Explain heating effect of current with examples

Question

Explain the heating effect of electric current. Derive Joule’s law and give three practical examples.

Solution — Step by Step

When an electric current flows through a conductor, the free electrons (charge carriers) move through the lattice of metal ions. These electrons collide with the ions as they drift through. Each collision transfers kinetic energy from the electrons to the ions, making the ions vibrate more — this increased vibration is what we perceive as heat.

The conductor resists the flow of electrons (this is resistance, $R$ ). More resistance = more collisions = more heat generated.

The work done in moving charge $Q$ through a potential difference $V$ is:

W = VQ

Since current $I = Q/t$ , we have $Q = It$ . Also, by Ohm’s law, $V = IR$ . Substituting:

W = V \cdot It = IR \cdot It = I^2 Rt

This work appears entirely as heat in a resistor:

\boxed{H = I^2 R t}

This is Joule’s Law of Heating: Heat generated is proportional to $I^2$ , $R$ , and $t$ .

Joule’s Law states:

$H \propto I^2$ (for constant $R$ and $t$ ) — doubling current quadruples heat
$H \propto R$ (for constant $I$ and $t$ ) — higher resistance generates more heat
$H \propto t$ (for constant $I$ and $R$ ) — longer time generates more heat

1. Electric bulb (incandescent): The tungsten filament has high resistance and a very high melting point. When current flows, heat builds up until the filament glows white-hot, producing light. The high melting point of tungsten (3422°C) prevents it from burning out quickly.

2. Electric heater / iron: The heating element (nichrome wire) has high resistance. When current flows, large amounts of heat are produced. Nichrome (nickel-chromium alloy) is used because it doesn’t oxidise easily at high temperatures.

3. Electric fuse: A thin wire of low-melting-point alloy (usually tin-lead) is used. When excess current flows (during a short circuit), $I^2Rt$ produces so much heat that the fuse wire melts, breaking the circuit and protecting appliances.

Why This Works

The fundamental reason for Joule heating is energy conservation. The electrical energy supplied by the source ( $VIt$ ) must go somewhere — in a pure resistor, it all goes into heat. There is no storage in a resistor.

The $I^2$ dependence is crucial: it means small increases in current cause large increases in heating. This is why transmission lines operate at very high voltage (and correspondingly low current for the same power $P = VI$ ) — reducing current by 10× reduces transmission line heating by 100×.

CBSE Class 7–10 and JEE frequently test: “Which bulb produces more heat — a 60W or a 100W bulb connected to the same supply?” Answer: 100W. Since $P = V^2/R$ , higher power means lower resistance, which means more current and more heating.

Alternative Method

Using power: $P = I^2R = V^2/R = VI$ .

Heat generated in time $t$ : $H = Pt = I^2Rt$ .

If you know power directly, use $H = Pt$ . This is the fastest approach for numerical questions.

Common Mistake

Students often write $H \propto I$ instead of $H \propto I^2$ . The current appears squared in Joule’s law — this is because both the number of collisions per second AND the energy per collision increase with current. Don’t confuse it with $P = VI$ , where $I$ appears to the first power (but $V$ also depends on $I$ through Ohm’s law, giving the $I^2$ when substituted).

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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