A servo voltage stabiliser limits its output voltage to 220 V 1 % . When an electric bulb rated at 220 V , 100 W is connected to this stabiliser, determine the minimum and maximum power consumed by the bulb.

Step-by-step solution

The power consumed by the bulb is given by P = V^2 R . For small variations, the fractional change in power is: P P = 2 V V Given that the voltage variation is 1 % , we have: P P = 2(1 %) = 2 % The nominal power of the bulb is P = 100 W . The variation in power is P = 2 % of 100 W = 2 W . Therefore, the minimum power consumed is 100 - 2 = 98 W and the maximum power consumed is 100 + 2 = 102 W .