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Current Electricity

Question

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About 5% of the power of a 100 W light bulb is converted to visible radiation. What is the average intensity of visible radiation at a distance of 1m from the bulb?

Solution

Given,
Percentage of power converted into visible radiation = 5 %
Power of bulb, P = 100 W

This implies,
Power converted into visible radiation,

P = \frac{5}{100} \times 100 W = 5 W

Therefore,

Intensity = \frac{Energy}{Area \times Time} = \frac{Power}{Area} = \frac{P}{4 {πr}^{2}}

Intensity, I =

\frac{5}{4 \times 3.14 \times 1 \times 1} = 0.4 {Wm}^{- 2}

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