Heating Effect of Current

On this page
  1. Direct answer
  2. What you must remember
  3. From kettle to electricity bill
  4. Where NEET sets the trap
  5. Frequently asked questions
  6. Related topics

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.

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