Although pioneers like Count Alessandro Volta, Andre Ampere, George Ohm, Joseph Henry, and Michael Faraday discovered the basics of electricity centuries ago, these fundamentals are more important today than ever before. Just look around you. Almost every activity you can think of relies on some sort of device powered by electrical energy. Each of us is practically hard-wired to an electrical source of some kind or another. We could even go as far as to say electricity has become a basic human need. With these thoughts in mind, the editorial staff of EC&M presents a review of the fundamentals of electricity. No matter how many advanced degrees or certifications you hold, it's always beneficial to review the basics. In fact, it's these concepts we usually tend to forget. And that's typically what leads to trouble.

Energy sources

We typically define sources of energy as either direct current (DC) or alternating current (AC). DC sources, such as batteries, deliver energy at a nearly constant voltage; whereas AC sources rely on a generator to produce a sinusoidal waveform. AC sources produce most of the energy generated, distributed, and used in the world today, while DC energy sources typically supply power to control circuitry.

Voltage and current

The potential for an electrical system to do work is known as voltage. We define voltage as the work to be done by an electrical system in moving an electric charge (coulomb) from one point to another in a circuit, divided by the charge. We measure voltage in joules per coulomb (J / C), or volts (V). (A joule equals 1W of power applied for 1 sec.)

An electrical conductor has mobile electrons that move in response to electric forces. So we describe current as simply charge in motion. We measure current in coulombs per second (C / sec), or amperes (A). The direction of the physical current in a conductor is opposite to the direction of electron motion. For example, with a battery the physical current flows out of the (`) terminal, through the circuit, and returns into the (1) terminal.

Resistors and resistance

As current flows through an electrical circuit, its path is impeded somewhat by the inherent properties of the elements of the circuit itself (wires, light bulbs, heater elements, and resistors). We refer to this property as resistance (R). Although resistance has the unit volts per ampere (V/A), we commonly recognize it in ohms (W).

The resistance of a piece of wire is directly proportional to its length, to a property of the metal called resistivity, and inversely proportional to its cross-sectional area.

R = l r / a Where "l" is the length of the wire (cm), "r" is the metal resistivity (W/cm), and "a" is the cross-sectional area of the wire (cm*cm). Resistivity is an inherent property of materials. The resistance of a wire also depends on its temperature, but this division is beyond the scope of this article.

A resistor is an electrical component designed to have a certain magnitude of resistance. Some resistors convert electric energy to heat energy. We use others to control circuits to modify electric signals and energy levels (i.e. motor-starting resistors). If you join resistors in series, then the effective resistance is the sum of the individual resistances. If you connect resistors in parallel, the effective resistance is the sum of the reciprocals of individual resistances.

Inductors and inductance

We call a device that stores electromagnetic energy in its magnetic field an inductor. In its basic form, winding a coil of wire (often copper) around a form, which may or may not contain ferromagnetic materials, forms an inductor. When current flows in the wire, moving charges are close together, and magnetic forces are large. This stores magnetic energy.

We call the property of the inductor that's useful in circuit analysis inductance (L). The inductance depends on coil dimensions and the number of turns of wire in the coil. The units of inductance are volt-seconds per ampere (V-sec/A), or more commonly known as the Henry (H). To increase inductance, you can connect inductors in series. Connect inductors in parallel to decrease inductance.

If you wind two coils on the same coil form or they exist in close proximity, then a changing current in one coil induces a voltage in the second coil. This mutual inductance serves as the basis for transformer theory.

Capacitors and capacitance

We define the capacitance between two conductors as the charge stored per unit potential difference between them. The amount of capacitance depends on the area of the conductors, spacing between them, and characteristics of the dielectric material or materials placed between the conductors. You measure the capacitance of a conductor in farads (F) or (mF) for microfarads; where one farad equals an electrical charge of one coulomb that produces 1V potential difference between the terminals of two parallel conductors. You may also see a farad expressed as coulombs per volt (C/V).

Just as inductance opposes any changes in current, capacitance opposes any change in voltage. Capacitance delays an increase or decrease of voltage due to the charge present on the conductors. For the voltage across a capacitor to change, its charge must change too. The charge on a capacitor cannot change instantly. Therefore, the voltage across the capacitor also cannot change instantly.

The capacitor is a device used to store electrons. Its construction consists of two conductive plates separated by a thin layer of insulation: a material called a dielectric. When you apply voltage to the plates, electrons are forced onto one plate and the other becomes deficient in electrons. The plate with excess electrons is negatively charged whereas the plate with a deficiency of electrons is positively charged.

Capacitors hold a specific quantity of electrons. This quantity relates to the area of the plates, distance between the plates, and type of material of the dielectric. A capacitor stores energy during periods of increased voltage as an electrical system follows the 60 Hz sine curve. Energy releases as the voltage decreases.

