JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Current Electricity

Two electric bulbs, rated at

(25 W, 220 V)

and

(100 W, 220 V)

, are connected in series across a 220 V voltage source. If the 25 W and 100 W bulbs draw powers

P_{1}

and

P_{2}

respectively, then:

Answer: (A)

Solution

A bulb rated

(P, V)

has resistance

R = \frac{V ^{2}}{P}

, so the

25 W

bulb is

R_{1} = \frac{22 0 ^{2}}{25} = 1936 Ω

and the

100 W

bulb is

R_{2} = \frac{22 0 ^{2}}{100} = 484 Ω

; notice the lower-wattage bulb has the larger resistance. In series they carry the same current

i = \frac{220}{1936 + 484} = \frac{220}{2420} = \frac{1}{11} A

. The power each actually dissipates is

i^{2} R

: for the

25 W

bulb

P_{1} = (\frac{1}{11})^{2} (1936) = 16 W

, and for the

100 W

bulb

P_{2} = (\frac{1}{11})^{2} (484) = 4 W

. In series the bulb rated for less power actually glows brighter.

JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Single correct option