Q: If two bulbs of power 25 W and 100 W respectively each rated at 220 V are connected in series with the supply of 440 V. Which bulb will fuse?
(a) 25 W bulb
(b) 100 W bulb
(c) none of these
(d) both
Click to See Answer :
Ans: (a)
Sol: Resistance of one bulb, R1 = (220)2/25 Ω ;
Resistance of other bulb, R2 = (220)2/100 Ω.
Total resistance when two bulbs are in series,
R = (220)2/25 + (220)2/100 = (220)2/20
current I = V/R = 440/((220)2/20) = 2/11 amp.
Pot. diff. across 25 watt bulb
= IR1 = 2/11 × (220)2/25 = 352 V
Pot. diff. across 100 watt bulb
= IR2 = 2/11 × (220)2/100 = 88 V
Thus, the bulb 25 W will be fused, because it can tolerate only 220 V while the voltage across it is 352 V.