A resistor turns electrical energy into thermal energy: that is how toasters, heaters and old-style light bulbs work. In Electric Power and Energy Use you met $P = IV$. Combined with $V = IR$ and what you know about resistance, it comes in three forms.
Why it works
Each coulomb that crosses a resistor drops through a potential difference $V$, so it gives up $V$ joules. If $I$ coulombs cross each second, the resistor receives $IV$ joules per second: $P = IV$. Substituting Ohm's law: with $V = IR$, $P = I^2R$; with $I = V/R$, $P = V^2/R$. They are one idea; pick the form that uses what you know.
Key idea
$$P = IV = I^2R = \frac{V^2}{R}$$ Same current (series): the bigger resistor uses more power. Same voltage (parallel): the smaller resistor uses more power. Energy $=$ power $\times$ time; $1\,\text{kWh} = 3.6 \times 10^{6}\,\text{J}$.
Worked examples
Worked example: A lamp
A lamp on $120\,\text{V}$ draws $0.50\,\text{A}$. Find its power and its resistance.
- $P = IV = (0.50)(120) = 60\,\text{W}$.
- $R = \dfrac{V}{I} = 240\,\Omega$. Check: $V^2/R = 14{,}400/240 = 60\,\text{W}$.
Answer: $60\,\text{W}$; $240\,\Omega$
Worked example: Series: bigger wins
A $2\,\Omega$ and a $4\,\Omega$ resistor are in series on $12\,\text{V}$. Find each one's power.
- $I = \dfrac{12}{2 + 4} = 2.0\,\text{A}$ through both.
- $P_2 = (2.0)^2(2) = 8.0\,\text{W}$ and $P_4 = (2.0)^2(4) = 16\,\text{W}$.
Answer: $8.0\,\text{W}$ and $16\,\text{W}$
Worked example: Parallel: smaller wins
The same resistors are put in parallel on $12\,\text{V}$. Find each one's power.
- Each has the full $12\,\text{V}$.
- $P_2 = \dfrac{12^2}{2} = 72\,\text{W}$ and $P_4 = \dfrac{12^2}{4} = 36\,\text{W}$.
Answer: $72\,\text{W}$ and $36\,\text{W}$
Common mistake
Concluding that a bigger resistor always uses more power. That is true only when the current is the same (series). With the same voltage (parallel), the smaller resistor lets more current through and uses more power, which is why a $1500\,\text{W}$ heater has a smaller resistance than a $60\,\text{W}$ lamp on the same outlet. Check: first decide what is shared, $I$ or $V$, then use $I^2R$ or $V^2/R$.