Electrical Power and Energy

After this lesson you should be able to compute electrical power from any two of voltage, current and resistance, rearrange the power formula to find a missing quantity, and turn power and time into energy in joules, watt-hours or kilowatt-hours. You will also estimate electricity cost, efficiency and battery runtime. These skills appear regularly in ASVAB Electronics Information questions.

Power ties together the quantities that every Electronics Information question is built from. A label on a soldering iron, a space heater or a phone charger states its power, and from that number you can work out how much current it pulls, whether a circuit or fuse can handle it, and how much it costs to run. The ASVAB rewards people who can move quickly between volts, amps, ohms, watts and watt-hours without mixing them up. Most power questions need only one formula and one or two steps of arithmetic, so they are good places to collect points if you know the relationships well.

Power tells you how fast energy is being transferred or converted, not how much energy exists. Think of a garden hose: the flow rate in gallons per minute is like power, while the total gallons that land in a bucket is like energy. A 100 W bulb converts energy at a steady 100 joules every second. Leave it on longer and it uses more energy, yet its power stays 100 W the whole time.

The joule (J) is the SI unit of energy. The watt (W) is the SI unit of power, and one watt means one joule transferred per second. So a device rated at 25 W moves 25 J of energy each second it runs. If a question asks for power, the answer belongs in watts; if it asks for energy, look for joules or watt-hours.

Voltage is the energy each coulomb of charge gives up as it passes through a component, measured in joules per coulomb. Current is how many coulombs pass each second. Multiply the two and the coulombs cancel, leaving joules per second, which is watts. That is why electrical power equals voltage times current. The relationship holds for any component at a given instant, whether it is a resistor, a motor or a charging battery. It is the starting point from which every other power formula in this lesson is built.

Here P is power in watts (W), V is the voltage across the component in volts (V), and I is the current through it in amperes (A). Example: a car dome light on 12 V that draws 0.5 A uses

When you know the power and the current, divide power by current:

Example: a heating pad uses 60 W and draws 0.5 A. Its voltage is

Check by multiplying back:

When you know the power and the voltage, divide power by voltage:

Example: a 1,200 W microwave on a 120 V outlet draws

This is the kind of figure you compare with a circuit breaker rating.

A quick memory aid is a triangle with P on top and V and I side by side underneath. Cover the quantity you want. Covering P leaves V beside I, so multiply. Covering V leaves P over I, so divide. Covering I leaves P over V, so divide.

Each of these quantities answers a different question. Voltage, in volts, is the push or energy per unit of charge. Current, in amperes, is the rate at which charge flows. Power, in watts, is the rate at which energy is converted. A 9 V battery can supply a tiny current to a smoke alarm or a much larger one to a toy motor; the voltage is the same, but the power is very different. Never treat a number in one unit as if it were another.

Because

power is directly proportional to voltage when current is held constant. If the current stays at 2 A and the voltage rises from 5 V to 10 V, power goes from 10 W to 20 W: double. Note the condition. In a plain resistor, raising the voltage also raises the current, so the result there is different, as later sections show.

With voltage held steady, power is directly proportional to current. A 24 V supply delivering 1.5 A provides 36 W; if the load draws 3 A instead, the supply delivers 72 W. Doubling current doubled power. This is a common exam item: at constant voltage, double the current and you double the power, not quadruple it and not leave it unchanged.

Ohm's law,

links voltage, current and resistance for an ohmic resistor. Substituting it into

produces two more forms of the power equation. Replacing V gives a formula using current and resistance. Replacing I with $V/R$ gives a formula using voltage and resistance. All three forms describe the same physics.

Substitute

into

P is in watts, I in amperes, R in ohms. Because current is squared, it has a strong effect. Example: 3 A through a 4 Ω resistor gives

Substitute

into

Example: 6 V across a 3 Ω resistor gives

Since voltage is squared, doubling the voltage on a fixed resistor makes the power four times larger.

The three formulas must agree, so you can use one to check another. Take 10 V across a 5 Ω resistor. Ohm's law gives 2 A. Then

If your answers disagree, an arithmetic step went wrong.

Pick the formula that uses the two values you were given, so you avoid an extra step. Given voltage and current, use

Given current and resistance, use

Given voltage and resistance, use

Finding Ohm's law values first also works, but each extra step is one more chance for an arithmetic slip under time pressure.

