The illumination at a point 5 meters below a lamp is 6 lux. Th... | Electrical MCQ | Examoogle

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The illumination at a point 5 meters below a lamp is 6 lux. The candle power of the lamp is

140

150

200

Correct Answer

⚙️ TE • Technical Direct FormulaElectricalUtilization Of Electrical Energy

Option C

150

Quick Summary:

Given: Illumination E = 6 lux, Distance h = 5 meters.

📐MAMath SolutionDirect Formula

📋 Given

Illumination E = 6 lux, Distance h = 5 meters.

🔢 Formula Used

E=ICPh2E = \frac{I_{CP}}{h^2}E=h2ICP​​

📊 Diagram / Illustration

🔢 Step-by-Step Solution

Identify the relationship

The illumination $E$ at a point directly below a lamp is related to its candle power $I_{CP}$ and the vertical height $h$ by the inverse square law.

E=ICPh2E = \frac{I_{CP}}{h^2}E=h2ICP​​

Rearrange the formula

To find the candle power $I_{CP}$ , rearrange the formula as $I_{CP} = E \times h^2$ .

ICP=E⋅h2I_{CP} = E \cdot h^2ICP​=E⋅h2

Substitute and solve

Substitute the given values $E = 6$ lux and $h = 5$ m into the formula to calculate $I_{CP}$ .

ICP=6×5°2=6×25=150I_{CP} = 6 \times 5°2 = 6 \times 25 = 150ICP​=6×5°2=6×25=150

✅

C is correct because the candle power is calculated as $6 \times 5°2 = 150$ candela.

Core Concepts Used

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Illumination Inverse Square Law Candle Power

💡 EXAM TIP

This concept is fundamental to lighting design in Electrical Engineering; remember that illumination at a point follows the cosine law ( $E = \frac{I_{CP}}{d^2} \cos\theta$ ) for points not directly underneath the lamp.

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