Electric current flowing in a power transmission line

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New member has been reminded to post schoolwork questions in the Homework Help forums and fill out the Template showing their work
A cable of resistance 12 Ohm carries electric power from a generator producing 250 kilowatt at 10,000 volts. Calculate the current in the cable.
Solution
using P=VI
I=25A
but using P=I^2R
250000=I^2×12
I =500/root 12 Amps
What is the reason for the different answers?
Thanks .
 

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  • #2
stockzahn
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Identical answers would mean that the entire power of the generator is dissipated and transformed into heat by the electrical resistance of the cable, so 100 % losses.
 
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Which is the correct answer to this question. I am confused.
 
  • #4
stockzahn
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Which is the correct answer to this question?
Have you draw a circuit diagram? Let's assume there is no other consumer, only the (resistance of) the cable - like a large heater. What would be the current in that case?
 
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The second answer would be correct in that case as per my understanding. Still I am confused.
 
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Please explain with the help of the correct circuit diagram.
 
  • #7
CWatters
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You are calculating two different powers...

If you use V= 50000 and P=IV that's the voltage and current of the generator so you are calculating the power of the generator.

If you use R=12 ohms and P= I^2R that's the resistance and current in the cable so you are calculating the power dissipated or lost in the cable as heat.

There is a third place you can calculate the power.... As an exercise try calculating...

a) the power delivered to the city at the far end of the cable.
b) the voltage delivered to the city at the far end of the cable.
c) the voltage drop down the cable.
 
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Thanks but I think a circuit diagram would clarify it even more.
 
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I am still confused as to which is the correct answer to my question
 
  • #10
stockzahn
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I think the situations looks like in the attached picture. Any thoughts?
 

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Inspite of the nice explanations I am not getting it. So I think I should revise my basics and come back. Thanks a lot.
 
  • #13
sophiecentaur
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Inspite of the nice explanations I am not getting it. So I think I should revise my basics and come back. Thanks a lot.
The way the question is phrased could account for part of your problem. If you know the power being supplied and you know the volts then you can easily work out the current flowing (same through the cable AND the load). With the information in the question, there is no need to know the cable resistance but if you had been told the generator volts, the cable resistance and the power dissipated in the load, the question becomes more interesting. OR, if you had been told the volts across the load OR a number of other combinations of information, the question would also be more complicated.
What is the reason for the different answers?
Your first result for current is correct. What would be the point of using that calculated value of current to re-calculate it another way? :wink: But that second calculation of yours is assuming that all the power is being dissipated by the cable and not by the (load plus cable in series).

What you call the 'correct' answer is presumably the answer given in the back of the book. Your first answer is the correct one and your book is wrong (not a rare occurrence). COMPLAIN to your teacher and present him / her with the right calculation. Go on, I dare you.
 
  • #14
stockzahn
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Is this the cable in the attached diagram
No, the line is just a auxiliary line to see how the different components are connected. The rectangle containing the symbol ##R_{cable}## corresponds to the cable itself. The rectangle insinuates that the cable is treated as a ohmic resistance in which, due to friction between the conducted electrons with the nuclei, part of the electrical energy is transformed into heat (therefore the cable's temperature increases when current is flowing). This friction results in a drop of voltage (very similar to the pressure drop in pipe). Do you know the rules about volatges and currents in a closed circuit?
 
  • #15
Dale
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I am still confused as to which is the correct answer to my question
The first is correct.

Once you have done the first then you can use Ohm’s law to calculate the voltage drop on the cable. That turns out to be 300 V, and the 9700 V is across the load
 
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Thanks for the prompt and wonderful replies. I read and re-read the posts and ultimately realised my mistake and got through.
 
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sophiecentaur
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Thanks for the prompt and wonderful replies. I read and re-read the posts and ultimately realised my mistake and got through.
Glad you're happy now. EE is far from intuitive and you can only ever get to grips with it by strict application of the 'rules'. After a while, it starts to become more intuitive/ :smile:
 
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