Electrical power dissipated in a resistor becomes heat. Every resistor has a power rating, such as 1/4 W, 1/2 W or 5 W, which is the most it can safely dissipate. A resistor forced to handle more than its rating can overheat, change value, discolor the circuit board or burn open. A good habit is to choose a part rated at roughly twice the expected dissipation. For example, a resistor expected to dissipate 0.3 W should get about 0.6 W of rating, so a 1/2 W part is marginal and a 1 W part is the sensible choice.

When the voltage is fixed, use

Resistance sits in the denominator, so smaller resistance means more power. On a 12 V supply, a 24 Ω load uses 6 W, while an 8 Ω load uses 18 W. This is why household heaters have low-resistance elements: on a fixed outlet voltage, low resistance draws more current and makes more heat.

When the current is fixed, as in resistors connected in series, use

Now resistance is in the numerator, so larger resistance means more power. With 0.5 A flowing, a 10 Ω resistor dissipates 2.5 W while a 40 Ω resistor dissipates 10 W. Whether more resistance means more or less power depends entirely on what is held constant, which is why exam questions state that condition.

In these problems, electrical energy is the total amount delivered to or used by a device over some time. It becomes heat, light, motion or sound, or it is stored chemically in a battery being charged. The power company bills you for energy, not power. A large motor running briefly can use less energy than a small fan running all day, because energy depends on both power and time.

E is energy, P is power and t is time. Watts times seconds gives joules; watts times hours gives watt-hours (Wh); kilowatts times hours gives kilowatt-hours (kWh). Example: a 25 W fan running 4 hours uses

Wh. The same rule, rearranged as

finds power from energy.

The watt measures a rate. The watt-hour measures an amount: the energy used by a 1 W load running for one hour. A 40 Wh label on a battery describes stored energy, while a 40 W label on a lamp describes how fast the lamp uses energy. The names look alike, but swapping them is like confusing miles per hour with miles. Check the units of the question before choosing an answer.

The metric prefixes work the same for power as for any unit. One kilowatt (kW) equals 1,000 W, and one megawatt (MW) equals 1,000,000 W. To convert kW to W, multiply by 1,000, so 3.5 kW is 3,500 W. To go from W to kW, divide by 1,000, so 750 W is 0.75 kW. Megawatts describe power plants and large industrial loads.

One kilowatt-hour is the energy used by a 1 kW load in one hour. A 1.5 kW space heater on for 2 hours uses

kWh. Utility meters and bills use this unit.

One hour contains 3,600 seconds, and one watt is one joule per second, so 1 Wh equals 3,600 J. Multiply watt-hours by 3,600 to get joules, or divide joules by 3,600 to get watt-hours. A 5 Wh battery stores 18,000 J. One kilowatt-hour is 3,600,000 J, or 3.6 MJ.

Time units must match the energy unit you want. For joules, put time in seconds. For watt-hours or kilowatt-hours, put time in hours. Turn minutes into hours by dividing by 60: 15 minutes is 0.25 h, 45 minutes is 0.75 h, and 90 minutes is 1.5 h. Turn minutes into seconds by multiplying by 60: 3 minutes is 180 s.

Convert watts to kilowatts first, then multiply by hours:

A 150 W television watched for 6 hours uses

kWh.

Energy cost equals energy in kilowatt-hours multiplied by the price per kilowatt-hour. A 2 kW electric oven used for 1.5 hours consumes 3 kWh; at a made-up rate of 16 cents per kWh, the energy charge is 48 cents. Real bills add fixed fees, taxes and changing rates, but exam questions usually include only the energy charge. Keep the units straight: kWh times cents per kWh gives cents.

Efficiency compares the useful output of a device with what you put in:

The Greek letter $\eta$ (eta) is efficiency, $P_{\text{out}}$ is useful output power and $P_{\text{in}}$ is input power, both in watts. Energies over the same time period work too. Example: a pump motor takes in 250 W and delivers 225 W of mechanical power, so

or 90 percent. To find required input instead, divide output by efficiency.

A power rating says how much power a device draws or delivers. Efficiency is a ratio with no units that says what fraction of the input ends up as useful output. A 1,000 W motor and a 50 W motor could both be 85 percent efficient. Efficiency of a real device is always below 100 percent, so an answer choice over 100 percent signals that the output and input were swapped.

Energy is conserved, so input power equals useful output power plus lost power. The losses are usually heat from wire resistance and friction, plus some sound or vibration. If a power supply draws 80 W from the wall and delivers 68 W to a computer, 12 W is lost as heat, which is why the supply feels warm. Its efficiency is

or 85 percent.

