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Current Electricity question

2016 · Q122
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Current Electricity question

2016 · Q122

NEETPhysicsCurrent ElectricityMCQ+4 / −1
A filament bulb (500 W, 100 V) is to be used in a 230 V main supply. When a resistance R is connected in series, it works perfectly and the bulb consumes 500 W. The value of R is
  1. A
    230 Ω\OmegaΩ
  2. B
    46 Ω\OmegaΩ
  3. C
    26 Ω\OmegaΩ
  4. D
    13 Ω\OmegaΩ
View written solutionFree

Correct answer: C

Resistance of bulb,
RB=V2P=(100)2500=20Ω{R_B} = {{{V^2}} \over P} = {{{{\left( {100} \right)}^2}} \over {500}} = 20\Omega RB​=PV2​=500(100)2​=20Ω

Power of the bulb in the circuit,
P = VI

I=PVBI = {P \over {{V_B}}}I=VB​P​

=500100=5A = {{500} \over {100}} = 5A=100500​=5A

VR = IR ⇒\Rightarrow⇒ (230 – 100) = 5 × R

∴\therefore∴ R = 26 Ω\Omega Ω

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