If the current in an electric bulb drops by 1 per cent, the power decreases, approximately by

The power (P) consumed by an electric bulb is given by the formula P = I²R, where I is the current and R is the resistance [c2][t1]. According to Joule's law of heating, power is directly proportional to the square of the current for a given resistance [c2]. For small percentage changes, the relative change in power can be approximated using differential calculus: ΔP/P ≈ 2(ΔI/I) [t2][t8]. If the current drops by 1% (ΔI/I = -0.01), the power decrease is approximately 2 × 1% = 2% [t5][t8]. Mathematically, if the new current is 0.99I, the new power is (0.99I)²R = 0.9801I²R, which represents a 1.99% decrease, approximately 2% [t1]. Thus, a 1% drop in current results in a 2% decrease in power output.

Sources

1. [1] Science , class X (NCERT 2025 ed.) > Chapter 11: Electricity > H = I2 > p. 189