Question

A bulb is marked 100 W – 200 V and an electric heater is marked 1000 W – 200 V. Answer the following questions.

1. What is the ratio of resistance of filament of the bulb and the element of heater?
2. How does the power voltage rating of an electric appliance help us to decide the type of connecting wire (leads) to be used for it?
3. In the above mentioned devices in (b) which of the devices needs thicker wire?

Answer 1

(i) Resistance is given by R = `V^2/P`

For bulb: `R_b = (200)^2/100 = 40000/100` = 400 Ω

For heater: `R_h = (200)^2/1000 = 40000/1000` = 40 Ω

Therefore, the ratio of the resistance of the bulb filament to the heater element = 400 : 40 = 10 : 1.

(ii) The power and voltage rating tell us the normal current drawn by the appliance using I = `P/V`.

This current helps us choose a connecting wire that can safely carry that current without overheating. A higher current requires a thicker wire or a wire of higher current-carrying capacity.

(iii) The electric heater needs a thicker wire because it draws more current.

Current drawn by bulb = `100/200` = 0.5 A

Current drawn by heater = `1000/200` = 5 A

Since the heater draws a much larger current, it needs a thicker connecting wire.