Question

A 1000 w heating unit is designed to operate from a 200 voltline . By what percentage will it's heat output drop if the line voltage drops by 40 v??

Answer 1

Since R = V^2 / P

=) R = (200)^2 / 1000

= 200 * 200 / 1000

= 40 ohms

In second case ;

V = 200-40 = 160 V

So P = V^2 /R

= 160 * 160 / 40

= 640 W

Since decrease% = decrease / original * 100%

= (1000-640) / 1000 * 100 %

= 360/ 1000 * 100 %

= 36%

Hope it helps uh!!

Answer 2

$\huge{hey}$


$\huge{ \mathfrak{here \: is \: answer}}$

$\bf{solution}$

given : power = 1000 watt

v= 200volt

R=v²/p


$\sf{ \frac{({200}^{2}) }{1000} }$


R=40 ohm

as voltage drops by 40 volt , the voltage of line

(200-40)v

160v

hence
$\sf{\frac{ ({160}^{2} )}{40} }$


640W

also , drop in heat production per second

(1000-640) watt


360 w

hence , percentage drop in heat production per second


$\sf{ \frac{360}{1000} \times 100}$


$\bf{36\%}$

$\huge \boxed{hope \: it \: helps}$


$\huge{ \mathbb{THANKS}}$