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Two electric bulbs are rated as (a) $100 W, 220 V$ , and (b) $40 W, 220 V$ . Calculate their resistances.

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474 Ω, 1200 Ω

464 Ω, 1230 Ω

484 Ω, 1210 Ω

454 Ω, 1220 Ω

answer is C.

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Detailed Solution

The resistance of bulb (a) is

484 Ω

and the resistance of bulb (b) is

1210 Ω

.
Given data in the problem,
For bulb (a)
Power,

P = 100 W

Voltage, V = 220 V

For bulb (b)
Power,

P = 40 W

Voltage,

V = 220 V

By using the formula of power,

Power = current \times voltage

\Rightarrow P = I \times V

As we know that,

I = \frac{V}{R}

, putting the value of I in the above equation,

\Rightarrow P = V \times \frac{V}{R}

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

So, for bulb (a)

\Rightarrow 100 = \frac{220^{2}}{R}

By rearranging the above equation,

R_{1} = 484 ohm

For bulb (b)

\Rightarrow 40 = \frac{220^{2}}{R}

By rearranging the above equation,

R_{2} = 1210 ohm

So, for the bulb (a) resistance is

484 ohms

.
And for the bulb (b) resistance is

1210 ohms

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