Power

Power is defined as work done over time. Power measures the rate that voltage sources, like batteries, convert energy into electrical energy. Power could also measure the rate that electrical components convert electrical energy in other forms.

$$P = \frac{W}{\Delta t}$$

\(P\) = power [W, J/s, watt]
\(W\) = work, change in energy [J, joules]
\(\Delta t\) = time period [s, second]

Power is analogous to velocity. Velocity measures how position changes over time. Power measures how energy changes over time.

Example: The label on my electric space heater says 1500 watts. This means it can convert 1500 joules of electrical energy into thermal energy every second. How many joules of energy can it add to a room in an hour?
solution $$P = \frac{W}{\Delta t}$$ $$W = P\Delta t$$ $$W = (1500\,\mathrm{W})(3600\,\mathrm{s})$$ $$W = (1500\,\mathrm{\tfrac{J}{s}})(3600\,\mathrm{s})$$ $$W = 5\,400\,000 \, \mathrm{J}$$

Power bills in America use a unit called kilowatt hour. This is actually a unit of energy, not power.

$$\text{(kilo)(watt)(hour)}$$ $$(1000) \left(\mathrm{\frac{J}{s}}\right)(3600\, \mathrm{s})$$ $$3\,600\,000\, \left(\mathrm{\frac{J}{s}}\right)(\mathrm{s})$$ $$1\, \mathrm{kWh} = 3\,600\,000\, \mathrm{J}$$ Example: How many kilowatt hours of energy would I consume if I ran my 1500 W electric heater at full power for 4 weeks?
solution $$\Delta t = 4 \, {\color{Tomato}\mathrm{week} } \left(\frac{7 {\color{LimeGreen}\,\mathrm{day}} }{1 \, {\color{Tomato}\mathrm{week}} }\right) \left(\frac{24 \, {\color{DarkTurquoise}\mathrm{hour}}}{1 \, {\color{LimeGreen}\mathrm{day}}}\right) \left(\frac{3600 \,\mathrm{s}}{1 \, {\color{DarkTurquoise}\mathrm{hour}}}\right)$$ $$\Delta t = 2\,419\,200 \, \mathrm{s}$$
$$P = \frac{W}{\Delta t}$$ $$W = P\Delta t$$ $$W = (1500 \, \mathrm{\tfrac{J}{s}})(2\,419\,200\,\mathrm{s})$$ $$W = 3\,628\,800\,000 \, \mathrm{J} $$ $$W = 3\,628\,800\,000 \, \mathrm{J} \left( \frac{1\, \mathrm{kWh}}{3\,600\,000\, \mathrm{J}} \right) = 1008 \, \mathrm{kWh}$$

Example: The average price in the United States for 1 kilowatt hour is $0.12. How much would the kilowatt hours from the previous problem cost in dollars?
solution $$W = 1008 \, \mathrm{kWh}$$ $$W = 1008 \, \mathrm{kWh} \left( \frac{\$0.12}{1\, \mathrm{kWh}} \right) = \$120.96$$

Two friends of mine installed solar panels on their roof in 2019. The solar panels produce their electricity when the sun is out, and my friends get electricity from the power grid when the sun isn't shining. If they produce more energy than they use, it is pushed back into the power grid. I wrote down the data from the panel's readout below.

Time Current Power Energy Today Energy Yesterday
8:45 am 583.36 W 513 Wh 20 kWh
Example: Use the energy produced yesterday to estimate the total energy production in kWh for one year.
solution $$\mathrm{\frac{ 20 \, kwh}{day} \left( \frac{365.25 \, day}{1 \, year} \right) } = \mathrm{\frac{ 7305 \, kwh}{year}}$$ $$ 7305 \, \mathrm{kwh} / \mathrm{year} $$

The data we used was from a sunny summer day. They will produce less in the winter.


The laws in California let you use extra energy produced by your solar panels to reduce your power bill to zero.

Example: My friends used 5500 kWh of energy last year. How much money could they save each year? Convert the energy they used last year into dollars with the California rate of $0.25 per kilowatt hour.
solution $$5500 \, \mathrm{kWh} \left( \mathrm{\frac{\$ 0.25}{kWh}} \right) = \$ 1375$$

Example: The solar panels cost $14 000, including installation. How long will it take to pay for the panels with reduced power bills?
solution $$ \$ 14\,000 \left( \frac{1 \, \mathrm{yr}}{ \$ 1375} \right) = 10.1 \, \mathrm{years}$$

They also received a tax break for 1/3 the price of the panels. They hope to pay the solar panels off in 6.8 years based on the reduced price.

