Heating Effect of Current: Joule's Law of Heating

Physics · Current Electricity · NEET

When current I flows through a resistor R for time t, electrical energy turns into heat. Joule's law of heating says the heat produced is H = I²Rt (also written as VIt or V²t/R). The key idea: heat depends on the SQUARE of current, so doubling the current gives four times the heat. Memory hook: "I-squared-R-t" — say it like a chant, because I², not I, decides the heat.
Joule heating: H = I²Rt = V²t/R = VItVR (gets hot)IheatSeries (same I): use I²Rtbigger R → hotterParallel (same V): use V²t/Rsmaller R → hotterHeat depends on I² — doubling current gives 4× the heat.
Current I through resistance R dissipates heat H = I²Rt. Match the formula to what is fixed: in series (same current) the bigger R heats more; in parallel (same voltage) the smaller R heats more.

Your doubts, answered

Is the heat formula I²Rt or V²t/R — which one do I use?

Both are correct because V = IR, so I²Rt = (V²/R²)Rt = V²t/R = VIt. They are the same law written three ways. The trick is to pick the form where the quantity that stays FIXED is on top. If the SAME CURRENT flows (series circuit), use H = I²Rt, so more R means more heat. If the SAME VOLTAGE is applied (parallel across a supply), use H = V²t/R, so more R means LESS heat. Choosing the right form for the situation is what NEET tests.

If two resistors have the same voltage, which one gets hotter?

When voltage V is the same across both (they are in parallel), use H = V²t/R. Heat is inversely proportional to R, so the SMALLER resistance gets hotter. Example (NEET 2022): 100 Ω and 200 Ω in parallel — heat ratio H₁/H₂ = R₂/R₁ = 200/100 = 2:1, so the 100 Ω resistor produces twice the heat.

If the same current flows, which resistor heats more?

When the SAME current I flows (resistors in series), use H = I²Rt. Heat is directly proportional to R, so the LARGER resistance gets hotter. This is the opposite of the same-voltage case. Series → bigger R hotter; Parallel → smaller R hotter. Mixing these two up is the single most common mistake in this topic.

Why does heat depend on I² and not just I?

Power dissipated is P = VI, and by Ohm's law V = IR, so P = (IR)·I = I²R. The I comes twice — once as the charge flow rate and once as the voltage drop it causes — so it appears squared. Because I² is always positive, heating happens whether the current is positive or negative (this is why AC still heats, and why the ampere is defined by heating effect).

A charge varies with time as Q = at − bt². How do I find total heat?

Heat needs current, not charge. First get i = dQ/dt. Then heat over the time the current flows is H = ∫i²R dt. You cannot use H = I²Rt directly because the current is not constant — you must integrate. This exact style appeared in NEET 2016.

⚠️ The NEET trap
Same formula for series and parallel — students blindly use H = V²t/R everywhere and conclude the bigger resistor always heats less.
Match the formula to what is constant. Series (same current) → H = I²Rt → bigger R is hotter. Parallel (same voltage) → H = V²t/R → smaller R is hotter. The answer flips depending on the connection.
🧠 Same CURRENT? Big R burns. Same VOLTAGE? Small R burns.

Real NEET questions

NEET 2016

The charge through a resistance R varies with time as Q = at − bt² (a, b are positive constants). The total heat produced in R is:

A · a³R/(6b)
B · a³R/(3b)
C · a³R/(2b)
D · a³R/b
Solution: Current i = dQ/dt = a − 2bt. Current stops when i = 0, i.e. at t = a/(2b). Heat H = ∫i²R dt from 0 to a/(2b) = R∫(a − 2bt)² dt. Let the integral evaluate: R·[(a−2bt)³/(−6b)] from 0 to a/(2b) = R·[0 − (a³/(−6b))] = a³R/(6b). Because current is not constant, you MUST integrate, not use I²Rt.
NEET 2022

Two resistors 100 Ω and 200 Ω are connected in parallel. The ratio of thermal energy developed in 100 Ω to that in 200 Ω in a given time is:

A · 1 : 2
B · 2 : 1
C · 1 : 4
D · 4 : 1
Solution: In parallel the voltage V is the same across both, so use H = V²t/R. Then H₁/H₂ = R₂/R₁ = 200/100 = 2 : 1. The smaller resistance (100 Ω) develops more heat. Trap: if you wrongly used I²Rt you would flip the answer to 1:2.
NEET 2026 (Phase 1)

A room heater is rated 400 W, 220 V. If the supply voltage drops to 200 V, the power consumed (approximately) is:

A · 200 W
B · 400 W
C · 331 W
D · 121 W
Solution: The heater's resistance R is fixed. With fixed R, P = V²/R, so P ∝ V². Thus P₂ = P₁·(V₂/V₁)² = 400·(200/220)² = 400·(0.909)² ≈ 400·0.826 ≈ 331 W. Do not assume power stays 400 W — the rating only holds at the rated voltage.

Solved Current Electricity NEET PYQs

Try the real previous-year questions from this chapter — each with the answer and a full solution.

See all 48 Current Electricity NEET PYQs ›
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Frequently asked

What is Joule's law of heating?

It states that the heat produced in a resistor equals H = I²Rt, where I is the current, R the resistance, and t the time. Equivalently H = VIt = V²t/R. Heat is directly proportional to the square of current, to resistance, and to time.

What are the three forms of the heating formula?

H = I²Rt (use when current is fixed, i.e. series), H = V²t/R (use when voltage is fixed, i.e. parallel across a supply), and H = VIt (general). All three are equal because V = IR.

What is the SI unit of heat produced?

The joule (J), since H is energy. Power of heating P = H/t is in watts (W). One kilowatt-hour (unit of electricity) = 3.6 × 10⁶ J.

Why does a fuse work on the heating effect?

A fuse is a thin wire with high resistivity and a low melting point. When current becomes too large, the I²Rt heating melts the fuse and breaks the circuit, protecting appliances. This was directly asked in NEET 2019.

Does the heating effect happen in AC too?

Yes. Since heat depends on I² which is always positive, current in either direction produces heat. This is why AC bulbs and heaters work, and why the rms value of AC current is defined through its heating effect.