If we want to transmit electic current for a long distance, we must minimize a heat that releases because of the resistanse. We cannot make a cable wide because it is expensive and it will be massive. So, we must decrease a current because $Q = I^2R\Delta t$. A power that we want to transmit is constant, $P = UI$, so we must increase the voltage. Nevertheless, $Q = \frac{U^2}{R}\Delta t$, so if we increase the voltage, the heat will increase too. And for addition, $U = IR$, so if we decrease the current, the voltage will decrease too.

Could you, explaine in details where I have a mistake.

P.S. sorry for my English, I am not a native speaker.

  • 3
    $\begingroup$ You've used U for two different voltages. The power transmitted from a source is the voltage of the source times the current; the power loss in the transmission line is the voltage drop across the line $\endgroup$ – DJohnM May 19 '13 at 18:40
  • $\begingroup$ @User58220 but what about Ohm's law? U refers to the voltage of the source. $\endgroup$ – cheremushkin May 19 '13 at 18:58
  • $\begingroup$ No Ohm's law refers to the voltage difference accros the resistor - in this case the transmission line $\endgroup$ – Martin Beckett May 20 '13 at 3:49
  • $\begingroup$ @User58220: Why don't you write up your comment as an answer (since it is)? $\endgroup$ – Art Brown May 20 '13 at 4:59

Suppose we have a source of electrical energy, say a battery, that puts out 100 Volts. It is connected through wires with a total resistance of 1 ohm to a heater with a resistance of 99 ohms.

The battery sees a total resistance of 100 ohms, and thus pushes 1 Ampere of current through the circuit. The battery is delivering energy at 100 Watts

The Power delivered to the heater is $I^2 \times R=1^2\times 99 =99$ Watts

The Power lost in the wiring is $I^2 \times R=1^2 \times 1=$ 1 Watt

A voltmeter would measure 100 V across the battery, and 99 V across the load.

Now, assume that this 1% loss is unacceptable. So we leave the wiring the same, and increase the battery output to 1000 Volts. We also increase the resistance in the heater to 10 000 ohms.

Now the calculations go like this:

The battery sees a total resistance of 10 001 ohms, and thus pushes 0.1 Ampere of current through the circuit. The battery is still delivering energy at 100 Watts

The Power delivered to the heater is $I^2×R=0.1^2×10 000=100$ Watts

The Power lost in the wiring is $I^2×R=0.1^2×1= 0.01 $Watt

The power loss in the wiring has been reduced to 1/100 of the previous amount, while the power delivered to the heater stays about the same...

  • $\begingroup$ Thank you for this answer. It is very clear! $\endgroup$ – user2705196 Mar 13 at 0:42

Resistance is simply (resistivity X length)/area
Since your resistivity is material dependent and length is also fixed, you can manipulate the area. Decreasing the cross section area of the wire does mean that you are effectively increasing the resistance. You have to optimize the parameters so you get the max out of it. And as the first answer points out, you are confusing the source voltage with load voltage. Also, the source voltage is different then the voltage between the source point and the end point. Say, source will generate x amount of power. In the end you will get (x-Ohmic loss). This ohmic loss will be due to this potential difference (voltage) between source and endpoint. Nothing more, nothing less (assuming no other loss in procedure/set-up is induced).


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