When a voltage source, such as a battery, creates current for a circuit, the majority of the energy is consumed to keep the current flowing. A portion of the energy, however, is wasted in the form of heat. The full energy in a circuit comprising simply a battery and resistance is dissipated as heat. This is the current’s heating impact.

The heating effect of an electric current depends on three factors:

- The resistance, R of the conductor. A higher resistance produces more heat.
- The time, t for which current flows. The longer the time the larger the amount of heat produced
- The amount of current, I. the higher the current the larger the amount of heat generated.

*Power = Work Done / Time TakenorP = W / t*

Since, *Potential Difference = Work Done / Charge**orV = W / Q*

*P = VQ / t*or *P = VI*, where *I (current) = Charge / Time Taken*

Energy supplied by power source to a circuit in time t is, *H = P x t = VIt*

Or from Ohm’s law, *H = I ^{2}Rt*

This is known as **Joule’s law of heating**. It states that heat produces in a circuit is directly proportional to the square of current flowing, resistance for current and time for which current flows.

Disadvantages of heating effect

- Loss of energy in the unwanted heat.
- Wear and tear of components.

There are four effects of electricity : **The ‘effects of electricity’** **are: **

- magnetic effect.
- heating effect.
- chemical effect.
- electric shocks.

**Example**

- An electrical bulb is labeled 100W, 240V. Calculate:

a)The current through the filament when the bulb works normally b)The resistance of the filament used in the bulb.

**Solution**

- I = P/V = 100/240 = 0.4167A
- R = P/I
^{2}= 100/ 0.4167^{2}= 576.04Ω or R = V^{2}/P =240^{2}/100 = 576Ω - Find the energy dissipated in 5 minutes by an electric bulb with a filament of resistance of 500Ω connected to a 240V supply. {
**ans. 34,560J**}

**Solution**

E = Pt = V2/R *t = (240^{2} *5*60)/500 = 34,560J.

- A 2.5 kW immersion heater is used to heat water. Calculate:

- The operating voltage of the heater if its resistance is 24Ω
- The electrical energy converted to heat energy in 2 hours.

{**ans. 244.9488V, 1.8*10 ^{7}J**}

**Solution**

- P=VI=I
^{2}R

I = (2500/24)^{1/2} =10.2062A

V=IR= 10.2062 * 24 = 244.9488V

- E = VIt = Pt = 2500*2*60*60 = 1.8 * 10
^{7}J

OR E= VIt = 244.9488 * 10.2062 * 2 * 60 * 60 = 1.8 * 10^{7}JAn electric bulb is labeled 100W, 240V. Calculate:

The current through the filament

The resistance of the filament used in the bulb.

**Solution**P = VI I = P/V = 100/240 =0.4167A

From Ohm’s law, V =IR R=V/I =240/0.4167 = 575.95Ω