JEE Main 2020 · Physics · Current Electricity · easy

Two bulbs rated 100 W, 200 V and 60 W, 200 V are connected in series across a 200 V supply. Assuming constant filament resistances, the total power…

Single correct option

Two bulbs rated

100\ \mathrm{W}

200\ \mathrm{V}

and

60\ \mathrm{W}

200\ \mathrm{V}

are connected in series across a

200\ \mathrm{V}

supply. Assuming constant filament resistances, the total power consumed by the pair is

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Answer: (D) $37.5\ \mathrm{W}$

Resistances:

R_1 = V^2/P_1 = 40000/100 = 400\ \Omega

and

R_2 = 40000/60 = 666.7\ \Omega

. In series

R = 1066.7\ \Omega

, so the power is

P = V^2/R = 40000/1066.7 = 37.5\ \mathrm{W}

. Shortcut:

\dfrac{1}{P} = \dfrac{1}{P_1} + \dfrac{1}{P_2} = \dfrac1{100}+\dfrac1{60}

gives

P = 37.5\ \mathrm{W}

. Adding the ratings (

160\ \mathrm{W}

) is the usual trap.

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