Question

By what percentage will the illumination of the lamp decrease if the current drops by 20%?

• A. \(46\%\)
• B. \(26\%\)
• C. \(36\%\)
• D. \(56\%\)

Answer 1

The Correct Option is C

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:

\[ 36\%. \]