Battery capacity is often stated in ampere-hours (Ah) or milliampere-hours (mAh). An ampere-hour is a quantity of charge: one ampere flowing for one hour. A 2 Ah battery could, in an ideal case, supply 1 A for 2 hours or 0.5 A for 4 hours. A 3,000 mAh phone battery holds 3 Ah. Ampere-hours alone are not energy, because they leave out the voltage at which that charge is delivered.

Multiply nominal voltage by capacity in ampere-hours to estimate stored energy in watt-hours:

The units work because volts times amperes is watts, and watts times hours is watt-hours. A 3.7 V, 2.5 Ah lithium cell holds about

Wh. A 12 V, 7 Ah backup battery holds about 84 Wh. Real usable energy is a bit lower.

Ideal runtime equals stored energy divided by load power: hours equals Wh divided by W. A 48 Wh pack running a 12 W light lasts about 4 hours. If a converter sits between the battery and the load, the battery must supply more than the load uses, so divide the load power by the converter efficiency first. Treat the result as an upper estimate, since batteries are rarely drained completely.

A label such as 60 W assumes a stated operating condition, usually the rated voltage. Run a heater rated 1,500 W at 120 V on a lower voltage and its power drops, because its resistance stays about the same while $V^2/R$ falls. Motors draw different power at light and heavy loads. Battery capacity depends on how fast it is discharged and on temperature. When an exam says "operates at its rated power," it is telling you to use the label value directly.

Compare a 1,500 W heater with a 10 W LED lamp, both on 120 V. The heater draws

the lamp draws about 0.083 A. In one hour the heater uses 1.5 kWh while the lamp uses only 0.01 kWh. The heater uses as much energy in one hour as the lamp would use in 150 hours.

When several loads run at the same time, their powers add. Total power equals the sum of the individual powers, regardless of how they are wired. A 900 W coffee maker and a 600 W toaster on together demand 1,500 W, which on 120 V is 12.5 A total. Their combined energy over 0.2 hours is

kWh. Adding powers like this is how electricians check whether a circuit will be overloaded.

In a series string, every resistor carries the same current, so

is the natural choice, and the largest resistor dissipates the most power. With 2 A flowing through 3 Ω and 7 Ω, the 3 Ω part dissipates 12 W and the 7 Ω part dissipates 28 W, for 40 W total. That total equals $I^2$ times the 10 Ω combined resistance.

In parallel, every branch has the same voltage, so use

The smallest resistor dissipates the most power because it carries the most current. Across 10 V, a 5 Ω branch dissipates 20 W and a 20 Ω branch dissipates 5 W, for 25 W total. Total parallel power is the sum of the branch powers.

In an ideal circuit with lossless wires, the power the source delivers equals the total power the loads absorb. This is conservation of energy. A 9 V source supplying 0.4 A delivers 3.6 W, so the loads together must dissipate 3.6 W. If a problem gives you all the load powers, add them to get source power; if it gives source power and all but one load power, subtract to find the missing one.

Household outlets put each appliance across the same voltage, which is a parallel arrangement. Each load then draws current according to its own power:

On 120 V, a 60 W lamp draws 0.5 A and a 360 W blender draws 3 A, and the circuit supplies the total of 3.5 A. A higher-power device on the same voltage always draws more current.

A wattmeter reads power, the rate of energy use at that moment, in watts or kilowatts. An energy meter, like the utility meter on a house, reads accumulated energy in kilowatt-hours; it keeps counting upward as long as anything is drawing power. A plug-in monitor often shows both: present watts on one screen and total kWh since reset on another. Watts tell you how hard something is working now; kWh tell you how much it has used so far.

An energy meter counts power multiplied by time, so a small load that never turns off still adds up. A 5 W standby device runs 24 hours a day:

Wh each day, or about 3.6 kWh in a 30-day month. Low power over a long time can match high power over a short time.

Instantaneous power is the value at one moment. Average power is total energy divided by total time. A refrigerator compressor may draw 150 W while running and nearly zero while off. If it runs 8 hours of a 24-hour day, it uses 1,200 Wh, and its average power over the day is

Use average power when finding energy over long periods for loads that cycle on and off.

When a device runs at different power levels, find the energy for each stage separately, then add. A blender at 300 W for 10 minutes and 600 W for 5 minutes uses

Wh. Do not average the two powers unless the stages last equal times.

