NDA Physics · Electricity and Magnetism

Electrical Power, Energy and Heating

Electrical power is the rate of energy delivery — P = VI = I²R = V²/R; energy is power × time (billed in kilowatt-hours), and current passing through a resistance dissipates that energy as heat (Joule heating, H = I²Rt).

All Electricity and Magnetism notes NDA Physics strategy

Why this matters

Ten PYQs spanning the three power formulas, the 'which expression is NOT power' recall trap, how an appliance's power changes when run at the wrong voltage (P ∝ V²), the cost of energy in kilowatt-hours, and Joule heating — including the HARD parallel-vs-series heat ratio. Knowing WHEN to use I²R versus V²/R is the whole game.

Concept 1 of 5

Electrical power — three equivalent forms

Intuition

Power is energy delivered per second. The base formula is P = VI (voltage × current). Substituting Ohm's law gives two more forms — P = I²R and P = V²/R — which are handy when you happen to know current-and-resistance or voltage-and-resistance instead.

Definition

Electrical power (rate of energy use), in watts: ** $P = VI = I^2 R = \dfrac{V^2}{R}$ ** — three equivalent forms via Ohm's law $V = IR$ . Use $P = VI$ always; $P = I^2R$ when current and resistance are known; $P = V^2/R$ when voltage and resistance are known. Expressions like $IR^2$ or $I^2/R$ are NOT power (wrong dimensions).

Electrical power

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

Ppower (watt)
Vvoltage (volt)
Icurrent (ampere)
Rresistance (ohm)

Worked example

A bulb connected to a 220 V supply draws a current of 600 mA. What is its power?

Use $P = VI$ ; convert 600 mA = 0.6 A.
$P = 220 \times 0.6 = 132$ W.

Answer:132 W.

Practice this conceptself-check · 3 quick reps

From the bank · past-year question

Example 1Electricity and MagnetismEASY

An electric bulb is connected to 220 V generator. The current drawn is 600 mA. What is the power of the bulb?

[Q66 · Sep · 2022]

I²R is power; IR² and I²/R are not

The bank tests this by dimensions. P = I²R ✓ and P = V²/R ✓. But IR² and I²/R are NOT power — check by substituting V = IR if unsure. Memorise the three valid forms and spot the impostor.

Drill 3 more on electrical power — three equivalent forms

Concept 2 of 5

Power rating and running at the wrong voltage

Intuition

A bulb stamped '220 V, 100 W' only delivers 100 W AT 220 V. Its resistance is fixed, so if you run it at a lower voltage the power drops — and because P = V²/R, it drops with the SQUARE of the voltage. Halving the voltage quarters the power.

Definition

An appliance rated $(V_0, P_0)$ has a fixed resistance $R = V_0^2 / P_0$ . Run at a different voltage $V$ , its actual power is $P = V^2 / R$ , so ** $P \propto V^2$ ** (resistance fixed): halving the voltage gives one-quarter the power.

Power vs voltage at fixed R

P = \dfrac{V^2}{R}, \quad R = \dfrac{V_0^2}{P_0} \;\Rightarrow\; \dfrac{P}{P_0} = \left(\dfrac{V}{V_0}\right)^2

Worked example

A bulb rated 60 W, 120 V is run on a 60 V supply. What power does it actually consume?

Its resistance is fixed: $R = V_0^2/P_0 = 120^2/60 = 240\,\Omega$ .
At 60 V: $P = V^2/R = 60^2/240 = 3600/240 = 15$ W.
Shortcut: voltage halved ⟹ power ×(1/2)² = 1/4 ⟹ $60/4 = 15$ W.

Answer:15 W.

Practice this conceptself-check · 3 quick reps

From the bank · past-year question

Example 2Electricity and MagnetismMODERATE

An electric bulb is rated as 220 V and 80 W. When it is operated on 110 V, the power rating would be :

[Q147 · Apr · 2023]

Power scales as V², not V

Running a bulb at half voltage does NOT halve the power — it quarters it. The resistance is fixed by the rating, so P = V²/R falls with the square of the voltage. Picking 40 W (half) instead of 20 W (quarter) is the standard trap.

Concept 3 of 5

Electrical energy and the cost of running appliances

Intuition

Energy is power kept up over time. Electricity bills measure it in kilowatt-hours: run 1 kW for 1 hour and you've used 1 'unit'. Multiply units by the price per unit to get the cost.

Definition

Electrical energy = power × time. The commercial unit is the kilowatt-hour (kWh): $1\text{ kWh} = 1\text{ kW} \times 1\text{ h}$ = one 'unit'. Energy in kWh = (power in kW) × (time in hours). Cost = (number of units) × (rate per unit).

Energy and cost

E\,(\text{kWh}) = P\,(\text{kW}) \times t\,(\text{h}), \qquad \text{Cost} = E \times \text{rate}

Worked example

A 2 kW geyser runs for 3 hours a day for 10 days. At ₹5 per unit, what is the cost?

Energy per day: $2 \times 3 = 6$ kWh.
Over 10 days: $6 \times 10 = 60$ units.
Cost: $60 \times 5 = ₹300$ .

Answer:₹300.

