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 = I^{2}R and P = V^{2}/R are well known equations for
power consumption in DC circuits. Similar equations apply for AC circuits:

P = I_{rms}^{2}R

P = V_{rms}^{2}/R

P in the above two equations refers to average power. The rms subscript is
usually left off so that the equations P = I^{2}R and P =
V^{2}/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 = I^{2}X

Q = V^{2}/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 P_{x} 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 = θ_{v} - θ_{i}

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.

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.

The relationships are just what you'd expect from a triangle:

S^{2} = P^{2} + Q^{2}

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

=I^{2}R

=V_{R}^{2}/R

where cosθ is the power factor (pf).

Reactive power in AC circuits is given by these equations:

Q = EIsinθ VARs

= I^{2}X

= V_{x}^{2}/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).

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

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:

Here's how we calculate the triangle values:

S = P/pf = 20×10^{3}/.6 = 33.3 kVA

Q = √S^{2} -
P^{2} = 26.7 kVARS

Recalling that S = EI, we can find I, the motor current:

I = S/E = 33.3 × 10^{3}/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 = I^{2}R = (278)^{2} × .05 = 3.86 kW

Q = I^{2}X = (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:

P_{T }= P_{motor} + P_{line} = 20 kW + 3.86 kW = 23.86
kW

Q_{T} = Q_{motor} + Q_{line} = 26.7 kVAR + 23.19 kVAR =
48.89 kVAR

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

S is found from S^{2} = P^{2} + Q^{2}. 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:

E_{g} = S/I = 55.3×10^{3}/278 = 199 volts

In the above example, a 22 kVAR capacitor is placed across the motor to
partially correct the pf. Find the new values of E_{g}, 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.

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:

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 = 20.54×10^{3}/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 = I^{2}R = 1.47 kW

Q = I^{2}X = 8.79 kVAR

Now we apply conservation of energy to get the totals:

P_{T} = 20 kW + 1.47 kW = 21.5 kW

Q_{T} = 4.7 + 8.79 = 13.5 kVAR

This allows us to draw the composite power triangle.

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

E_{g} = S/I = 25.4×10^{3}/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.

- Draw a power triangle for each device.

The devices on the right have negative values of Q. - Find the total real power by adding all the Ps.
- Find the total reactive power by adding all the Qs.
- Make a new power triangle using the total P and the total Q.
- The total current that flows can be found from S = EI.
- 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**.

(*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.

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).

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.

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.

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.

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.

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 (V_{rms} × I_{rms}) 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:

- Setting the demand charge based on apparent power rather than real power, or
- Rewarding the customer by lowering the demand charge for power factor above a specified point and penalizing the customer by raising the demand charge for power factor below that specified point.

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.

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

- Base Service Charge: This is a basic monthly fee of perhaps $300.
- Demand charge
- Power factor adjustment to demand charge

**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.

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

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.* R_{L} =
R_{TH}). For complex circuits, the result is similar:

Z** _{L}** = Z

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.

For Z** _{L}** to consume the most
power, it must have an impedance of (8 - j7)* = 8 + j7 Ω.

Now let's calculate the power consumed in Z** _{L}**. 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:

I_{rms} = I_{m}/√2 = 6.25/√2 = 4.42 A

We then calculate the power in the load as follows:

P = I_{rms}^{2}R_{L} = (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 **Z _{L}** as this
would give us reactive power.

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