Question

An electric iron consumes energy at a rate of 840 watt When heating is at the maximum rate and 360 W when the heating is at the minimum. The voltage is 220 Volt. what are the current and the resistence in each cade?

Plzz need answer step by step

Answer 1

$\textbf{Hey There!}$

=>Here is ur Solution<=

As we know that power input is P = V I
Thus the current I = P/V

(a) When heating is at the maximum rate
I = 840 W/220 V = 3.28 A;
and the resistence of the electric iron is
R = V/I = 220 V/ 3.82 A = 57.60

(b) When heating is at minimum rate,
I = 360 W/ 220 V = 1.64 A;
and resistence of the electric iron is
R = V/I = 220 V/1.64 A = 134.15

|HopE__It__HelpS| :-|

Answer 2

heya...

Here is your answer...

We know that the power input is (P) = VI

Thus the current (I) = P/V

When heating is at the maximum rate,

I = 840W/220v

= 3.82A

and the resistance of the electric iron is

R = V/ I

= 220v/3.82A

= 57.59

When heating is at the minimum rate

I = 360W/220v

= 1.64A

and the resistance of the electric iron is

R = V/ I

= 220v/1.64A

= 134.15

It may help you...☺☺