When you apply alternating voltage to the terminals of a capacitor, the dielectric is subject to cyclic stress based on the AC frequency (60 Hz). If the material were perfectly elastic, you wouldn't lose any energy. What happens is as the applied voltage's amplitude increases, the movement of the molecules in the dielectric also increases, continuing as the applied voltage wanes. This movement causes friction of the molecules, which generates heat. We often refer to this feature as dielectric loss. The energy lost is proportional to the square of the applied voltage. Thus, a capacitor does not return the full amount of energy put into it during discharge.

You can place capacitors in an electrical system with either series and/or parallel connections. Connecting capacitors in series decreases total capacitance. The effect is like increasing the space between the plates. The rule for parallel resistance applies to series-connected capacitors. For example, you calculate the capacitance of three capacitors connected in series as 1/C1 + 1/C2 + 1/C3 = 1/CT.

You determine the capacitance for capacitors connected in a parallel configuration simply by adding the rated capacitance of each capacitor. Thus, for three capacitors connected in parallel, the total capacitance will be C1 + C2 + C3 = CT. An important characteristic of capacitors in an AC circuit is that they cause the current to lead the voltage. In a purely capacitive circuit, the current will lead the voltage by 90°. However, all electrical circuits have some inherent amount of resistance and often a certain amount of inductance. When induction motors are present in a circuit, the voltage may actually lead the current; even if capacitors are in the circuit.

Capacitors reduce the amount of inductance in a circuit (usually caused by induction motors) and reduce harmonics by serving as a filter.


When analyzing AC circuits, we represent voltages and currents as phasors. These phasors are complex numbers representing the amplitudes and phases of the waveforms in the sinusoidal steady state. Impedance refers to the complex number relating the phasor voltage and phasor current of an element in an AC circuit. The complex number has a real (resistive) part and imaginary (reactive) part. Admittance is the reciprocal of impedance.

For an inductor, the ratio of the phasor voltage to the phasor current is j w L. We call this quantity the reactance (X) of the inductor. Its reciprocal is susceptance (B). For a capacitor, the ratio of the phasor voltage to the phasor current is 1 / jwC = - j / wC.

We define impedance (Z) as the phasor voltage divided by the phasor current, Z = V / I. Note: The letters Z, V, and I represent vector quantities.

Power and energy

Basically, electric power (the rate of doing electric work) equals the amount of electromotive force (EMF) applied times the number of electrons it sets flowing; all divided by time. We know the number of electrons flowing per unit time is actually the current, as measured in amperes. Therefore, electric power equals EMF (in volts) times current (in amperes) or E x I.

The unit of electric power is the watt, defined as the power used when an EMF of 1V causes 1A to flow through a conductor. In most instances, such as in large electrical loads and services, the watt is too small of a unit to handle easily. Here, we use the kilowatt (kW), which equals 1000W.

In any AC circuit, power is again a function of current and voltage. But, it's also a function of the phase angle (if any) between the current and voltage waveforms. (Obviously, this is different from DC circuits, where power is simply equal to the current value (I) times the voltage value (E), with the answer in watts (W).)

Speaking of phase angles, AC circuits containing only resistive loads have current and voltage waveforms that are in phase with each other. In other words, the phase angle is zero. Here, power is equal to the instantaneous current value times the instantaneous voltage value, with the answer expressed in watts.

There's an easier way to express power in this context: Just average the power over a half cycle. So, the average power in each half cycle of a purely resistive AC circuit is equal to the rms current value times the rms voltage value. This basically yields the same equation for finding power as that for DC circuits: P = E x I.

The product of instantaneous current (i) and instantaneous voltage (e) gives us the power (P) at that instant. Note that when both "e" and "i" are negative, their product (power) is positive. As you can see, you're expending power throughout the cycle, delivering it to the load.

If an AC circuit contains a reactance in addition to a resistive load, how is the phase angle affected? Basically, the current and voltage waveforms will have a phase difference between them. This means the current will either lead or lag the voltage, depending on the nature of the reactance. Here, the circuit's power is not the product of voltage times current.

Consider a situation where current is out of phase with voltage and you can see the waveforms on a scope. Looking closely, you'll see there are some regularly occurring periods where the voltage is positive while the current is negative, and vice versa. A positive value times a negative value results in a negative value. So, multiplying voltage times current during these periods gives us a negative value for power. In other words, you're delivering power back to the source during these periods. Because reactive loads behave as above, they're basically exchanging power back and forth with the source.In AC circuits with both resistive and reactive loads, we use a different term to denote the product of voltage times current: apparent power. It's expressed in voltamperes (VA).

So, what's the amount of actual or useful power expended in a resistive-plus-reactive circuit? Let's use vectors to show this. The actual useful power (let's call it true power) relates to the apparent power by the cosine of the angle u. This angle is the angle by which the current either leads or lags the voltage. So, true power (watts) equals apparent power (VA) times cosine u.

When the angle u is zero (no reactive load), current and voltage are in phase with each other. The cosine of zero equals one. So, true power and apparent power are equal. With a purely reactive load, angle u is 90 degrees. The cosine of 90 degrees is zero. So, P = e x I x 0, which shows the reactive load consumes no power at this point in the cycle.

