# 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...

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?

*print*Print*list*Cite

### 1 Answer

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