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

Answer:

When they are connected in series

     The  50 W bulb glow more than the 100 W bulb

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  

        $R_1 = \frac{V^2}{P_1 }$

substituting values

        $R_1 = \frac{V^2}{100}$

Thus the resistance of the second bulb would be evaluated as

       $R_2 = \frac{V^2}{50}$

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