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 / ChargeorV = W / Q
P = VQ / tor 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 = I2Rt
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/I2 = 100/ 0.41672 = 576.04Ω or R = V2/P =2402/100 = 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 = (2402 *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*107J}
Solution
- P=VI=I2R
I = (2500/24)1/2 =10.2062A
V=IR= 10.2062 * 24 = 244.9488V
- E = VIt = Pt = 2500*2*60*60 = 1.8 * 107J
OR E= VIt = 244.9488 * 10.2062 * 2 * 60 * 60 = 1.8 * 107JAn electric bulb is labeled 100W, 240V. Calculate:
The current through the filament
The resistance of the filament used in the bulb.
SolutionP = VI I = P/V = 100/240 =0.4167A
From Ohm’s law, V =IR R=V/I =240/0.4167 = 575.95Ω