By what percentage will the illumination of the lamp decrease if the current drops by 20%?
- \(46\%\)
- \(26\%\)
- \(36\%\)
- \(56\%\)
The Correct Option is C
Approach Solution - 1
The question requires us to find the percentage decrease in illumination of a lamp when the current decreases by 20%. The key concept to use here is that illumination (intensity of light) is proportional to the square of the current flowing through the lamp. This is based on the relationship given by the formula for power in terms of electric current:
\(P \propto I^2\)
Here, \(P\) represents power, and \(I\) represents current. Since illumination is related to the power of the lamp, we use this concept to find the change in illumination.
- If the original current is \(I\), the initial illumination can be represented as proportional to \(I^2\).
- Given that the current decreases by 20%, the new current is: \(I_{\text{new}} = I - 0.2I = 0.8I\)
- The new illumination is proportional to: \((0.8I)^2 = 0.64I^2\)
- The percentage change in illumination can be calculated as:
\(\text{Percentage Decrease} = \left(\frac{I^2 - 0.64I^2}{I^2}\right) \times 100\% = (1 - 0.64) \times 100\% = 0.36 \times 100\% = 36\%\)
Thus, the illumination of the lamp decreases by 36%.
Therefore, the correct answer is \(36\%\).
Approach Solution -2
The power dissipated in a resistive circuit is given by:
\[ P = I^2R. \]Let the initial power be \( P_{\text{initial}} = I_{\text{initial}}^2 R \).
If the current drops by 20%, the new current \( I_{\text{final}} \) is:
\[ I_{\text{final}} = 0.8 I_{\text{initial}}. \]The new power \( P_{\text{final}} \) is:
\[ P_{\text{final}} = I_{\text{final}}^2 R = (0.8 I_{\text{initial}})^2 R = 0.64 I_{\text{initial}}^2 R. \]The percentage change in power is:
\[ \frac{P_{\text{initial}} - P_{\text{final}}}{P_{\text{initial}}} \times 100 = (1 - 0.64) \times 100 = 36\%. \]Thus, the illumination decreases by:
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