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
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
We’ve answered 334,346 questions. We can answer yours, too.Ask a question