Question 26.67: A 500 W heating unit is designed to operate with an applied ......
A 500 W heating unit is designed to operate with an applied potential difference of 115 V. (a) By what percentage will its heat output drop if the applied potential difference drops to 110 V? Assume no change in resistance. (b) If you took the variation of resistance with temperature into account, would the actual drop in heat output be larger or smaller than that calculated in (a)?
(a) We use P = V²/R ∝ V², which gives ΔP ∝ ΔV² ≈ 2V ΔV. The percentage change is roughly
ΔP/P = 2ΔV/V = 2(110 – 115)/115 = –8.6%.
(b) A drop in V causes a drop in P, which in turn lowers the temperature of the resistor in the coil. At a lower temperature R is also decreased. Since P ∝ R^{-1} a decrease in R will result in an increase in P, which partially offsets the decrease in P due to the drop in V. Thus, the actual drop in P will be smaller when the temperature dependency of the resistance is taken into consideration.