
#1
Feb2509, 11:30 PM

P: 183

1. The problem statement, all variables and given/known data
A power station delivers 890 kW of power at 12 kV to a factory through wires with total resistance 5.0 [tex]\Omega[/tex]. How much less power is wasted if the electricity is delivered at 50 kV rather than 12 kV? 2. Relevant equations eq1) P = IV eq2) P = I^{2}R Ohm's Law: V = IR 3. The attempt at a solution I know the problem can be solved by solving for the current value in each case from eq1, and plugging it in to eq2 to calculate power loss. My real question is why can't we apply Ohm's law to calculate I, and then plug it into eq2? Why must we use eq1 to get the current? Doesn't Ohm's law apply for all ohmic conductors, and since we have both the resistance and voltage of and across the conductor, can't we get the current that way? Thanks! 



#2
Feb2509, 11:51 PM

HW Helper
P: 3,394

If you apply Ohm's Law to the line resistance, you would need to know the voltage drop in the line.
If you apply it to the load plus the line, you would have to use the combined resistance of line and load. You could use the power formula to find the load resistance, then use Ohm's Law to find the current. 



#3
Feb2609, 12:13 AM

P: 183

So when you have a 12kV transmission line, what is the 12 kV drop across? The negative terminal is obviously at the powerplant, the positive, after passing through the load, back at the powerplant? 



#4
Feb2609, 03:21 PM

HW Helper
P: 3,394

Power loss in transmission linesVirtually all transmission lines are AC, so there is no negative or positive terminal. 



#5
Feb2609, 03:26 PM

Sci Advisor
HW Helper
P: 8,961

This confuses everybody who first sees this question  do a search here and you will see dozens of threads on it! 



#6
Feb2609, 03:37 PM

P: 183

Thank you!
I have a much better understanding of the concept now. 



#7
Feb2609, 04:10 PM

Sci Advisor
HW Helper
P: 6,564

AM 



#8
Nov1211, 01:03 AM

P: 2

KW KV R I=KW/KV Loss=I*I*R Loss reduction
890 12 5 74.17 27503.47 890 50 5 17.8 1584.2 25919.27 PF=1 



#9
Nov1211, 01:11 AM

P: 2

KW KV R I=KW/KV Loss=I*I*R Loss reduction
890 12 5 74.17 27503.47 890 50 5 17.8 1584.2 25919.27 PF=1 890 kw 12kv current= 890/12=74.17 Amp assuption pf=1, loss= I*I*R=27503 W 890kw, 50kv, current=890/50=17.8 amp, loss= 1584.2 w, loss reduction= 25919.27 w, As the voltage increases, current decreases for same power. current decreases loss also decreases. current has a squred relation with loss. 


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