Question

If two bulbs of power 60w and 100w respectively each rated 110v are connected in series with the supply of 220v then

Answer 1

Given :

Bulb A power= 60w

Bulb B power= 100w

$V_A=\ 110V\\ V_B=220v$

To find:

which bulb is fuse=?

Solution :

Following are the attachment of diagrammatic representation of the given question  

We know that

$P =I V=\frac{V^2}{R}$

For bulb A

$R_A=\frac{V^2}{P}=\frac{110^2}{60}=201.6\ ohm$

For bulb B

$R_B=\frac{110^2}{100}=\frac{110^2}{6100}=121\ ohm$

Total resistance

$R=R_A\ +\ R_B\\\R=201.6+121\\\=322.6\ ohm$

Therefore current

$I=\frac{220}{322.6}=0.68\ A$

The voltage at bulb A

$V_A=0.68\times 201.6=137.48\ V$

The voltage at bulb B

$V_B=0.68\times 121=82.28\ V$

Therefore bulb A will be fuse.

Answer 2

Hope it is helpful...

60W bulb gets fused.