Effect of an Electric Current

1. Current (I) — Heat produced increases with the square of the current.
2. Resistance (R) — Heat produced increases with resistance.
3. Time (t) — The longer the current flows, the more heat is produced.

Mathematically:

$H \propto I^2Rt \Rightarrow H = KI^2Rt$

Where K is a constant (K = 1 in SI units), so:

$H = I^2Rt$

By Ohm's Law, $V = IR$, so:

$H = VIt \quad \text{or} \quad H = \frac{V^2t}{R}$

This is known as Joule's Law of Heating.

One joule is the work done when a charge of one coulomb flows through a conductor with a potential difference of one volt in one second.

An electric kettle has a wire of resistance 5Ω. 1kg of water is to be heated from 300K to 373K in 10 minutes. What current must flow through the wire?

Given:

$C = 4200 \, J/kg·K$

$\Delta T = 373 - 300 = 73K$

$M = 1kg$

$R = 5Ω$

$t = 10 \text{min} = 600 \text{s}$

Step 1: Thermal energy required

$H = MCQ = 1 \times 4200 \times 73 = 306600 \, J$

Step 2: Use Joule's law:

$H = I^2Rt \Rightarrow 306600 = I^2 \times 5 \times 600$

$I^2 = \frac{306600}{3000} = 102.2 \Rightarrow I = \sqrt{102.2} = 10.1 \, A$

Power (P) is the rate at which electrical energy is delivered:

$P = \frac{\text{Electrical Energy}}{\text{Time}} = \frac{I^2Rt}{t} = I^2R$

Also, since $V = IR$:

$P = IV$

An electric iron is rated 240V, 2000W. Find:

(a) Current:

$I = \frac{P}{V} = \frac{2000}{240} = 8.33A$

(b) Resistance:

$P = \frac{V^2}{R} \Rightarrow R = \frac{V^2}{P} = \frac{240^2}{2000} = \frac{57600}{2000} = 28.8Ω$

The commercial unit of electrical energy is the Kilowatt-hour (kWh).

$1 \, \text{kWh} = 1000 \, W \times 1 \, \text{hour} = 1000 \times 3600 = 3.6 \times 10^6 \, J$

An electric iron rated 240V, 2000W is used for 10 hours.

Cost of electrical energy = Tshs 100 per unit (1 unit = 1kWh).

Energy used:

$2000W \times 10hr = 20000Wh = 20kWh$

Cost:

$1 \, \text{kWh} = 100 \, \text{Tsh} \Rightarrow 20 \, \text{kWh} = 20 \times 100 = 2000 \, \text{Tsh}$

Cost of using the appliance = 2000 Tsh.