# 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, like resistors or LEDs, convert electrical energy in other types of energy, 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]
derivation of electrical power

We can convert power into electrical terms with some substitution

$$P = \frac{E}{\Delta t} \ \ \ E = Vq$$ $$P = \frac{Vq}{\Delta t}$$ $$P = \frac{q}{\Delta t}V \ \ \ I = \frac{q}{\Delta t}$$ $$P = IV$$

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

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

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

$$P$$ = power [W, Watt, J/s]
$$I$$ = current [A, amp]
$$V$$ = voltage [V, volt]

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

Example: Calculate the power dissipated by each resistor.
solution $$I = 0.02A$$ $$P=(0.02)^2 (6000)$$ $$P=2.4W$$
$$P=(0.02)^2 (20000)$$ $$P=8.0W$$
$$P=(0.02)^2 (5000)$$ $$P=2.0W$$
Example: Find the power dissipated by the resistor?
solution $$P=I^2R$$ $$P=(0.05)^2(300)$$ $$P=0.75W$$
Example: Find the power output of the battery.
solution $$P=IV$$ $$P=(0.095)(9)$$ $$P=0.855W$$
Example: Find the total energy converted to electrical energy over 5 minutes by the battery?
solution $$P=IV$$ $$P=(0.024)(1.5)$$ $$P=0.036W$$
$$P = \frac{E}{\Delta t}$$ $$E = P\Delta t$$ $$E = (0.036 \frac{J}{s})(5min)\left(\frac{60s}{1min}\right)$$ $$E = (0.036 \frac{J}{s})(300s)$$ $$E = 10.8J$$

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

$$(kilo)(Watt)(hour)$$ $$(1000)\left(\frac{J}{s}\right)(3600s)$$ $$(3,600,000)\left(\frac{J}{s}\right)(s)$$ $$1kWh = 3,600,000J$$ Example: The average price in America for 1 kiloWatt hour is \$0.12. How much would it cost to run a 800W microwave for 3 hours?
solution $$800W \left(\frac{k}{1000}\right) = 0.8kW$$ $$(0.8kW)(3h)\left(\frac{\0.12}{kWh}\right) = \20$$