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Current Electricity question

2003 · Q130
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Current Electricity question

2003 · Q130

NEETPhysicsCurrent ElectricityMCQ+4 / −1
Two 220 volt, 100 watt bulbs are connected first in series and then in parallel. Each time the combination is connected to a 220 volt a.c. supply line. The power drawn by the combination in each case respectively will be
  1. A
    50 watt, 100 watt
  2. B
    100 watt, 50 watt
  3. C
    200 watt, 150 watt
  4. D
    50 watt, 200 watt
View written solutionFree

Correct answer: D

R=V2P=220×220100=484ΩR = {{{V^2}} \over P} = {{220 \times 220} \over {100}} = 484\Omega R=PV2​=100220×220​=484Ω

In series, Req = 484 + 484 = 968 Ω\Omega Ω

∴Peq=V2968=220×220968=50 watt \therefore {P_{eq}} = {{{V^2}} \over {968}} = {{220 \times 220} \over {968}} = 50\,watt∴Peq​=968V2​=968220×220​=50watt

In parallel, Req=242Ω{R_{eq}} = 242\Omega Req​=242Ω

∴Peq=V2242=220×220242=200 watt \therefore {P_{eq}} = {{{V^2}} \over {242}} = {{220 \times 220} \over {242}} = 200\,watt∴Peq​=242V2​=242220×220​=200watt

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