12 min readConcept sheetSection A MCQ (1 mark) and match-the-columns parts on units; Section B short answer (3 marks) numerical on heat, power or the units of electricity used

Heating effect, electric power and the kilowatt hour

Current heats a resistor by H = i²Rt (Joule's law). Power is P = Vi = i²R = V²/R, and bills count energy in kilowatt hours (units).

Joule's law of heatingH = i²Rt; heat in joules (4.18 J = 1 cal)

Electric powerP = Vi = i²R = V²/R; unit watt (W); 1 W = 1 V × 1 A

Commercial unit of energy1 kWh = 1 unit = 3.6 × 10⁶ J

Other units of power1 kW = 1000 W; 1 MW = 10⁶ W; 1 horse power (hp) = 746 W

Why a current heats a wire: Joule's law

1
A potential difference V across a resistor R drives a current i for a time t.
2
The charge that moves is Q = it. The work done on it is W = VQ = Vit.
3
By Ohm's law V = iR, so W = (iR) × it = i²Rt.
4
This work appears as heat, so H = i²Rt. This is Joule's law of heating.
5
Inside the wire, the drifting electrons keep colliding with the atoms. The atoms vibrate more, so the wire gets hotter.
6★
So the heat is proportional to the square of the current, to the resistance and to the time. Double the current and the heat is four times.

★ key step

Activity: heating effect with heater coils

You need two pieces of heater coil (10 turns and 20 turns), two dry cells and connecting wires. Touch the coil only after the current is off.

1
Take two short pieces of a heater element, one with 10 turns and one with 20 turns. Fix connecting wires to the ends of the 10-turn coil.
2
Touch the wires to the terminals of a dry cell for 10 seconds, then let go. Carefully feel the coil.
3
Repeat, passing the current for 20 seconds. Observe: the coil is warmer.
4
Join two dry cells in series and repeat for 10 seconds. Observe: a larger p.d. gives a hotter coil.
5
Repeat the steps with the 20-turn coil.
With the same cells, the longer coil gets less hot, because its larger resistance lets less current flow.
6
Conclude.
A current heats a conductor. More heat is made with a larger p.d., a larger current or a longer time.

Appliances that use the heating effect

Appliance or use	How it uses the heat
Electric iron, kettle, toaster, oven, cooking range, room heater	A coil of high resistance (usually nichrome) gets very hot
Immersion rod and electric geyser	A heating element in the water warms it
Electric bulb	A thin filament gets white-hot and gives out light
Electric fuse	A thin wire of low melting point melts and breaks the circuit when the current is too large
Electric welding and the electric arc	Very large currents make enough heat to melt and join metals

Electric power and electrical energy

Electric powerThe rate at which electrical energy is used: P = W/t = Vi. Using V = iR, also P = i²R = V²/R.

Watt1 W is the power used when a current of 1 A flows under a p.d. of 1 V.

Bigger units1 kW = 1000 W; 1 MW = 10⁶ W; 1 GW = 10⁹ W; 1 hp = 746 W.

Energy usedEnergy = power × time. A joule is too small for daily use, so we use the watt hour and the kilowatt hour.

1 kilowatt hour (kWh)The energy used by a 1 kW appliance in 1 hour: 1000 W × 3600 s = 3.6 × 10⁶ J.

Unit on the bill1 kWh is the commercial unit of electrical energy, also called the Board of Trade unit, or simply 1 unit.

Mains supplyHomes in India get AC at 230 V and 50 Hz (its peak is about 325 V). Your NIOS lesson and older books say 220 V.

Heat and power

Problem

(a) A current of 3 A flows through a heating coil of resistance 20 Ω for 2 minutes. Find the heat produced. (b) An electric iron is rated 1150 W, 230 V. Find the current it draws and its resistance.

Working

(a) Given: i = 3 A, R = 20 Ω, t = 2 min = 120 s. Formula: H = i²Rt.
H = (3 A)² × 20 Ω × 120 s = 9 × 20 × 120 J = 21,600 J = 2.16 × 10⁴ J.
(b) Given: P = 1150 W, V = 230 V. Formula: P = Vi, so i = P/V.
i = 1150 W ÷ 230 V = 5 A.
R = V²/P = (230 × 230) ÷ 1150 = 52,900 ÷ 1150 = 46 Ω. Check: R = V/i = 230 ÷ 5 = 46 Ω.

Answer

(a) 2.16 × 10⁴ J; (b) 5 A and 46 Ω

Counting units of electricity

Problem

A family uses a 1.5 kW geyser for 40 minutes a day and four 15 W LED bulbs for 6 hours a day. Find the units used in 30 days and the cost at an example rate of ₹ 6 per unit.

Working

Geyser: 1.5 kW × (40/60) h = 1.5 × 2/3 = 1.0 kWh per day.
Bulbs: 4 × 15 W × 6 h = 360 Wh = 0.36 kWh per day.
Total per day = 1.0 + 0.36 = 1.36 kWh.
In 30 days: 1.36 kWh × 30 = 40.8 kWh = 40.8 units.
Cost = 40.8 units × ₹ 6 per unit = ₹ 244.80.

Answer

40.8 units; ₹ 244.80 at the example rate of ₹ 6 per unit

In real life

An immersion rod, a kettle and a toaster all turn electrical energy into heat in a high-resistance coil. The meter at home counts units: a 1000 W iron used for one hour uses exactly 1 unit (1 kWh).

Exam angle

1 marks: “What is the commercial unit of electrical energy? Write its value in joules.” The kilowatt hour (kWh). 1 kWh = 3.6 × 10⁶ J.

2 marks: “How does the heat produced in a wire change if the current through it is made three times, all else staying the same?” H = i²Rt, so the heat depends on the square of the current. Three times the current gives 3² = 9 times the heat.

3 marks: “An electric kettle rated 2 kW is used for 15 minutes every day. Find the units of electricity it uses in 30 days, and the energy in joules used each day.” Each day: 2 kW × 0.25 h = 0.5 kWh. In 30 days: 0.5 × 30 = 15 units. Daily energy = 0.5 × 3.6 × 10⁶ J = 1.8 × 10⁶ J.

Remember it

Joule's heat is 'i squared R t': double the current, four times the heat. 1 unit = 1 kWh = 3.6 million joules.