Electrical power and energy

Look at any electricity bill and you are reading a receipt for electrical energy, priced not in joules but in a friendlier unit called the kilowatt-hour. Behind that number sits a chain of simple ideas: power is the rate at which a circuit element converts electrical energy into heat, light, or motion, and multiplying power by time gives the total energy used. Because $V$ , $I$ , and $R$ are linked by Ohm's law, the same power can be written three different ways depending on what a problem gives you. This section ties those formulas together, shows where the heat in a heater or the "units" on a bill actually come from, and works through the numbers examiners love to test — heater ratings, Joule heating, and monthly appliance costs.

The core insight

Electrical power $P = VI = I^2R = V^2/R$ multiplied by time gives energy $W = Pt$ , billed commercially in kilowatt-hours and, for pure resistive heating, expressed by Joule's law $H = I^2Rt$ .

The core idea

Power is a rate. Whenever charge flows through a circuit element under a potential difference, energy is converted every second — into heat in a resistor, light in a bulb, or motion in a motor. Electrical power is defined as

P = VI

where: $P$ is power in watts (W), $V$ is potential difference in volts, and $I$ is current in amperes.

Three equivalent faces. Substituting Ohm's law $V = IR$ turns this single formula into three interchangeable ones, so you can always pick the version that matches the quantities a question gives you:

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

Use $P=VI$ when $V$ and $I$ are both known, $P=I^2R$ when $I$ and $R$ are known, and $P=V^2/R$ when $V$ and $R$ are known.

Energy is power times time. Since power is energy per second, the total electrical energy converted in a time $t$ is

W = Pt

where: $W$ is energy in joules when $P$ is in watts and $t$ is in seconds. Because a joule is too small a unit for household consumption, bills use the kilowatt-hour (kWh), or "unit," equal to $3.6\times10^6$ J — the energy used by a 1 kW appliance running for one hour.

Joule's law of heating. For a purely resistive device, all the electrical energy delivered to it turns into heat, so the heat produced over time $t$ follows directly from $P=I^2R$ multiplied by time:

H = I^2Rt

where: $H$ is heat in joules, $I$ is current in amperes, $R$ is resistance in ohms, and $t$ is time in seconds. This is exactly why devices built to produce heat — electric heaters, geysers, toasters, fuse wires — are designed with a resistive element carrying current, and why the wattage printed on such an appliance already tells you the rate at which it generates heat.

Reading a household bill. An energy meter simply totals up $\sum Pt$ for every appliance running in a home over a billing period, converts the sum to kilowatt-hours, and multiplies by the per-unit tariff. Any appliance with a stated power rating and a known daily running time can be folded into this sum one term at a time, which is exactly the pattern examiners test in multi-appliance billing numericals.

Figure

Three equivalent power formula boxes VI I squared R and V squared over R all pointing into one power output box

Derivation AThe three equivalent forms of the power formula

Given a resistor of resistance $R$ carrying current $I$ under potential difference $V$ , show algebraically that the electrical power it dissipates can be written in the three equivalent forms $P=VI$ , $P=I^2R$ , and $P=V^2/R$ , using nothing beyond the definition of power and Ohm's law.

Step 1. Start from the definition of electrical power delivered to a circuit element: $P = VI$ , where $V$ is the potential difference across it and $I$ is the current through it. This is the fundamental form — it holds for any device, resistive or not.

Step 2. Restrict attention to a purely resistive element, for which Ohm's law applies: $V = IR$ . Substitute this expression for $V$ directly into $P = VI$ :

P = (IR)\, I = I^2R

This form is useful whenever the current and the resistance are the given quantities, and it shows power growing with the square of the current.

Step 3. Alternatively, solve Ohm's law for the current, $I = V/R$ , and substitute into $P = VI$ instead:

P = V\left(\frac{V}{R}\right) = \frac{V^2}{R}

This form is useful whenever the voltage and resistance are given directly, such as an appliance's rated voltage and its resistance.

Step 4. All three expressions describe the same physical power — they are algebraic rearrangements of one another via Ohm's law, not three separate laws. Choosing the right one is purely a matter of which two quantities a problem hands you.

Worked example

An electric heater has resistance $40\ \Omega$ and is connected to a $220\ \text{V}$ supply.

Step 1. Find the power rating using $P = V^2/R$ since $V$ and $R$ are given: $P = \dfrac{220^2}{40} = \dfrac{48400}{40} = 1210\ \text{W}$ .

Step 2. Find the heat produced in 5 minutes. Convert time: $5\ \text{min} = 300\ \text{s}$ . Since a heater converts all electrical energy to heat, $H = Pt = 1210 \times 300 = 3.63\times10^5\ \text{J}$ .

Step 3. If this heater runs for 2 hours a day and electricity costs ₹6 per unit, find the daily cost. Energy used $= 1.21\ \text{kW} \times 2\ \text{h} = 2.42\ \text{kWh}$ , so cost $= 2.42 \times ₹6 = ₹14.52$ per day.

The heater is rated 1210 W, produces $3.63\times10^5$ J of heat in 5 minutes, and costs ₹14.52 per day when run for 2 hours.

Competitive stretch

A household separately runs a $2\ \text{kW}$ geyser for $1.5$ hours and five $60\ \text{W}$ bulbs for $4$ hours every day, at the same ₹6 per unit. Step 1. Daily geyser energy: $2\ \text{kW}\times1.5\ \text{h}=3\ \text{kWh}$ . Step 2. Daily bulb energy: total bulb power $=5\times60=300\ \text{W}=0.3\ \text{kW}$ , so $0.3\ \text{kW}\times4\ \text{h}=1.2\ \text{kWh}$ . Step 3. Total over 30 days: $(3+1.2)\ \text{kWh}\times30=126\ \text{kWh}$ , costing $126\times₹6=₹756$ . This confirms the same billing method scales cleanly to several appliances at once: convert each to kW and hours, sum the kWh, then multiply by the rate.

Common traps

Mixing units when computing kWh — power must be in kilowatts and time in hours before multiplying; watts-with-hours or kilowatts-with-seconds give wrong "unit" counts.
Picking a power formula that doesn't match the given quantities, e.g. reaching for $P=I^2R$ when only $V$ and $R$ are known instead of using $P=V^2/R$ .
Applying $H=I^2Rt$ to a device that does mechanical work, like a motor — that formula gives only the heat lost to resistance, not the total energy supplied.
Confusing joules with kilowatt-hours — remember $1\ \text{kWh} = 3.6\times10^6\ \text{J}$ , and never quote a bill in joules or plug kWh into a formula expecting SI seconds.
Forgetting to add up power ratings correctly for identical appliances — five $60\ \text{W}$ bulbs draw a combined $300\ \text{W}$ , not $60\ \text{W}$ , so always multiply rated power by the count of identical devices before converting to kW.

Ask the tutor: Want to try a mixed household-billing problem with two or three appliances, or double-check which power formula fits a tricky numerical?

Why do we need three different power formulas instead of just one?

All three describe the same power — they are just Ohm's law substituted into $P=VI$ in different ways. A real numerical rarely gives you all of $V$ , $I$ , and $R$ at once, so having three equivalent forms lets you compute power directly from whichever two quantities the problem actually states, without an extra algebra step to find the missing one first.

Why does a motor not fit Joule's law the way a heater does?

Joule's law $H=I^2Rt$ gives only the heat produced by resistance in the wire, which is one part of the electrical energy supplied to a motor. A motor also converts most of its input into mechanical work (turning a shaft), so $I^2Rt$ underestimates the total energy — it only accounts for the wasted resistive heating, not the useful output.

Why does my electricity bill use kilowatt-hours instead of joules?

A joule is a tiny amount of energy — running a 1000 W heater for just one second uses 1000 J, so a month's household consumption would be a number in the billions of joules, unwieldy to print on a bill. The kilowatt-hour packages a much larger, everyday-sized chunk of energy (3.6 million joules) so that a typical month's usage comes out as a manageable two- or three-digit number.

Electrical power and energy

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads