When a 40 watt lamp is connected in series with a 100 watt lamp across a 220 V supply, which one glows brighter 

Dear Student,
Answer to your question is given below:
Assumption: V is taken 200 volts here, but the answer will be same for this type of questions for even V=220 volts
We know that, $P=\frac{V^2}{R}$
For the 40 W bulb:
     $40= \frac{{200}^2}{R}$
$$\begin{gathered} \Rightarrow R = \frac{40000}{40} \\ \Rightarrow R = 1000 \Omega \end{gathered}$$


For the 100 W bulb:
   $100 = \frac{{200}^2}{R}$
$$\begin{gathered} \Rightarrow R = \frac{40000}{100} \\ \Rightarrow R = 400 \Omega \end{gathered}$$

The total current produced by the resistor in series:
     $I = \frac{V}{R}$
$$\begin{gathered} \Rightarrow I = \frac{220}{(1000+400)} \\ \Rightarrow I = \frac{220}{1400} \\ \Rightarrow I =0.157 A \end{gathered}$$

Now we will calculate the power in each bulb by using voltage divider.
For the 40 W bulb:
$$\begin{gathered} V = \frac{1000}{(100+400)}\times 220 \\ \Rightarrow V = \frac{1000}{1400}\times 220 \\ \Rightarrow V = 157.14 volts \end{gathered}$$

Hence, P = V x I = 157.14  x 0.157 = 24.67 W

For the 100 W bulb:
$$\begin{gathered} V = \frac{400}{(1000+400)}\times 220 \\ \Rightarrow V = \frac{400}{1400}\times 220 \\ \Rightarrow V = 62.86 volts \end{gathered}$$

Hence, P = V x I = 62.86  x 0.157 = 9.86 W

Therefore, the 40W bulb glows brighter.

Regards,