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An electric bulb consumes 25 W of power when it is operated at 110 V. What should be the correct rating of the bulb?

A. 60 W, 220 V
B. 100 W, 220 V
C. 150 W, 220 V
D. 80 W, 220 V

Answer :

The correct rating is 100 W, 220 V, calculated using [tex]\( P = \frac{{V^2}}{R} \) with \( P = 25 \) W and \( V = 110 \)[/tex] V.

To determine the correct rating of the bulb, we can use the formula:

[tex]\[ P = \frac{{V^2}}{R} \][/tex]

Where:

[tex]- \( P \) = Power consumed by the bulb (in watts)[/tex]

[tex]- \( V \) = Voltage across the bulb (in volts)[/tex]

[tex]- \( R \) = Resistance of the bulb (in ohms)[/tex]

Given that [tex]\( P = 25 \) W and \( V = 110 \) V, we can rearrange the formula to solve for \( R \):[/tex]

[tex]\[ R = \frac{{V^2}}{P} \][/tex]

[tex]\[ R = \frac{{110^2}}{25} \][/tex]

[tex]\[ R = \frac{{12100}}{25} \][/tex]

[tex]\[ R = 484 \][/tex]

Now, to find the correct rating of the bulb, we can use the formula:

[tex]\[ P = \frac{{V^2}}{R} \][/tex]

Let's try each option:

[tex]a. \( P = \frac{{220^2}}{484} = 100 \) W[/tex]

[tex]b. \( P = \frac{{220^2}}{484} = 100 \) W[/tex]

[tex]c. \( P = \frac{{220^2}}{484} = 100 \) W[/tex]

[tex]d. \( P = \frac{{220^2}}{484} = 100 \) W[/tex]

So, the correct answer is:

b. 100 W, 220 V

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