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If two bulbs, whose resistances are in the ratio of 1:2 are connected in series, the power dissipated in them has the ratio of

Asked in CBSE AIPMT 1997 · Rated bulbs and appliances

(1) 2:1
(2) 1:4
(3) 1:1
(4) 1:2

Answer: (4) 1:2

Step-by-step solution

In series the same current I flows through both bulbs.

P=I²R, so P∝ R.

P₁:P₂=R₁:R₂=1:2.

Why the other options are wrong

(1) Uses P=(V²)/R as if both bulbs had the same voltage (a parallel connection); in series it is the current that is common.
(2) Squares the resistance ratio, as if P∝ R².
(3) Assumes equal current means equal power; the power also depends on each bulb's resistance.

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