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

Answer: (C)

$36\%$

Answer 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:

$36\%.$