Question

Two bulbs 100W,250V and 200W,250V are connected in series across a 500V line. Then what will happen?? a)100W will fuse b)200W will fuse c)both will fuse d)no bulb will fuse

Answer 1

The 100W bulb will fuse soon.

Explanation:

Power =`V^2/R`, so Resitance R= `V^2/P`

The 100W bulb has a resistance =`(250 *250)/100`=625 ohms
The 200 W bulb resistance will be half above = 312.5ohms
Total resistance in series - 937.5 ohms

So total series current = `V/R` = `500/937.5` =0.533A

Power dissipated in Bulb 1: `I^2*R` = `0.533^2*625` =177.5W
Power dissipated in Bulb 2 wil be half above: 88.5 W

Bulb1 , a 100W unit, will eventually burnout.