SOLUTION: An electric radiator which is an ordinary constant resistor emits a heating power of 1.2 KW at a mains supply voltage of 230 V. what is the total heating power in each respective

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Question 378919: An electric radiator which is an ordinary constant resistor emits a heating power of 1.2 KW at a mains supply voltage of 230 V.
what is the total heating power in each respective instance when two such radiators are connected
a) in parallel, and
b) in series to the 230 V supply voltage ?

Found 2 solutions by rfadrogane, Alan3354:
Answer by rfadrogane(214)   (Show Source): You can put this solution on YOUR website!
An electric radiator which is an ordinary constant resistor emits a heating power of 1.2 KW at a mains supply voltage of 230 V.
what is the total heating power in each respective instance when two such radiators are connected
a) in parallel, and
b) in series to the 230 V supply voltage ?
Try to clarify your 230-V, its an AC or DC?, if AC, then state the frequency
Assuming that the radiator is a non-inductive load,
let us solve first the resistance of the radiator
from: P = E^2/R
R = E^2/P, E = 230-V & P = 1,200-W
= (230)^2/1,200
R = 44.083-ohms
a) when parallel,the total power is the summation of its rating, thus it is
Pt = 1,200W + 1,200W = 2,400W ----answer
b) when series,the total power is the summation of its rating, thus it is
Pt = 1,200W + 1,200W = 2,400W ----answer


Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
Power = E^2/R
In series, R is doubled, Power is 1/2 = 600 watts (0.6 KW)
In parallel, it's just 2 heaters --> 2x = 2.4 KW
-------------
Since it's resistive, AC or DC is not relevant.

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