NEET Physics Alternating Current Questions Solved

NEET - 2016

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 

(a) 230Ω          (b) 46Ω

(c) 26Ω            (d) 13Ω

(c) If a rated voltage and power are given,

 then   Prated=Vrated2R

Current in the bulb, i=PV

                            i=500100=5A

Resistance of bulb, Rb=100×100500=20Ω

Resistance R is connected in series.

 Currenti=ERnet=230R+R0

 R+20=2305=46   R=26Ω 

Difficulty Level:

  • 22%
  • 35%
  • 39%
  • 6%
Crack NEET with Online Course - Free Trial (Offer Valid Till September 21, 2019)