Electrical power

Current through a resistance delivers energy every second — heat in a heater, light in a bulb. The rate is the power.

Source: Serway & Jewett, Physics for Scientists and Engineers, 7th ed., §27.6.

Before you start

What you need first

Topic 16 — current $I$.
Topic 17 — Ohm's law $V = IR$.
Topic 9 — energy = charge × voltage.

What you'll be able to do

Use $P = VI$ and the watt.
Use the three forms $P = VI = I^2R = V^2/R$.
Find energy used, $E = Pt$.
Work in kilowatt-hours and find a running cost.

The rate of energy

Electric power

Power is the energy delivered each second. For a current $I$ driven through a voltage $V$:

$$ P = VI $$

The unit is the watt (W = J/s). A "60 W" bulb turns 60 joules of electrical energy into light and heat every second.

A battery drives current $I$ through a resistor $R$; the resistor dissipates power $P = I^2R$ as heat.

where:
Symbol	Meaning	SI unit
P	power (energy each second)	W = J/s
V	voltage across the device	V
I	current through it	A

Three ways to write power

Substituting Ohm's law ($V = IR$) into $P = VI$ gives three equal forms:

$$ P = VI = I^2R = \frac{V^2}{R} $$

Pick the one that matches what you're given. The $I^2R$ form is the heat dissipated in a resistor; $V^2/R$ is handy when the voltage is fixed (like the mains).

📐 Worked example 1

Power two ways

(a) A device runs at $V = 12$ V and draws $I = 2.0$ A — find the power. (b) A current $I = 3.0$ A flows through $R = 4.0\ \Omega$ — find the heat power.

1From $P = VI$:

$$ P = (12)(2.0) = 24\ \text{W} $$

2From $P = I^2R$:

$$ P = (3.0)^2(4.0) = 36\ \text{W} $$

Energy used — and the electricity bill

Energy is power times time:

$$ E = Pt $$

In joules if $t$ is in seconds. But bills use the kilowatt-hour (kWh): running 1 kW for 1 hour is 1 kWh. The cost is just energy (kWh) × price per kWh.

📐 Worked example 2

Running cost of a bulb

A 100 W (= 0.10 kW) bulb runs for 5.0 hours. Electricity costs 4 baht per kWh. Find the energy used and the cost.

1Energy in kWh, $E = Pt$:

$$ E = (0.10\ \text{kW})(5.0\ \text{h}) = 0.50\ \text{kWh} $$

2Cost = energy × price:

$$ \text{cost} = 0.50 \times 4 = 2\ \text{baht} $$

✏️ Try it yourself

A hair dryer is rated 1500 W on a 230 V supply.

(a) Find the current it draws.
(b) Find its resistance.
(c) Find the energy used (in kWh) and the cost of running it 5 minutes at 4 baht/kWh.

(a) $I = P/V = 1500/230 \approx 6.5\ \text{A}.$ (b) $R = V^2/P = 230^2/1500 \approx 35\ \Omega$ (or $V/I$). (c) $E = (1.5\ \text{kW})(5/60\ \text{h}) \approx 0.125\ \text{kWh}$; cost $= 0.125 \times 4 = 0.50$ baht.

Common mistakes

Mistake	Fix
Forgetting to square in $I^2R$ or $V^2/R$.	The current (or voltage) is squared — a common slip.
Mixing the three forms.	$VI$, $I^2R$, $V^2/R$ are equal; don't combine pieces of different ones.
Treating a kWh as a power.	A kWh is energy (power × time), not power.
Leaving time in minutes for $E=Pt$ in joules.	Use seconds for joules (or kW and hours for kWh).

Recap — the whole topic on one screen

Idea	What you own now
Power	$P = VI = I^2R = V^2/R$; watt = J/s.
Energy	$E = Pt$ (joules), or kWh for bills.
Cost	energy (kWh) × price per kWh.

Next topic

EMF & internal resistance

A real battery isn't perfect — it has its own internal resistance, so the voltage you actually get drops as you draw more current. Next we separate the ideal EMF from that internal loss.

→ Topic 19 · EMF & internal resistance

Mistake	Fix
Forgetting to square in \(I^2R\) or \(V^2/R\).	The current (or voltage) is squared — a common slip.
Mixing the three forms.	\(VI\), \(I^2R\), \(V^2/R\) are equal; don't combine pieces of different ones.
Treating a kWh as a power.	A kWh is energy (power × time), not power.
Leaving time in minutes for \(E=Pt\) in joules.	Use seconds for joules (or kW and hours for kWh).

Idea	What you own now
Power	\(P = VI = I^2R = V^2/R\); watt = J/s.
Energy	\(E = Pt\) (joules), or kWh for bills.
Cost	energy (kWh) × price per kWh.