# 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) 230$\Omega$          (b) 46$\Omega$

(c) 26$\Omega$            (d) 13$\Omega$

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

then   ${P}_{rated}=\frac{{V}_{rated}^{2}}{R}$

$\therefore$Current in the bulb, $i=\frac{P}{V}$

$i=\frac{500}{100}=5A$

$\therefore$Resistance of bulb, ${R}_{b}=\frac{100×100}{500}=20\Omega$

$\because$Resistance R is connected in series.

$\therefore$ Current$i=\frac{E}{{R}_{net}}=\frac{230}{R+{R}_{0}}$

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