Question

When bulb $1$ is screwed in, but bulb $2$ is unscrewed, the power generated in bulb $1$ is:

 $$\begin{gathered} \left(A\right)4watts \\ \left(B\right)12watts \\ \left(C\right)36watts \\ \left(D\right)48watts \end{gathered}$$

Answer 1

Hint :In order to solve this question, we are going to calculate the power generated in the bulb $1$ by taking the lower loop that is the screwed in one, for that we need the value of voltage developed across the bulb $1$ and the resistance of the bulb $1$ . Using these, we can easily calculate the power generated.
The power generated for any circuit is given by
 $P_1={\displaystyle \frac{V^2}{R_1}}$

Complete Step By Step Answer:
As it is given in this question, that the lower loop bulb is screwed in, this means that it is the lower loop only that is active, which shows that the power generated will be only due to the bulb $1$ , now that the power generated is due to the bulb $1$ , we must know the value of the resistance of the bulb $1$ and the voltage across the lower loop
The power generated for any circuit is given by
 $P_1={\displaystyle \frac{V^2}{R_1}}$
Now the power supply has maintained the voltage equal to $12V$ across the lower loop $1$ , now if the resistance of the bulb $1$ is $3\mathrm{\Omega }$
Then, the power generated for this circuit can be calculated as
  $$\begin{gathered} P_1={\displaystyle \frac{{\left(12\right)}^2}{3}}={\displaystyle \frac{144}{3}} \\ \Rightarrow P_1=48W \end{gathered}$$
Hence the power generated in bulb $1$ is $48Watts$
Thus, option $\left(D\right)48watts$ is the correct answer.

Note :
It is to be noted that as the upper loop with the bulb $2$ is the unscrewed one, that means that this circuit is an open one while as the bulb $1$ is screwed in, it is a closed circuit that supplies a proper amount of current and maintains a proper voltage across the bulb $1$ . That’s why bulb $1$ is the only active one and the power generated for it is calculated.