Two electric bulbs A and B rated 200V~100W and 200V~60W are connected in series to a 200V line.Then the potential drop across
1)Each bulb is 200V
2)100W bulb is greater than that across 60W bulb
3)100W bulb is smaller than that across 60W bulb
4)Each bulb is 100V

Dear Student,

Please find below the solution to the asked query:

Given information is,

first bulb rating is 200 V - 100 W. So, the resistance of the first bulb is,

$$\begin{gathered} P_1=\frac{V_1^2}{R_1} \Rightarrow R_1=\frac{V_1^2}{P_1}=\frac{{200}^2}{100}=\frac{40000}{100} \\ \Rightarrow R_1=400 \Omega \end{gathered}$$

second bulb ratings is 200 V - 60 W. Therefore,

$$\begin{gathered} P_2=\frac{V_2^2}{R_2} \Rightarrow R_2=\frac{V_2^2}{P_2}=\frac{{200}^2}{60}=\frac{40000}{60} \\ \Rightarrow R_2=\frac{2000}{3} \Omega \end{gathered}$$
 

Therefore, when two bulbs are connected in series, the effective resistance is,


$$\begin{gathered} R_{eff}=R_1+R_2 \Rightarrow R_{eff}=400+\frac{2000}{3}=\frac{1200+2000}{3} \\ \Rightarrow R_{eff}=\frac{3200}{3} \Omega \end{gathered}$$

So, current in the circuit is,


$$\begin{gathered} i=\frac{V}{R_{eff}}=\frac{200}{{\displaystyle \raisebox{1ex}{$3200$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}=\frac{600}{3200} \\ \Rightarrow i=\frac{3}{16} A \end{gathered}$$

The potential drop across each bulb is,

across 100 W bulb is,

$$\begin{gathered} V_1=i{R}_1=\frac{3}{16}\times 400=3\times 25 \\ \Rightarrow V_1=75 V \end{gathered}$$

across 60 W bulb is,

$$\begin{gathered} V_2=i{R}_2=\frac{3}{16}\times \frac{2000}{3} \\ \Rightarrow V_2=125 V \end{gathered}$$

So, the potential drop across 100 W bulb is less than that of the potential drop across 60 W.

 

Hope this information will clear your doubts about the topic.

If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.

Regards