Answer
The correct answer is:
C. $~~P \gt S = T \gt Q = R$
Work Step by Step
The power dissipated by each bulb is: $P = \frac{V^2}{R} = I^2~R$
The brightness depends on the power that is dissipated.
Since bulb Q and bulb R are in parallel, the equivalent resistance is less than the resistance of bulb T. Therefore, more current flows through the section with bulb P than the section with bulb S.
Therefore, bulb P is brighter than bulb S and bulb T, while bulb S and bulb T are equally bright with each other since they are connected in series.
Then the potential difference across bulb P is greater than the potential difference across bulb S. Then the potential difference across bulb both bulb Q and bulb R is less than the potential difference across bulb T.
Therefore, bulb Q and bulb R are less bright then bulb T.
The correct answer is:
C. $~~P \gt S = T \gt Q = R$