DC Electric Power

Circuit components get their energy to do work from dropping voltage. When the voltage decreases it converts electrical energy into heat. Heat is defined as energy exiting a system. Heat could exit the system in forms like light, sound, or thermal energy.

Heat
derivation of electrical power

We can convert power into electrical terms with some substitution.

$$P = \frac{W}{\Delta t} \quad \quad \quad W = {\color{DarkTurquoise}Vq} \quad \quad \quad I = \color{Tomato} \frac{q}{\Delta t}$$ $$P = \frac{{\color{DarkTurquoise}Vq}}{\Delta t}$$ $$P = {\color{Tomato}\frac{q}{\Delta t}}V$$ $$P = IV$$

Electrical power is the rate of energy entering or exiting a system. In a circuit it can be calculated for any circuit element that changes the voltage, like a battery or resistor.

$$P = IV$$

\(P\) = power [W, watt, J/s]
\(I\) = current [A, amp]
\(V\) = voltage drop [V, volt]

Watt is the unit of power? The wattage of an electrical appliance gives you an idea of how much electrical energy the appliance converts. An 800 W microwave uses 800 J per second. A 1600 W microwave will heat food much faster because it converts twice the electrical energy into heat each second.

When used properly electricity is safe, but a circuit with both high voltage and high current can be dangerous. Be careful.

9 V 0.095 A Example: Calculate the electric power the battery is adding to the circuit.
solution $$P=IV$$ $$P=(0.095)(9)$$ $$P=0.855\, \mathrm{W}$$
Question: A battery and a resistor in a simple looping circuit will drain the chemical energy of the battery. Energy can't be created or destroyed, but it can change forms. What form does the battery's energy change into?
answer

The battery converts chemical energy into electric potential energy.

The resistor converts the electric potential energy into thermal energy.

$$P=IV \quad \quad V=IR$$

Combine electric power with ohm's law to derive two more equations.

Derive: Find a relationship between Power, Current, and Resistance.
strategy

Use substitution.

V=IR means we can substitute V with IR in the power equation.

derivation $$V={\color{Tomato}IR}$$ $$P=IV$$ $$P=I({\color{Tomato}IR})$$ $$\boxed{P=I^2R}$$
Derive: Find a relationship between Power, Voltage, and Resistance.
strategy

Use substitution.

V=IR also means that I = V/R. Substitute I for V/R in the power equation.

derivation $$V=IR \quad I = {\color{DarkTurquoise}\frac{V}{R}}$$ $$P=IV$$ $$P={\color{DarkTurquoise}\frac{V}{R}}V$$ $$\boxed{P=\frac{V^2}{R}}$$
6 kΩ 20 kΩ 5 kΩ 20 mA Example: Calculate the power dissipated by each resistor.
solution $$I = 0.02\, \mathrm{A}$$ $$P=(0.02)^2 (6000)$$ $$P=2.4 \, \mathrm{W}$$
$$P=(0.02)^2 (20\,000)$$ $$P=8.0\, \mathrm{W}$$
$$P=(0.02)^2 (5000)$$ $$P=2.0\, \mathrm{W}$$ 2.4 W 8.0 W 2.0 W 20 mA
P = ? 300 Ω 0.05 A Example: Find the power used by the resistor.
solution $$P=I^2R$$ $$P=(0.05)^2(300)$$ $$P=0.75\, \mathrm{W}$$

How much energy is used by the resistor over 10 seconds.
solution $$P=0.75\, \mathrm{W}$$
$$P = \frac{W}{\Delta t}$$ $$W = P \Delta t$$ $$W = (0.75\, \mathrm{W})(10\, \mathrm{s})$$ $$W = 7.5\, \mathrm{J}$$
1.5 V 0.024 A Example: How much chemical energy does the battery convert into electrical energy over 5 minutes?
solution $$P=IV$$ $$P=(0.024)(1.5)$$ $$P=0.036\, \mathrm{W}$$
$$P = \frac{W}{\Delta t}$$ $$W = P\Delta t$$ $$W = (0.036 \, \mathrm{\frac{J}{s}})(5\, \mathrm{min})\left(\frac{60\, \mathrm{s}}{1\, \mathrm{min}}\right)$$ $$W = (0.036 \frac{J}{s})(300s)$$ $$W = 10.8\, \mathrm{J}$$

Practice printout.pdf

In case you wanted more practice I used AI to make some more problems. The rest of the site I made by hand, but generating endless problems seemed safe. I did find mistakes in the AI generated problems, and there are probably some I didn't find. Let me know if something could be fixed. I also added a practice problem on each page with no solution. That's intentional. Have fun!

