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.

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?

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.

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.

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 $$P=IV$$ $$P=(0.024)(1.5)$$ $$P=0.036\, \mathrm{W}$$
$$P = \frac{W}{\Delta t}$$ $$W = P\Delta t$$ $$W = (0.036 \, \mathrm{\frac{J}{s}})(5\, \mathrm{min})\left(\frac{60\, \mathrm{s}}{1\, \mathrm{min}}\right)$$ $$W = (0.036 \frac{J}{s})(300s)$$ $$W = 10.8\, \mathrm{J}$$