Question

An electric iron consumes energy at a rate of 840 W when heating is
at the maximum rate and 360 W when the heating is at the minimum.
The voltage is 220 V. What are the current and the resistance in each
case?​

Answer 1

Answer:

When heating at max rate,

power, P = 840W

V = 220V

We know P = VI

⇒ 840 = 220 × I

⇒ I = 840/220

⇒ I = 4 A

R = V/I = 220/4 = 55 Ω

So current = 4A and resistance = 55Ω

When heating at minimum rate,

power, P = 360W

V = 220V

We know P = VI

⇒ 360 = 220 × I

⇒ I = 360/220

⇒ I = 1.636 A

R = V/I = 220/1.636 = 134.45 Ω

So current = 1.636A and resistance = 134.45Ω

PLZ MARK AS BRIANLIEST,FLW ME AND THX FOR THE SUPERB QUESTION

Answer 2

Explanation:

I=P/V ( formula )

I=840 W/ 220 V ( at maximum rate )

I= 4A

I=360 W / 220 V ( at minimum rate )

I= 1.636363636 A

R= V/I ( formula)

R=220/4 ( at maximum rate )

R=55

R=220/1.6 ( at minimum rate )

R=137.5