For home energy problems, remember that kilowatt hours measure energy, not power. Convert watts to kilowatts and seconds or minutes to hours when the answer should be in kWh.

Question: A power bill charges for kilowatt hours. Is a kilowatt hour a unit of power or a unit of energy?
answer A kilowatt hour is a unit of energy. A kilowatt tells the rate of energy transfer, and multiplying by time gives the total energy transferred.
Example: A small oil-filled radiator heater is warming a room. This type of heater takes longer to heat up, but it provides quiet, steady warmth after it gets hot. If it transfers 1800 J of energy to the air in 3.0 s, what is its power use?
solution $$P = \frac{W}{\Delta t}$$ $$P = \frac{1800 \, \mathrm{J}}{3.0 \, \mathrm{s}}$$ $$P = 600 \, \mathrm{W}$$
Example: A phone charger with a 1.8 m cable transfers energy at 12 W for 15 minutes while a phone is below 20% battery. How many joules of energy does it transfer?
solution Convert minutes to seconds. $$\Delta t = 900 \, \mathrm{s}$$ $$P = \frac{W}{\Delta t}$$ $$W = P\Delta t$$ $$W = (12 \, \mathrm{W})(900 \, \mathrm{s})$$ $$W = 10800 \, \mathrm{J}$$
Example: A 1500 W space heater runs under a desk for 2.5 hours on a cold morning. The room is 3.0 m wide, but the energy use depends on power and time. How much energy does it use in kWh?
solution Convert watts to kilowatts. $$P = 1.5 \, \mathrm{kW}$$ $$W = P\Delta t$$ $$W = (1.5 \, \mathrm{kW})(2.5 \, \mathrm{h})$$ $$W = 3.75 \, \mathrm{kWh}$$
Example: Electricity costs 18 cents for each kWh. How much would the 3.75 kWh from the heater in the previous example cost?
solution $$\mathrm{cost} = (3.75 \, \mathrm{kWh})(18 \, \mathrm{cents/kWh})$$ $$\mathrm{cost} = 67.5 \, \mathrm{cents}$$ The cost is about 68 cents.
Example: A roof solar panel is about 1.7 m tall and 1.0 m wide. On a clear day it averages 420 W for 6.0 hours. How much energy does it produce in kWh?
solution Convert watts to kilowatts. $$P = 0.420 \, \mathrm{kW}$$ $$W = P\Delta t$$ $$W = (0.420 \, \mathrm{kW})(6.0 \, \mathrm{h})$$ $$W = 2.52 \, \mathrm{kWh}$$
Example: A battery adds energy to a small circuit at 9.0 V while the current is 0.095 A. The circuit board is labeled for a science fair display, but the label does not affect the calculation. What power does the battery provide?
solution $$P = IV$$ $$P = (0.095 \, \mathrm{A})(9.0 \, \mathrm{V})$$ $$P = 0.855 \, \mathrm{W}$$
Example: A device runs for 12 minutes on a 9.0 V battery. How much energy does it use?
answer This cannot be solved from the information given. Energy use needs power and time. The voltage alone is not enough because the current is missing.
Example: A 12 V LED strip under a shelf draws 0.75 A. The strip is 2.0 m long and has many small LEDs, but treat it as one device. How much power does it use, and how much energy does it convert in 5.0 minutes?
solution First find the power. $$P = IV$$ $$P = (0.75 \, \mathrm{A})(12 \, \mathrm{V})$$ $$P = 9.0 \, \mathrm{W}$$ Convert minutes to seconds. $$\Delta t = 300 \, \mathrm{s}$$ $$W = P\Delta t$$ $$W = (9.0 \, \mathrm{W})(300 \, \mathrm{s})$$ $$W = 2700 \, \mathrm{J}$$
Example: A 60 W light bulb is connected to 120 V. What current does it draw?
solution $$P = IV$$ $$I = \frac{P}{V}$$ $$I = \frac{60 \, \mathrm{W}}{120 \, \mathrm{V}}$$ $$I = 0.50 \, \mathrm{A}$$
Example: A laptop charger label says it can accept 100-240 V from the wall. On its output side, it provides 65 W at 20 V. What current does it provide to the laptop?
