Homework Help

An iron with power P1=600W and a radiator with P2=1100W, both with nominal voltage...

user profile pic

catbyshe | Student, College Freshman | eNoter

Posted June 30, 2009 at 12:35 AM via web

dislike 1 like

An iron with power P1=600W and a radiator with P2=1100W, both with nominal voltage Un=120V are in serial link, at the network voltage U=220V.

What power develop each appliance? What voltage would be at the polarities of each appliance?

1 Answer | Add Yours

user profile pic

neela | High School Teacher | Valedictorian

Posted July 1, 2009 at 12:56 PM (Answer #1)

dislike 0 like

The power rating P , the potetential difference V, and the current drawn I by an eqipment is given by:

P = VI.

Therefore,  for the iron , 6000 = 120V*I, I the current to be found. So,

I1 = 600/120 =5 A , and the the resistance R1 = 120/5 = 24  Ohm.

The radiator's  drawing of current I2, and its resistance R2 are given by:

I2 = 1100/120 =9.1667..A   and R2 = 120/9.16673.. = 13.0909 ohm

The total resistance  of the sries circuit= 24+13.0909 =37.0909 Ohm.

So, the current flow in the ciruit in 220 V is 220/37.0909 =5.9314A

The voltage across the series of iron and radiators are:

(220/37.0909)24 and (220/37.0909)13.0909 or

142.35 V and  77.65 V 

The power consumtion = I^2*R .

Power consumption iron = 5.9314*142.15 = 844.33 watts.

Power consumtion of Radiator =  460.57 watts

 

 

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes