# Considering power loss in a simple series connection due to cable

A friend and I are struggling on a simple electricity exercise:

A heater is connected to a $$\U = 230V\$$ power network over a copper cable. The cable has a cross-section of $$\A = 1.5mm^{2}\$$ and a length of $$\l = 200m\$$. The heater has a net voltage and net power of $$\ U_{n} = 230V, P_{n} = 2200W \$$. Calculate the actual net power of the heater and the power loss due to the cable. Also calculate the total power which is drawn from the network. Put the net power and the loss power into relationship with the total power.

This is our way to our solution: We considered the cable as a separate resistance which is connected to the heater in series. With the data of the cable and a specific conductivity of $$\58.11\$$ we get a resistance of $$\2.29\Omega\$$ for the cable.

The great question for us is how the heater behaves now that it won't get the voltage of $$\230 V\$$. It was not explained in the lecture how it behaves in such a case. We assume that it has a constant resistance. Assuming $$\230 V\$$ and taking the net power consumption of the heater we get a drawn current of $$\\frac{2200W}{230V} = 9.57A\$$. This yields a resistance of $$\R_{Heater} = \frac{230V}{9.57A} = 24.03\Omega\$$.

Hence the total resistance in the circuit is $$\R_{total} = 24.03\Omega + 2.29\Omega = 26.32\Omega\$$. With that we can calculate the total drawn current: $$\I_{total} = \frac{230V}{26.32\Omega} = 8.74A\$$

Calculating the voltages on the single components yields: $$\U_{Heater} = 24.03\Omega * 8.74A = 210.02V, U_{cable} = 20.01V\$$

The actual power consumption of the heater thus is $$\P_{Heater} = 210.02V * 8.74A = 1835.57W\$$; and for the cable (power loss): $$\P_{cable} = 20.01V * 8.74A = 174.93W\$$ The total power consumption should then be: $$\P_{total} = 2010,5W\$$

The solution says we have a total power $$\P_{total} = 1835.57W\$$ which is exactly the power the heater uses in our solution. This was then divided in actual net power of around 1500 and power loss of around 335 (so they add up). This hints that we are not completely wrong but made a mistake somewhere in our thinking process. We were confused because we don't know how the heater behaves. Reverse Engineering the provided solution gave no clue on what could be wrong.

I hope someone can help us to solve this.

Thanks.

• How did you arrive at 1.5mm2 ? given? Also 24 Ω is correct at 2.3kW but only 23Ω at 200V ~21Ω @ 190V What makes you think any answer is correct when you say "actual"? – Tony Stewart Sunnyskyguy EE75 Dec 31 '18 at 0:06
• I get suspicious with any answer given that has 6 sig figs shown when there are only 2 for this type of problem. Too many assumptions were left out. Although your calculations are good otherwise. Return resistance and heater resistance and assumptions of copper lead me to say the power supplied at 230V is 2.1kW total including 366W in undersized wire – Tony Stewart Sunnyskyguy EE75 Dec 31 '18 at 0:16
• Reducing the voltage ~10% also reduces the resistance ~5% thus reducing the power only 7.5% (est.). Was the 200m the total loop length given? If so, then Pt=2.19kW incl 200W wire heat. – Tony Stewart Sunnyskyguy EE75 Dec 31 '18 at 0:24
• You are correct if wire heat is shared in the same space as the heater. Otherwise assume Delivered power to load as Net Power. Good work. – Tony Stewart Sunnyskyguy EE75 Dec 31 '18 at 0:32
• It was the most basic EE course you can imagine in a first semester and the first part of this lesson. So we couldn't know about change of R when U changes :) The double conductor length was more obvious with just common sense and worked for us :) Nevertheless thanks for your effort – birdfreeyahoo Dec 31 '18 at 13:45