Device labels show several kinds of numbers, and the unit tells you which is which. A power adapter might read "Input 100–240 V, 1.5 A" and "Output 19 V, 3.42 A, 65 W." The output power checks:

A battery label might show "11.1 V, 4,400 mAh, 48.84 Wh." The Wh value is energy; the mAh value is charge. Volts are never power.

Writing units alongside numbers catches many errors. Watts times hours gives watt-hours. Joules divided by seconds gives watts. Watt-hours divided by watts gives hours. Volts times ampere-hours gives watt-hours. If your setup ends in a strange unit, such as watts per hour or watt-squared, the setup is wrong even if the arithmetic is right.

Power delivered is what a source supplies to the circuit. Power dissipated is what a component converts, usually into heat. In a simple resistor circuit with ideal wiring, they are equal. In a real circuit, the source delivers slightly more than the main load receives, because the wires and connections dissipate a little. For example, a source delivering 100 W through long extension cords might put 96 W into the load and lose 4 W heating the cords. ASVAB questions usually assume ideal wiring unless they say otherwise.

An incandescent bulb is a resistor, and all the power it dissipates becomes heat and light, but most of it is heat. A traditional 60 W bulb turns only a few percent of its power into visible light. An LED produces similar brightness at around 8 to 10 W because more of its input becomes light. The dissipated power is still computed with the usual formulas; efficiency describes how much of it is useful.

For DC circuits and purely resistive AC loads,

works directly when the AC voltage and current are effective (RMS) values. Household 120 V is an RMS value. Motors and other loads with coils or capacitors can draw current that is out of step with the voltage, so their true power in watts is less than volts times amps. That product is then called apparent power, measured in volt-amperes (VA). Unless a question mentions this, treat loads as resistive.

Real circuits can injure or kill, so all hands-on work must follow proper procedures and training. The habits that matter for this lesson are general: high power means high current or high voltage, both of which create heat and shock hazards. Overloaded outlets and extension cords overheat. Devices should be de-energized and unplugged before anyone touches internal parts. On the test, treat the calculations as models of ideal circuits, never as instructions for trying things on live equipment.

Before any arithmetic, write down three things: what is given, with units; what is asked for, with units; and which formula connects them. Then check whether a conversion is needed, such as minutes to hours or watts to kilowatts. This ten-second plan prevents most mistakes on multi-step power questions.

Rough estimates eliminate wrong choices quickly. If a 1,000 W heater runs for a bit less than an hour, the energy must be a bit less than 1 kWh; a choice of 50 kWh is impossible. If a device on 120 V uses about 600 W, its current must be about 5 A. Round first, then pick the choice near your estimate.

A portable air compressor runs from a 12 V vehicle outlet and draws 8 A. What power does it use?

Given V = 12 V and I = 8 A. Use the basic power formula:

The compressor uses 96 W.

A laptop draws 45 W at a current of 2.25 A. What voltage does it run on?

The answer is 20 V.

A 36 W fluorescent fixture runs on a 120 V circuit. What current does it draw?

It draws 0.3 A, or 300 mA.

A current of 0.2 A passes through a 150 Ω resistor. How much power does it dissipate?

The resistor dissipates 6 W, so a 1/2 W part would overheat.

A 50 Ω resistor has 15 V across it. Find its power.

It dissipates 4.5 W.

An 8 W aquarium pump runs nonstop for 30 days. How much energy does it use?

Hours:

h. Energy:

Wh.

Divide by 1,000: the pump uses 5.76 kWh.

A small flashlight battery stores 3.5 Wh. Express this in joules.

The battery stores 12,600 J, since each watt-hour is 3,600 J.

A 1,800 W hair dryer runs 20 minutes a day for 30 days. At an assumed 15 cents per kWh, what is the energy cost?

Daily energy:

kWh. Monthly:

kWh.

Cost:

A drill motor takes 600 W from the outlet and is 70 percent efficient. How much useful mechanical power does it deliver, and how much is lost?

Lost power:

mostly as heat in the windings and bearings.

The motor delivers 420 W of useful power.

A battery charger draws 50 W from the wall while delivering 40 W into the battery. What is its efficiency?

The charger is 80 percent efficient; the other 10 W heats the charger. A quick check: efficiency must be below 100 percent, and it is.

A 12 V, 10 Ah battery powers a 30 W camping fridge directly. About how long will it run?

