Suppose a electric power station produces 200W of electricity @ 200V. Now instead of stepping it up, it decides to transmit it as 200V to a city 20 km away. The transmission cable has resistance per unit length of 5 ohm/m. As is straightforward, power loss should be 100000W which is way more than what is produced. So final power loss should be 200W only. Now my question is: 1.Where will this 200W heat be dissipated? (Whether through the whole length of the wire or a part or any other) And why so? 2.What will be the current in the wire? (Since 1A will generate more heat than supplied energy )

Or in an alternative way, suppose a specific power is being generated by a power station (like say 200V, 10A ) which is stepped down to a lower voltage like 20V using a step down transformer. Since power across primary and secondary coil is constant, current in secondary coil should be 100A. But suppose the wire in the secondary coil has a resistance of 10ohm. In that case it cant support a 100A current (since its impossible both by ohms law and energy conservation) and hence the value of current in the secondary will be less than 100A. Hence the power in the secondary will also be less than the primary (less than 2kW). So what happens to the excess power generated? (I.e. the power in the primary coil in excess over the maximum power carried by the secondary coil)


closed as unclear what you're asking by Danu, user36790, Sebastian Riese, Daniel Griscom, Gert Feb 19 '16 at 0:06

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ how do you get your 100000W figure for power loss? $\endgroup$ – Andrea Feb 17 '16 at 11:46
  • 1
    $\begingroup$ $5 \Omega /m$ is the resistance of gauge 42 wire! That would be about as thin as a human hair or so. The max. power you could get into this $100k\Omega$ wire is 0.4W. $\endgroup$ – CuriousOne Feb 17 '16 at 11:47
  • 1
    $\begingroup$ 0.4W would be the power lost through heating the wire actually. The key is that you can only get as much as 2 mA through the wire. $\endgroup$ – Andrea Feb 17 '16 at 11:48
  • 1
    $\begingroup$ This is a straightforward example of using the word "straightforward" for a totally not straightforward (and possibly faulty) context. $\endgroup$ – DarioP Feb 17 '16 at 12:27

You cannot have a pd of 200 V at the source and the same pd at the consumer end unless the connecting wires had no resistance.
In your example the resistance of the supply cables is $10^5 \;\Omega$ and so if there was a dead short across the supply at the consumer end the current which would flow in the cables would be $\dfrac {200}{10^5} = 2$ mA and the power loss in the cables would be $200 \times 2 \times 10^{-3} = 0.4 $ W and that would be the power delivered at the power station.

If the consumer connected a $100 \; \Omega$ device across the supply then the voltage across the device would be 100 V the other 100 V being lost across the cables.
The current in the circuit would be be 1 mA.

The power station would deliver 0.2 W with 0.1 W dissipated in the cables and the consumer using 0.1 W.

  • $\begingroup$ But what happens to the rest of the power generated? I understand that you are applying ohm's law to calculate current in the wire. But shouldn't the current be in accordance with the voltage and power of the source? (I=P/V) I assumed a 1A current from that formula. And hence the power dessipated is becoming greater. But if we apply ohm's law for current, we are getting only a part of the power generated. Please help me understand this. $\endgroup$ – Sporadic eccentricity Feb 17 '16 at 15:19
  • $\begingroup$ The power is never generated. If you have a voltage source that means it will have a constant voltage irrespective of the current delivered by it. At the power station they generators producing electricity at a constant voltage. If customers want more power then the generators deliver more power because of greater input of mechanical power. $\endgroup$ – Farcher Feb 17 '16 at 16:15
  • $\begingroup$ Okay, I get it now. Thanks. But after this, a similar doubt now comes to my mind. That is, what would happen if the power is generated at 2kV and 10A (using appropriate wiring ofcourse), then it is stepped down using a step down transformer to 20V (and hence 1000A, considering power constant across secondary and primary coil). But the wire of secondary has a resistance of 10ohm. What happens then? Or simply, what happens when the power in primary coil of a transformer is more than max power that the secondary coil can carry? $\endgroup$ – Sporadic eccentricity Feb 19 '16 at 18:45
  • $\begingroup$ In general the people who design the transformers look at the worst case scenario and then how often such a scenario will happen The they build into the design of the transformer a safety margin. $\endgroup$ – Farcher Feb 19 '16 at 23:01
  • $\begingroup$ Yes that's obvious. In real life it will no way come even near this kind of situation. But I'm talking about it's conceptual reasoning. What happens in theory if such a setting was applied $\endgroup$ – Sporadic eccentricity Feb 20 '16 at 10:58

You MUST be familiar with Ohms law and basic power equations.
There is so much on the internet about them that an explanation here is superfluous.

Ohms law results in the following equation.
The following is the same equation arranged 3 ways.

for V = volts, I = current , R = resistance then.

R = V/I
I = V/R
V = I x R

Calculate the unknowns from the knowns using this formula.

Energy dissipation in a resistor (derivation on internet) is (3 rearrangements)using Ohms law to convert between variables.

For P = Power

P = V^2/R
P = I^2 x R P = V x I


Not the answer you're looking for? Browse other questions tagged or ask your own question.