Practice this conceptself-check · 3 quick reps

From the bank · past-year question

Example 3Electricity and MagnetismEASY

The cost of energy to operate an industrial refrigerator that consumes 5 kW power working 10 hours per day for 30 days will be (Given that the charge per kW.h of energy = ₹ 4)

[Q90 · Apr · 2020]

Keep power in kW and time in hours

kWh = kW × hours. Don't convert power to watts or time to seconds for billing problems — that buries you in 10⁶ factors. Units × rate per unit gives the cost directly.

Drill 1 more on electrical energy and the cost of running appliances

Concept 4 of 5

Joule heating — current heats a resistor

Intuition

Whenever current flows through a resistance, electrical energy turns into heat. The amount depends on three things together: the voltage applied, the current driven, and how long it runs. Heaters, irons, and geysers all live on this effect.

Definition

Joule's law of heating: heat produced $H = I^2 R t = VIt = \dfrac{V^2}{R}\,t$ (joules). It depends on the current, the resistance/voltage, AND the time — all three. For a heating coil on a fixed supply, the temperature rise grows with the supply voltage, the current, and the time the voltage is applied.

Joule heating

H = I^2 R\,t = V I t = \dfrac{V^2}{R}\,t

Hheat produced (joule)
Icurrent (A)
Rresistance (Ω)
ttime (s)

Worked example

A 5 Ω heating coil carries 4 A for 2 minutes. How much heat does it produce?

Use $H = I^2 R t$ ; convert time: 2 min = 120 s.
$H = 4^2 \times 5 \times 120 = 16 \times 5 \times 120$ .
$H = 80 \times 120 = 9600$ J.

Answer:9600 J.

Practice this conceptself-check · 3 quick reps

From the bank · past-year question

Example 4Electricity and MagnetismMODERATE

Water is heated with a coil of resistance R connected to domestic supply. The rise of temperature of water will depend on 1. Supply voltage. 2. Current passing through the coil. 3. Time for which voltage is supplied. Select the correct answer from among the following:

[Q82 · Sep · 2019]

Heat depends on all of V, I, and t

Multi-statement questions try to drop one factor. H = VIt contains voltage, current, AND time — the temperature rise needs all three. Don't select 'current and time only'.

Concept 5 of 5

Heat dissipation in series vs parallel

Intuition

Connect resistors across the SAME voltage and the one with the SMALLER resistance burns more power (P = V²/R). So a parallel combination (small equivalent R) dissipates more heat than the same resistors in series (large equivalent R). For two equal resistors at the same voltage, parallel beats series by a factor of 4.

Definition

At a fixed voltage, power dissipated $P = V^2/R$ — so SMALLER equivalent resistance ⟹ MORE heat. Parallel gives a smaller equivalent than series, so a parallel combination dissipates more. For two equal resistors at the same applied voltage: $R_\parallel = R/2$ , $R_\text{series} = 2R$ , so $\dfrac{P_\parallel}{P_\text{series}} = \dfrac{R_\text{series}}{R_\parallel} = \dfrac{2R}{R/2} = 4$ .

Worked example

Two equal resistors are connected across the same battery, first in parallel and then in series. What is the ratio of heat produced (parallel : series)?

Same voltage V across each arrangement; $P = V^2/R_\text{eq}$ .
Parallel equivalent $= R/2$ ; series equivalent $= 2R$ .
$\dfrac{P_\parallel}{P_\text{series}} = \dfrac{V^2/(R/2)}{V^2/(2R)} = \dfrac{2R}{R/2} = 4$ .

Answer:4 : 1 (parallel produces four times the heat).

Practice this conceptself-check · 3 quick reps

From the bank · past-year question

Example 5Electricity and MagnetismHARD

Two conducting wires of the same material and of equal lengths and equal diameters are first connected in parallel and then in series in a circuit across the same potential difference. The ratio of heat produced in parallel and series combinations is

[Q128 · Apr · 2025]

Same VOLTAGE → use V²/R; don't reach for I²R

When the two arrangements share the same applied voltage, power is V²/R, so smaller resistance means more heat (parallel wins). Reaching for I²R here misleads, because the current is different in each arrangement. The ratio is 4 : 1, and picking the inverted 1 : 4 is the dominant trap.

Drill 1 more on heat dissipation in series vs parallel

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (4)

Electrical power — three equivalent forms
Electrical power
$P = VI = I^2 R = \dfrac{V^2}{R}$
Power rating and running at the wrong voltage
Power vs voltage at fixed R
$P = \dfrac{V^2}{R}, \quad R = \dfrac{V_0^2}{P_0} \;\Rightarrow\; \dfrac{P}{P_0} = \left(\dfrac{V}{V_0}\right)^2$
Electrical energy and the cost of running appliances
Energy and cost
$E\,(\text{kWh}) = P\,(\text{kW}) \times t\,(\text{h}), \qquad \text{Cost} = E \times \text{rate}$
Joule heating — current heats a resistor
Joule heating
$H = I^2 R\,t = V I t = \dfrac{V^2}{R}\,t$

Watch out for (5)

I²R is power; IR² and I²/R are not
→ Electrical power — three equivalent forms
Power scales as V², not V
→ Power rating and running at the wrong voltage
Keep power in kW and time in hours
→ Electrical energy and the cost of running appliances
Heat depends on all of V, I, and t
→ Joule heating — current heats a resistor
Same VOLTAGE → use V²/R; don't reach for I²R
→ Heat dissipation in series vs parallel

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