Energy:

Wh.

About 4 hours, ignoring losses and the usual advice not to drain the battery fully.

A 1.5 A current flows through 2 Ω and 10 Ω resistors in series. Find each power.

Total: 27 W. The larger resistor heats more.

A 20 V supply feeds 10 Ω and 40 Ω resistors in parallel. Find each power.

Total: 50 W. The smaller resistor heats more.

A space heater runs at 1,200 W for 2 hours, then at 600 W for 1 hour. Find total energy and average power.

Stage energies:

kWh and

kWh. Total: 3.0 kWh.

Average power:

which is 1,000 W.

Name the quantity each value describes.

250 W? Answer: power.

7.2 kWh? Answer: energy.

5 Ah? Answer: electric charge (battery capacity).

3,600 J? Answer: energy.

120 V? Answer: voltage, or energy per unit of charge.

15 A? Answer: current.

Pick the most direct formula.

Given 24 V and 2 A, find power. Answer:

Given 0.5 A and 100 Ω, find power. Answer:

Given 9 V and 18 Ω, find power. Answer:

Given 600 J over 20 s, find power. Answer:

Convert each value.

4.2 kW to W? Answer: 4,200 W.

850 W to kW? Answer: 0.85 kW.

2 Wh to J? Answer: 7,200 J.

40 minutes to hours? Answer: 2/3 h, about 0.67 h.

1,500 mAh to Ah? Answer: 1.5 Ah.

Decide how power changes.

Fixed voltage, resistance halved? Answer: power doubles.

Fixed current, resistance tripled? Answer: power triples.

Fixed resistance, current doubled? Answer: power becomes four times larger.

Fixed resistance, voltage halved? Answer: power drops to one-fourth.

Assume 20 cents per kWh.

A 100 W lamp on 10 hours? Answer: 1 kWh, 20 cents.

A 2.5 kW heater on 4 hours? Answer: 10 kWh, 2 dollars.

A 50 W router on for 24 hours? Answer: 1.2 kWh, 24 cents.

A 3 kW dryer for 45 minutes? Answer: 2.25 kWh, 45 cents.

Apply the idea that input equals output plus losses.

A source delivers 60 W to three loads using 15 W and 20 W and an unknown amount. The third? Answer: 25 W.

A motor takes 500 W and delivers 400 W. Losses? Answer: 100 W, so it is 80 percent efficient.

A 90 percent efficient converter must output 45 W. Input? Answer: 50 W.

The mistake: answering an energy question in watts, or a power question in watt-hours. Watts are a rate; watt-hours are an amount. The fix: if time is involved and the question asks how much was used, the answer must be in Wh, kWh or J.

The mistake: writing

Only current or voltage is squared, never resistance. The fix: recall the derivation. Substituting

into

gives $I\times IR$, so the current appears twice and the resistance once.

The mistake: saying more resistance always means more power, or always less. Both claims fail without knowing what is held fixed. The fix: at fixed voltage, use $V^2/R$, so more resistance gives less power. At fixed current, use $I^{2}R$, so more resistance gives more power. Read the question to see which quantity is fixed before choosing.

The mistake: comparing batteries by ampere-hours alone, as if Ah were energy. A 3.7 V, 4 Ah cell and a 12 V, 4 Ah battery hold very different energy. The fix: multiply by voltage. The first holds about 14.8 Wh; the second about 48 Wh.

Start by reading the last line to see what is asked and in which unit: watts, watt-hours, kilowatt-hours, joules, volts, amps or ohms. Next, list the given values and convert anything that needs it, such as kilowatts to watts, milliamps to amps, or minutes to hours or seconds. Choose the single formula that links the given values to the unknown. Calculate, then check the answer by size and units. Is the current reasonable for the voltage? Is the efficiency below 100 percent? Does energy grow with time? Watch for choices that come from multiplying instead of dividing, from adding the given numbers, or from forgetting to square. If a ratio question asks "by what factor," set up the formula as a proportion rather than plugging in made-up numbers.

Power is the rate of energy transfer, in watts, and

applies to any component. For resistors,

give the same answer. Energy is power times time, measured in joules, watt-hours or kilowatt-hours, with 1 Wh equal to 3,600 J. Cost is kWh times the rate. Efficiency is useful output divided by input. Battery energy is about volts times ampere-hours. Next you will apply these ideas to series circuits, where the same current flows through every part and the resistances and voltage drops add up.