### Chapter 3: Electricity

Q
##### Electricity

Question:

An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be –

(a) 100 W      (b) 75 W       (c) 50 W       (d) 25 W

$P = VI$

$P = \frac{V^2}{R} -----(1)$

The resistance of the electric bulb remains constant, so we will first find the resistance of the bulb to find the power consumed by the bulb at 110 W.

Given:

$P = 100\ W$

$V = 220\ V$

Using relation (1), we get

$R = \frac{V^2}{P}$

$R = \frac{(220)^2}{100}$

$R = \frac{220 \times 220}{100}$

$R = 22 \times 22$

$R = 484\ \Omega ---------(2)$

Now, when the bulb is operated at 110 V, the electric power consumed by the bulb is given as

$P = \frac{V^2}{R}$

$R = 484\ \Omega$ {From (2)}

$V = 110\ V$ {Given}

$P = \frac{(110)^2}{484}$

$P = \frac{110 \times 110}{484}$

$P = 25\ W$

So, the correct option is (d).

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