Parallel branch reconfiguration increases P's current, decreases total current

Steps:

• Initially, P and Q in parallel (equiv. r/2) in series with R (r); total resistance 1.5r; battery current 2V/(3r).
• P current V/(3r); R current 2V/(3r).
• Q breaking opens its path; now P (r) in series with R (r); total resistance 2r; battery current V/(2r).
• P current V/(2r) > previous, brighter; R current V/(2r) < previous, dimmer.

Why B is correct:

• By Ohm's law, Q's failure routes all parallel-branch current through P alone (higher voltage drop across P) while total current falls, per increased equivalent resistance.

Why the others are wrong:

• A: Total current drops (resistance rises from 1.5r to 2r), so R cannot brighten.
• C: P's current rises from V/(3r) to V/(2r), so it brightens.
• D: P's current increases, so it does not dim.

Final answer: B