Question

A bulb is marked 100W, 220V and an electric heater is marked 2000 W, 220 V.
(i) What is the ratio between the resistances of these two devices?
(ii) How does the power-voltage rating of a device help us to decide about the type of leads (connecting wires) to be used for it?
(iii) In which of the above two devices, a thicker connecting wire of lead is required?

Answer 1

(i) For bulb, let R₁ be the resistance of its filament wire

P = V × I or W = V × I

100 = 200 × I = `(220 xx "V")/"R"_1 = (220 xx 220)/"R"_1`

∴ R₁ = `(220 xx 220)/100` Ω = 484 Ω

For heater, let its resistance be R₂ 

`2000 = 220 xx "I" = (220 xx 220)/"R"_2`

∴ R₂ = `(220 xx 220)/2000` Ω = 24.2 Ω

i.e., R₁ : R₂ = 20 : 1

(ii) In the case of bulb current I₁ = `"W"/"V" = 100/200 = 5/11` A = 0.45 A 

We have (i) and (ii)

In the case of heater current =`"W"/"V" = 2000/220` = 9.09 A.

Hence power-voltage rating help us in this case. The current through the bulb is only 0.45 A while through the heater it is 9.09 A. Hence a heavy lead (to avoid power loss due to heating effect) is needed for the heater which for a bulb an ordinary thin connecting wire is required.

(iii) Electric heater requires a thicker wire of lead.