage source is (8 - j7) + (8 + j7) = 16 Ω .  The impedance seen by the voltage source is totally real; the imaginary parts cancel out.  We can now calculate the current:

I = 100/25° /16 = 6.25/25° A

To calculate the power consumed in the load, we just want the magnitude of I (i.e. 6.25 A).  Recall that phasor currents are based on the maximum value of the current, not the rms value.  To calculate power, we need the rms value:

Irms = Im/√2 = 6.25/√2 = 4.42 A

We then calculate the power in the load as follows:

P = Irms2RL = (4.42)2 × 8 = 156.25 W

Note that we use only the real part of the load impedance in this calculation.  We do not use the imaginary part of ZL as this would give us reactive power.


At this time you should complete Tutorial 12A on maximum power transfer in complex circuits.