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.
Bulbs P and Q are in parallel, and this combination is in series with bulb S.
Identical bulbs P and Q share current equally. Their parallel resistance is less
than the resistance of a single bulb.
Let R be the resistance of each bulb.
The equivalent resistance of P and Q in parallel is:
RPQ = R / 2
Total circuit resistance:
Rtotal = RS + RPQ = R + R/2 = 3R/2
The current splits equally between P and Q. Since brightness is proportional to
power (P = I²R), bulbs P and Q glow equally and brighter than S.
Brightness comparison:
P = Q > S
Case (ii): Key K opened
The circuit path through K is broken. Bulbs P and Q are now in series, and this
series combination is in parallel with bulb S.
Resistance of P and Q in series:
RPQ = R + R = 2R
This series branch is in parallel with bulb S (resistance R).
The total current divides into:
Branch 1: P + Q (2R)
Branch 2: S (R)
Since bulb S has lower resistance, more current flows through it, making it glow
brighter. Bulbs P and Q carry less current and glow dimmer.
Brightness comparison:
S > P = Q
