In the given figure, three identical bulbs P, Q, and S are connected to a battery. [(i)] Compare the brightness of bulbs P and Q with that of bulb S when key K is closed.
[(ii)] Compare the brightness of the bulbs S and Q when the key K is opened.
Justify your answer in both cases.
Show Hint
In circuits with identical bulbs, brightness is proportional to power \( P = I^2 R \). For comparison, check how current divides using equivalent resistances.
Case (i): When key K is closed
- Bulbs P and Q are in parallel with each other. This parallel combination is in series with bulb S.
- Since P and Q are identical and in parallel, they share the current equally, and the equivalent resistance of this parallel part is less than the resistance of a single bulb.
- Let \( R \) be the resistance of each bulb. The parallel combination of P and Q gives \( R_{\text{PQ}} = \frac{R}{2} \).
- The total resistance of the circuit becomes:
\[
R_{\text{total}} = R_S + R_{\text{PQ}} = R + \frac{R}{2} = \frac{3R}{2}
\]
- Current divides equally through P and Q, and since brightness is proportional to power \( P = I^2R \), both P and Q glow equally and brighter than S (as more current flows through them). Brightness comparison:
P = Q \(>\) S Case (ii): When key K is opened
- The path through K is broken, so P and Q are now in series. This series combination is in parallel with bulb S.
- Total resistance of P and Q in series = \( R + R = 2R \)
- This combination is in parallel with S, so:
\[
\text{Total current from battery is divided between two branches:}
\text{Branch 1: P + Q (2R)}, \quad \text{Branch 2: S (R)}
\]
- Since S has lower resistance than the PQ branch, more current flows through S, making it brighter.
- P and Q have less current (as they are in series and in higher resistance path), and hence glow dimmer. Brightness comparison:
S \(>\) P = Q Final Answer:
\begin{itemize}
[(i)] When key K is closed: P = Q \(>\) S [(ii)] When key K is opened: S \(>\) P = Q
\end{itemize}
