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