# Heating Effect of Current

> Joule's law H = I²Rt, the three power formulas, series-parallel bulb logic and kWh units for NEET Physics.

- Canonical URL: https://prepelephant.com/topics/neet-ug/physics/current-heating-neet
- Exam / course: NEET-UG · Subject: Physics
- Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team
- First published: 2026-10-02
- Last updated: 2026-10-02
- How to cite: "Heating Effect of Current", PrepElephant, https://prepelephant.com/topics/neet-ug/physics/current-heating-neet

## Direct answer

Joule heating, H = I²Rt (joules), is the electrical face of friction: drifting electrons collide with the lattice, and the work done appears as heat — NCERT's microscopic picture of resistance. Three interchangeable power forms follow from V = IR: P = VI = I²R = V²/R, and choosing among them is the whole skill — current-fixed situations (series) make P ∝ R, voltage-fixed situations (parallel across mains) make P ∝ 1/R, a reversal that decides most bulb questions. Commercial energy is billed in units of 1 kilowatt-hour = 3.6 × 10^6 J, the "unit" of every Indian electricity bill. Numericals typically couple electrical power to heat absorbed, as in a kettle raising water to boiling.

## What you must remember

- **Joule's law:** H = I²Rt = VIt = V²t/R; in calories, divide by 4.18 (about 4.2) — mechanical-equivalent-of-heat bookkeeping.
- **Power trio:** P = VI = I²R = V²/R — same current through elements (series) rewards high R; same voltage across elements (parallel) rewards low R.
- **Series-parallel bulbs:** in series the higher-resistance (lower-wattage-rated) bulb glows brighter; in parallel each gets full mains voltage and the higher-rated (lower R) bulb dominates.
- **Rating arithmetic:** a bulb marked 100 W/220 V has R = V²/P = 484 Ω and draws I = P/V ≈ 0.45 A; run at half voltage it dissipates a quarter of the power.
- **The billing unit:** 1 kWh = 1 unit = 3.6 × 10^6 J; a 1 kW appliance for one hour consumes exactly one unit.
- **Material logic:** heating elements use nichrome (high resistivity, high melting point); filaments use tungsten, running near 3000 K; fuse wires melt at low melting points to protect circuits.
- **Heat-coupled problems:** time to heat a mass of water, t = mcΔθ/P, ignoring losses — the standard kettle numerical template.

## From kettle to electricity bill

An electric kettle rated 1 kW heats 2 kg of water from 25°C to 100°C. Heat needed: Q = mcΔθ = 2 × 4200 × 75 = 6.3 × 10^5 J. Time = Q/P = 630 s — about 10.5 minutes — consuming 1 kW × 0.175 h ≈ 0.175 units on the meter. Now audit a household: five 40 W lamps burning 5 hours a day use 5 × 40 × 5 = 1000 Wh = 1 unit daily; a 1 kW geyser for 2 hours adds 2 more; over 30 days, 90 units. Every billing numerical is this multiplication with different furniture. The subtler step is ratings: the "100 W" on a bulb means 100 W only at its rated 220 V; connected elsewhere, its true power is V²/R with R approximately fixed — hence the quarter-power rule at half voltage.

## Where NEET sets the trap

The series-bulb question is eternal: a 100 W and a 60 W bulb in series across mains — the 60 W glows brighter because its resistance is larger and the current is common; in parallel the 100 W wins. Options are engineered for whoever applies P = V²/R to the series case. The unit trap: energy in kWh versus joules (factor 3.6 × 10^6) and time in hours versus seconds; a "unit per day" question is pure bookkeeping. Fuse questions are conceptual: a 5 A fuse carries up to 5 A indefinitely and melts above it, protecting the circuit, not the appliance. Kettle numericals punish those who forget that only the useful heat counts — NEET states "neglect heat losses" precisely so t = mcΔθ/P stays exact.

## Frequently asked questions

### State Joule's law of heating.

The heat produced in a conductor is proportional to I², R and time: H = I²Rt joules, equivalently VIt or V²t/R through Ohm's law.

### A 100 W and a 60 W bulb are wired in series. Which glows brighter?

The 60 W bulb: with the same current in both, P = I²R favours the higher resistance, and lower wattage rating means higher resistance at rated voltage.

### How much energy is one unit of electricity?

One kilowatt-hour = 3.6 × 10^6 J — a 1000 W appliance running for one hour consumes one unit.

### How long will a 1 kW kettle take to heat 2 kg of water by 75°C?

t = mcΔθ/P = (2 × 4200 × 75)/1000 = 630 s ≈ 10.5 minutes, neglecting losses to the surroundings.

### Why is nichrome preferred for heating elements?

Its high resistivity gives ample I²R heating per length and its high melting point survives red-hot operation — copper would pass current but never serve as an element.