Question

You obtain a 100-W light bulb and a 50-W light bulb. Instead of connecting them in the normal way, you devise a circuit that places them in series across normal household voltage. If each one is an incandescent bulb of fixed resistance, which statement about these bulbs is correct?

Answer 1

When they are connected in series

The 50 W bulb glow more than the 100 W bulb

Step-by-step explanation:

From the question we are told that

The power rating of the first bulb is
`P_1 = 100 \ W`

The power rating of the second bulb is
`P_2 = 50 \ W`

Generally the power rating of the first bulb is mathematically represented as

`P_1 = V^2 R`

Where
`V` is the normal household voltage which is constant for both bulbs

So

`R_1 = (V^2)/(P_1 )`

substituting values

`R_1 = (V^2)/(100)`

Thus the resistance of the second bulb would be evaluated as

`R_2 = (V^2)/(50)`

From the above calculation we see that

`R_2 > R_1`

This power rating of the first bulb can also be represented mathematically as

`P_ 1 = I^2_1 R_1`

This power rating of the first bulb can also be represented mathematically as

`P_ 2 = I^2_2 R_2`

Now given that they are connected in series which implies that the same current flow through them so

`I_1^2 = I_2^2`

This means that

`P \ \alpha \ R`

So when they are connected in series

`P_2 > P_1`

This means that the 50 W bulb glows more than the 100 \ W bulb