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 $Ω$ (b) 46 $Ω$

then $P_{r a t e d} = \frac{V_{r a t e d}^{2}}{R}$

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

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

$∴$ Resistance of bulb, $R_{b} = \frac{100 \times 100}{500} = 20 Ω$

$∵$ Resistance R is connected in series.

$∴$ Current $i = \frac{E}{R_{n e t}} = \frac{230}{R + R_{0}}$

$\Rightarrow R + 20 = \frac{230}{5} = 46 ∴ R = 26 Ω$

Difficulty Level:

(a)
22%
(b)
35%
(c)
39%
(d)
6%

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