Question

A 504 W heating unit is designed to operate with an applied potential difference of 124 V. (a) By what percentage will its heat output drop if the applied potential difference drops to 115 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)?

Answer 1

(a) 13.98%

(b) Larger

Solution:

As per the question:

Power consumed the heating unit, P = 504 W

Potential difference at which it consumes 504 W, V = 124 V

Potential difference after the drop, V' = 115 V

Now,

(a) The drop in heat output can be calculated as:

`P = (V^(2))/(R)`

`R = (124^(2))/(P) = (124^(2))/(504) = 30.5\ \Omega`

Now, the power, P' after the drop in potential is given by:

`P' = (V'^(2))/(R)`

`P' = (115^(2))/(30.5) = 433.49\ W`

Now, the percent drop in heat output is given by:

`(P - P')/(P)* 100 = (504 - 433.49)/(504)* 100` = 13.98%

(b) If the change in resistance is to be considered, then we know that resistance varies as a function of temperature and with an increase in temperature, resistance increases.

the increase in resistance thus increases the potential drop.

Therefore, if this is the case, heat output would be greater than calculated in (a)