Question

An electric iron has a rating of 750W; 200V. Calculate (a) the current required (b) the resistance of its heating element (c) energy consumed by the iron in 2 h.

Answer 1

Given,
Power = 750W
Potential Difference = 200V
a) P=VI
=> 750=200*I
I=750/200
Therefore, I=3.75A
b) P=I2R
=> R= 750/(3.75)2
R=53.33 Ohm
c) E=Pt
t=2 hrs= 7200 secs
E=750*7200
E= 5400000 J=5400 kJ

Answer 2

Given :

Power (P) = 750 W

Potential Difference (V) = 200 V

To Find :

(a) The current required

(b) The resistance of its heating element

(c) Energy consumed by the iron in 2 h.

Solution :

a) Let 'V' be the potential and 'I' be the current. The power (P) can be given as -

     P   =  V × I

⇒   I    =  $\frac{P}{V}$

⇒   I    =  $\frac{750}{200}$

∴    I    =  3.75 A

Therefore, current required is 3.75 A

b)  We know,

       Resistance (R) =  $\frac{V^2}{P}$

⇒         R                  =  $\frac{200 * 200}{750}$

⇒         R                  =  53.33 ohm

Therefore, Resistance of the heating element is 53.33 ohm.

c) Energy (E) =  Power (P) × Time (T)

Given, Time (T) = 2 hours = 2×60×60 sec;  Power (P) = 750 W

∴ Energy (E)  =  750 × 2 × 60 × 60

                     =  5400000

                     =  5400 KJ

Therefore, Energy consumed by the iron in 2 hours is 5400 KJ.