Mangal's room heater is marked as 1,000 W -200 V. Find the percentage change in the power of the heater, if the voltage drops to 160 V.

Explanation

Case - I :
P = 100 W
V = 200 V
`R = V^2/P = (200xx200)/1000=40Omega`
Case II :
`R = 40 Omega`
V. = 160 V
`P.= ((V.)^2)/(R) = (160xx160)/40 =640W`
percentage change in power
`= (P-P)/(P) xx100`
`= (1000-640)/1000 xx100=36%`