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My answer to this question was:

"The brightness of each lamp would increase due to more voltage available to spread across the now 39 lamps (since one burns out)"

This, however, is incorrect. Instead, the answer is given as "the entire circuit of lamps do not light"

Could someone explain why this is so?

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When an incandescent bulb burns out, the filament inside breaks, so it is essentially an open switch, i.e. it "fails open". Since the bulbs are in series, once any of them fails, no current can flow and no bulb will light.

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  • $\begingroup$ Thanks, Puk. The second part of the problem wants us to describe what happens to the brightness of the lamps when in the alternative set. Following on from the reasoning you've given in your answer, I said for much of the same reason, the lamp filaments cannot light as charge cannot flow along that branch. This however is wrong. Instead they've said the lamp will "light more dimly". Could you explain why this is the case pls? $\endgroup$
    – LearningCHelpMeV2
    Commented May 6, 2022 at 17:45
  • $\begingroup$ @LearningCHelpMeV2 The resistor in parallel with the broken filament still allows current through, so the other lamps will light. Knowing that, you should think of how the broken filament impacts the equivalent resistance of the whole circuit and how this impacts the brightness of the other lamps. $\endgroup$
    – Puk
    Commented May 6, 2022 at 18:03

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