solution $$P = IV$$ $$I = \frac{P}{V}$$ $$I = \frac{65 \, \mathrm{W}}{20 \, \mathrm{V}}$$ $$I = 3.25 \, \mathrm{A}$$
Question: The resistance of a resistor stays the same. If the voltage across it doubles, what happens to the power dissipated by the resistor?
answer The power becomes four times larger. For a constant resistance, use P = V²/R, so doubling voltage makes the voltage squared four times larger.
Example: A 30 Ω resistor has 0.20 A through it. How much power does it dissipate?
solution $$P = I^2R$$ $$P = (0.20 \, \mathrm{A})^2(30 \, \Omega)$$ $$P = 1.2 \, \mathrm{W}$$
Example: A 240 Ω resistor has 12 V across it during a lab test. The resistor is rated for 0.25 W, but first calculate the power it would dissipate.
solution $$P = \frac{V^2}{R}$$ $$P = \frac{(12 \, \mathrm{V})^2}{240 \, \Omega}$$ $$P = 0.60 \, \mathrm{W}$$ This is more than 0.25 W, so that resistor would get too hot.
Example: A 2.0 kΩ resistor has 15 mA through it. How much power does it dissipate?
solution Convert the units first. $$R = 2000 \, \Omega$$ $$I = 0.015 \, \mathrm{A}$$ $$P = I^2R$$ $$P = (0.015 \, \mathrm{A})^2(2000 \, \Omega)$$ $$P = 0.45 \, \mathrm{W}$$
Example: A small heating element for a cup warmer should use 48 W when connected to 24 V. The metal plate is 8 cm across, but the resistance is set by the electrical values. What resistance should it have?
solution $$P = \frac{V^2}{R}$$ $$R = \frac{V^2}{P}$$ $$R = \frac{(24 \, \mathrm{V})^2}{48 \, \mathrm{W}}$$ $$R = 12 \, \Omega$$
Example: A 12 V motor rated for 36 W spins a small fan in a desk cooler. The fan has three blades, but model the motor as one electrical device. What current does it draw, and what is its effective resistance while running?
solution Find the current from electric power. $$P = IV$$ $$I = \frac{P}{V}$$ $$I = \frac{36 \, \mathrm{W}}{12 \, \mathrm{V}}$$ $$I = 3.0 \, \mathrm{A}$$ Now use Ohm's law. $$V = IR$$ $$R = \frac{V}{I}$$ $$R = \frac{12 \, \mathrm{V}}{3.0 \, \mathrm{A}}$$ $$R = 4.0 \, \Omega$$
Example: A 150 Ω resistor is connected across 9.0 V for 10 minutes inside a simple timing circuit. The plastic case is 6 cm long, but ignore temperature changes in the resistor. How much energy does it convert to heat?
solution First find the power. $$P = \frac{V^2}{R}$$ $$P = \frac{(9.0 \, \mathrm{V})^2}{150 \, \Omega}$$ $$P = 0.54 \, \mathrm{W}$$ Convert minutes to seconds. $$\Delta t = 600 \, \mathrm{s}$$ $$W = P\Delta t$$ $$W = (0.54 \, \mathrm{W})(600 \, \mathrm{s})$$ $$W = 324 \, \mathrm{J}$$

Reading (7 minutes): Read Explainer: What is the electric grid? from Science News Explores. Then answer these questions.

Why must grid operators closely match the amount of electrical power supplied with the amount people are using?
answer

Too little supplied power can lead to outages, while too much can overheat wires or damage equipment. The grid must stay balanced moment by moment.


Why can a heat wave make operating an electric grid more difficult?
answer

More people use air conditioners during a heat wave, so demand rises sharply. Grid operators must find enough generation to meet that extra demand.


The article says many outages occur locally. Give one local problem that can interrupt electrical service.
answer

A storm can knock down a power line, equipment can fail, or damage to a local wire can interrupt service. These problems can happen even when the larger grid is working.