Question

Compact fluorescent bulbs are much more efficient at producing light than are ordinary incandescent bulbs. They initially cost much more, but last far longer and use much less electricity. According to one study of these bulbs, a compact bulb that produces as much light as a 100 W incandescent bulb uses only 23.0 W of power. The compact bulb lasts 1.00×104 hours, on the average, and costs $ 12.0 , whereas the incandescent bulb costs only 76.0 ¢, but lasts just 750 hours. The study assumed that electricity cost 9.00 ¢ per kWh and that the bulbs were on for 4.0 h per day.

Answer 1

Answer:

The resistance is 626.0 Ω.

Explanation:

Given that,

Power of compact bulb= 100 W

Power of incandescent bulb = 23.0 W

Time $t= 1.00\times10^{4}\ hours$

What is the resistance of a “100 W”fluorescent bulb? (Remember the actual rating is only 23W of powerfor a 120V circuit)

We need to calculate the resistance of the bulb

Using formula of power

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

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

Where, P = power

R = resistance

V = voltage

Put the value into the formula

$R=\dfrac{120^2}{23}$

$R=626.0\ \Omega$

Hence, The resistance is 626.0 Ω.