Electric power

"60 W," "1200 W," "220 V" — the numbers stamped on every bulb, fan, heater and iron are a claim about electric power: the rate at which a device converts electrical energy into heat, light or motion. This is the closing quantitative idea of the CBSE Class 10 chapter Electricity, built directly on current, potential difference, resistance and the heating effect already covered. Here you meet the definition of power, its three interchangeable formulas $P = VI = I^2R = V^2/R$ , the SI unit watt, and the commercial unit kilowatt-hour that turns up on every electricity bill. The same appliance-rating numerical — "given 220 V and the wattage, find the current or resistance" — is one of the most repeated question types in the whole electricity chapter, so getting comfortable switching between the three formulas pays off directly in exams.

The core insight

Electric power is energy delivered per unit time, $P = VI = I^2R = V^2/R$ , measured in watts, with household energy billed in kilowatt-hours.

The core idea

Definition. If a device does electrical work $W$ (consumes energy $W$ ) in time $t$ , its power is the rate of that energy conversion:

P = \dfrac{W}{t}

where: $P$ is power (watt, W), $W$ is electrical energy consumed (joule, J), and $t$ is time (second, s).

From energy to the working formula. Electrical energy delivered in time $t$ is $W = VIt$ , so substituting gives the form used constantly in numericals:

P = VI

where: $V$ is potential difference across the device (volt, V) and $I$ is current through it (ampere, A).

Two more equivalent forms. Ohm's law, $V = IR$ , lets $P = VI$ be rewritten two more ways, so the same power can be reached from whichever pair of quantities is known:

P = VI = (IR)I = I^2R \qquad\qquad P = VI = V\left(\dfrac{V}{R}\right) = \dfrac{V^2}{R}

where: $R$ is resistance (ohm, Ω). All three forms — $P = VI$ , $P = I^2R$ , $P = V^2/R$ — describe the same physical power; only the algebra used to reach it changes with what data a question gives you. This is the same relationship behind Joule's law of heating, $H = I^2Rt = VIt = \dfrac{V^2}{R}t$ , since power is simply heat (or energy) delivered per unit time: $H = Pt$ .

Units. The SI unit of power is the watt: $1\ \text{W} = 1\ \text{J/s}$ — a 1 W device converts 1 joule of electrical energy every second. Larger appliances are rated in kilowatts, $1\ \text{kW} = 1000\ \text{W}$ .

Why the joule is too small for billing. Household energy consumption runs into millions of joules, so bills use a commercial unit, the kilowatt-hour (kWh) — informally a "unit" of electricity — defined as the energy used by a 1 kW device running for 1 hour:

1\ \text{kWh} = 1000\ \text{W} \times 3600\ \text{s} = 3.6\times10^{6}\ \text{J}

To find energy consumed by any appliance in kWh, use $\text{Energy (kWh)} = \dfrac{P(\text{W}) \times t(\text{h})}{1000}$ , equivalently $P(\text{kW}) \times t(\text{h})$ . The figure below traces this path: from an appliance's printed rating, through the three power formulas, to the energy bill.

Figure

A flow diagram showing a bulb rated 220 volts 100 watts feeding into three equivalent power formulas P equals V I, P equals I squared R, P equals V squared over R, which combine with time to give energy in kilowatt hours for the electricity bill

Appliance labels such as "220 V, 60 W" let you find the rated current or resistance directly from $P = VI$ or $P = V^2/R$ — this is the single most frequently asked numerical type in the chapter.

Derivation AGetting resistance and current straight from a bulb's printed rating

The goal is to show, starting only from the definition $P = W/t$ and Ohm's law $V = IR$ , exactly how an appliance's printed rating (voltage and wattage) yields both its rated current and its filament resistance, without ever measuring the resistance directly — this is the algebraic backbone behind almost every appliance-rating numerical in the electricity chapter.

Step 1. Start from the energy definition. In time $t$ , a device carrying current $I$ across potential difference $V$ moves charge $Q = It$ , and the work done moving that charge is $W = VQ = VIt$ . Dividing by $t$ gives power, $P = W/t = VI$ — this is the master formula, valid for any device regardless of whether it obeys Ohm's law.

Step 2. Bring in Ohm's law for a resistive appliance (a bulb filament, heater coil). Since $V = IR$ , substitute $V$ in $P = VI$ to eliminate $V$ : $P = (IR)I = I^2R$ . Rearranging for current gives $I = \sqrt{P/R}$ , but more usefully, if $P$ and $V$ are both known (as on a rating label), rearrange $P = VI$ directly for current: $I = P/V$ — no resistance needed yet.

