Question

Question 34:
An electric iron is connected to the mains power supply of 220 V. When the electric iron is adjusted at ‘minimum heating’ it consumes a power of 360 W but at ‘maximum heating’ it takes a power of 840 W. Calculate the current and resistance in each case.

Lakhmir Singh Physics Class 10

Answer 1

First case

i=p/v=360/220=1.63A and
R =v/i = 220/1.63 =134.96 ohm

Second case

i=p/v=840/220=3.81A
R=v/i=220/3.81=57.74ohm

Answer 2

"The resistance of the electric iron is $134.15 \Omega$

Solution:

We know that P = VI

P = power; V = voltage and I = Current

Therefore, ${ I }=\frac { P }{ V }$

Given at heating is maximum rate$I=\frac { 840W }{ 220V }$

Maximum current = 840W and minimum current = 220W

Then, I = 3.82A

The resistance of electric iron is ${ R }={ \frac { V }{ I } }=\frac { 220V }{ 3.82A } =57.59\Omega$

When heating is at minimum rate,${ I }=\frac { 360W }{ 220{ V } } =1.64{ A }{ R }=\frac { 220V }{ 1.64A } =134.15\Omega$"