A bulb is rated 150W-220V and a heater is rated 1500W-200V. (i) Which device needs a thicker wire? (ii) Find the ratio of their resistances. (iii) Which device will use a fuse of 1A?
Answers
Explanation:
(i) For bulb, let R1 be the resistance of its filament wire
P = V x I or W = V x I
100 = 220 x I = 220 x V/R1 = 220 x 220/R1
∴ R1 = 220 x 220/100Ω = 484Ω
For heater, let its resistance be R2
2000 = 220 x I = 220 x 220/R2
∴ R2 = 220 x 220/2000Ω = 24.2Ω
i.e., R1 : R2 = 20 : 1.
(ii) In the case of bulb current I1 = W/V = 100/200 = 5/11A = 0.45 A
We have (i) and (ii)
In the case of heater current = W/V = 2000/220 = 9.09A
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.