Step 3. Now find resistance from the same label. Substitute $I = V/R$ into $P = VI$ instead: $P = V(V/R) = V^2/R$ . Rearranging gives $R = V^2/P$ — the filament resistance is recovered purely from the printed voltage and wattage, without any direct resistance measurement.

Step 4. Read off the two working results that solve every "given V and P, find I and R" question: $I = P/V$ and $R = V^2/P$ . Both trace back to the same pair of substitutions into $P = VI$ , so a single definition and Ohm's law generate all three power formulas and both rating-label results.

Worked example

An electric bulb is rated "220 V, 100 W." Find (a) the current drawn at rated voltage, (b) the filament resistance, and (c) the cost of running it for 5 hours a day for 30 days at ₹6 per kWh.

Given: $V = 220\ \text{V}$ , $P = 100\ \text{W}$ , $t = 5\ \text{h/day} \times 30\ \text{days} = 150\ \text{h}$ , rate = ₹6/kWh.

Step 1 — current. Using $I = P/V$ : $I = \dfrac{100}{220} \approx 0.4545\ \text{A} \approx 0.45\ \text{A}$ .

Step 2 — resistance. Using $R = V^2/P$ : $R = \dfrac{(220)^2}{100} = \dfrac{48400}{100} = 484\ \Omega$ .

Step 3 — energy consumed. Convert power to kilowatts, $P = 0.1\ \text{kW}$ , then $E = P \times t = 0.1 \times 150 = 15\ \text{kWh}$ .

Step 4 — cost. $\text{Cost} = 15\ \text{kWh} \times ₹6/\text{kWh} = ₹90$ .

Answer: the bulb draws about 0.45 A, has a filament resistance of 484 Ω, and costs ₹90 to run for 5 hours a day over 30 days.

Competitive stretch

If the same household also runs a 2 kW heater for 3 hours daily alongside the bulb, the heater alone adds $E = 2 \times 3 = 6\ \text{kWh/day} = 180\ \text{kWh}$ over 30 days, costing $180 \times 6 = ₹1080$ — showing how a single high-wattage appliance dwarfs a bulb's contribution to the bill, even though both obey the very same $P = VI$ relationship.

Common traps

Forgetting to convert watts to kilowatts (divide by 1000) before multiplying by hours — the most common arithmetic slip in bill numericals.
Using $P = I^2R$ or $P = V^2/R$ with mismatched units, such as plugging current in milliampere directly without converting to ampere.
Confusing energy (joules or kWh, total work done) with power (watts, the rate of doing it) — a "60 W" bulb left on longer consumes more energy, but its power rating itself never changes.
Misremembering $1\ \text{kWh} = 3.6\times10^{6}\ \text{J}$ , sometimes written as $3.6\times10^{3}$ or left unconverted entirely.

Ask the tutor: Want to work through your own household's monthly bill using these formulas, or need help deciding which power formula fits a tricky numerical?

Why are there three different formulas for power instead of just one?

All three, $P = VI$ , $P = I^2R$ and $P = V^2/R$ , describe the exact same physical power — they are just algebraic rearrangements of $P = VI$ using Ohm's law $V = IR$ . Having three forms is a convenience: a numerical might give you current and resistance, or voltage and resistance, or voltage and current, and picking the matching formula avoids an extra rearranging step. Always check which two quantities a question actually gives you before choosing which form to use.

Is the wattage printed on an appliance always the exact power it uses?

It is the power the appliance draws when connected at its rated voltage, usually 220 V in Indian households. If the actual supply voltage differs even slightly, the power drawn changes too, since $P = V^2/R$ means power depends on the square of voltage for a fixed resistance. This is also why voltage fluctuations can make bulbs unusually bright and short-lived, or dim — the printed wattage is a rating at the standard voltage, not a fixed, unchangeable quantity.

Why does the heater cost so much more than the bulb even though both use the same electricity?

Cost depends on total energy consumed, which is power multiplied by time, not on power alone. A 2 kW heater running for even a short while consumes far more energy per hour than a 100 W bulb, because its power rating is twenty times larger. Two appliances can be "on" for the same number of hours and still produce vastly different bills, because it is the wattage, not the running time, that usually makes the biggest difference to energy consumed.

Electric power

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads