# Power

Power is defined as the rate that energy changes 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, like heat or light.

# $$P = \frac{E}{\Delta t}$$

$$P$$ = power [W, J/s, watt]
$$E$$ = energy [J, joules]
$$\Delta t$$ = time period [s, second]

## Electrical Power

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 uses up twice that much energy per second.

derivation of electrical power

We can convert power into electrical terms with some substitution

$$P = \frac{E}{\Delta t} \quad \quad \quad E = {\color{#00f}Vq} \quad \quad \quad I = \color{#f00} \frac{q}{\Delta t}$$ $$P = \frac{{\color{#00f}Vq}}{\Delta t}$$ $$P = {\color{#f00}\frac{q}{\Delta t}}V$$ $$P = IV$$

If we combine P=IV with Ohm's law we can get another power equation

$$P=IV \quad \quad \quad V=IR$$ $$P=I(IR)$$ $$P=I^2R$$

# $$P = IV$$ $$P = I^2R$$

$$P$$ = power [W, watt, J/s]
$$I$$ = current [A, amp]
$$V$$ = voltage [V, volt]
$$R$$ = resistance [Ω, ohm]
Example: Calculate the power dissipated by each resistor.
solution $$I = 0.02\, \mathrm{A}$$ $$P=(0.02)^2 (6000)$$ $$P=2.4 \, \mathrm{W}$$
$$P=(0.02)^2 (20000)$$ $$P=8.0\, \mathrm{W}$$
$$P=(0.02)^2 (5000)$$ $$P=2.0\, \mathrm{W}$$
Example: Find the power dissipated by the resistor.
solution $$P=I^2R$$ $$P=(0.05)^2(300)$$ $$P=0.75\, \mathrm{W}$$

Find the heat dissipated by the resistor in 10 seconds.
solution $$P=0.75\, \mathrm{W}$$
$$P = \frac{E}{\Delta t}$$ $$E = P \Delta t$$ $$E = (0.75\, \mathrm{W})(10\, \mathrm{s})$$ $$E = 7.5\, \mathrm{J}$$
Example: Find the power output of the battery.
solution $$P=IV$$ $$P=(0.095)(9)$$ $$P=0.855\, \mathrm{W}$$
Example: Find the total energy converted to electrical energy over 5 minutes by the battery?
solution $$P=IV$$ $$P=(0.024)(1.5)$$ $$P=0.036\, \mathrm{W}$$
$$P = \frac{E}{\Delta t}$$ $$E = P\Delta t$$ $$E = (0.036 \, \mathrm{\frac{J}{s}})(5\, \mathrm{min})\left(\frac{60\, \mathrm{s}}{1\, \mathrm{min}}\right)$$ $$E = (0.036 \frac{J}{s})(300s)$$ $$E = 10.8\, \mathrm{J}$$

## Kilowatt Hour

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: The average price in America for 1 kiloWatt hour is \$0.12. How much would it cost to run an 800 Watt microwave for 3 hours?
solution $$800\, \mathrm{W} \left( \frac{\mathrm{k}}{1000} \right) = 0.8\, \mathrm{kW}$$ $$(0.8\, \mathrm{kW})(3\, \mathrm{h})\left(\frac{\0.12}{\mathrm{kWh}}\right) = \0.28$$