The amount of electrical energy used by an appliance equals the amount of power it requires (in watts) times the length of time (in sec) it continues to use this power. The unit of energy is the watt-second, or more commonly known as joule (J). But since this unit is so small, we use the unit kilowatt-hour (kWh), which is 1000W applied for 1 hr.

We're all familiar with the kWh unit, since our electric utility calculates our electric bills on it. For example, if we keep five 60W lamps lit for 4 hr, we consume 1.2kWh of energy (5 x 60W x 4hr). If we do this every day for 30 days, we consume 36kWh of energy (0.3kW x 4 hrs/day x 30 days). If our utility rate is $0.09/kWh, our bill is $3.24 (36kWh x $0.09/kWh).

Of course, there are the inevitable power and energy losses in electrical systems. We know, for example, that current flowing over a conductor produces an unavoidable drop in voltage from the source to the load. And, as long as this current flows to the load, there are power losses proportional to the volts dropped in the conductor and the amount of current flowing.

Also, for any period of time we apply the power, there's an expenditure of energy (energy loss) equal to power loss times the time involved. Both of these losses (power and energy) are inherent in any power system. They cost money due to the charge for kWh expended as well as the energy expended in the form of heat.

To minimize these effects, we just address energy and power losses in our system design. For example, we make sure conductors have sufficient circular mil area to limit voltage drop to NEC-required maximums.

Power factor

Power factor (PF) measures how effectively your AC electrical system operates. A high PF (considered to be above 0.95) means the apparent power requirements for an electrical system, as expressed in kilovolt-amperes (kVA), are very close to the actual power requirements, as expressed in kilowatts (kW). The actual power requirement represents the power needed for useful work as well as normal line losses. The apparent power relates to the total electrical system requirements, including reactive power (which in itself performs no useful work). The reactive power requirements of an electrical system are the volt-amperes necessary to provide the magnetizing power for inductive loads.

Power factor can be expressed in two ways: PF = actual power (kW) divided by apparent power (kVA) or kW = kV x PF (where you use a wattmeter to measure kilowatts and a voltmeter and ammeter to measure kilovolt amperes); and as Cos u. The apparent power is the hypotenuse, the actual power is the neutral base (along the abscissa or time element line), and the vertical leg of the triangle is the reactive power (kvar). As you can see, the magnitude of the reactive power is a direct reflection of the power factor. Note: PF = Cos u = actual power (kW) divided by apparent power (kVA). To improve the power factor (reduce the kVA), introduce leading reactive power into the circuit.

In an ideal situation, voltage and current (amperes) are in phase, and the voltage and current cross the neutral horizontal line (along the abscissa) concurrently. When dealing with realistic AC electrical systems, the current usually crosses the neutral after the voltage (current lags voltage causing phase displacement); a condition caused by inductive-type loads. The amount of phase displacement relates to the power factor.

If an electrical system has a low PF (the average uncorrected industrial PF is around 0.8), it means the apparent power requirements are somewhat larger than the actual power requirements. A system with low power factor can result in several penalties.

For example, you may need to purchase additional equipment and energy from the serving utility (for nonproductive purposes). Ideally, the actual power equals the apparent power (PF = 1). A low power factor can result from poor equipment design or from equipment operating conditions (e.g. lightly loaded induction motors, which are probably the worst offenders).

The power factor of an electric motor reaches its maximum rated value when under a full load, and the power factor decreases rapidly as the load decreases. The power factor associated with a transformer reacts to load variation in a similar manner. The reactive power used by a transformer at full load may be as high as 8% to 12% of the rated power of the transformer. When unloaded, the amount of reactive power may remain as high as 4% to 6% of the rated power. Thus, a transformer consumes power even though it may not serve a load.

You can improve PF by using synchronous motors, usually in very large sizes, to drive mechanical loads. The motors also provide a leading PF to help overcome an electrical system with a lagging PF (which is almost always the case).

Sometimes, people use synchronous motors (they aren't powering any loads) solely to provide leading kvars into an electrical system to counteract inductive loads. Automatic controls can adjust the field excitation in the rotor windings of a synchronous motor, which in turn determines the amount of leading kvars produced. These motors are usually referred to as synchronous condensers.

Adding power factor capacitors to compensate for the consumption of reactive energy is another method used to improve PF. You can calculate the total kvar rating of capacitors required to increase the power factor to any desired value easily by using the tables published by leading power factor capacitor manufacturers. Determining the kvar value is a two step process: First, determine the actual power (kW) based on the power factor in your system. For a single-phase system:

KW = (volts x amps x PF) / 1000

For a 3-phase system, multiply by the square root of 3 (which is 1.732):

KW = (volts x amps x PF x 1.732) / 1000 Where voltage is measured line-to-line.

Next, determine the corrective (leading) reactive power (kvar) needed for obtaining a desired PF by multiplying the kW value, as found above, by a selected value from a power-factor multiplier table.