In the series, the current is the same in the bulb and the extra resistance, which is given by,
I=VP=100500=5A.
But, Vbulb+VR=Vnet.
So, voltage across resistance R will be,
VR=Vnet−Vbulb
⇒VR=230−100=130V.
From Ohm's law,
R=IVR=5130=26Ω.
NEET UG 2016 — Physics Electromagnetism
A filament bulb (500W,100V) is to be used in a 230V main supply. When a resistance, R is connected in series, it works perfectly and the bulb consumes 500W. The value of R is
Held on 30 Apr 2016 · Verified 9 Jul 2026.
230Ω
46Ω
26Ω
13Ω
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