3
$\begingroup$

My textbook states that power is transmitted at high voltage and low current since $P=I^2R$ and as the current has a small magnitude, the heat dissipated across the transmission lines is less than when we carry it out at high current and low voltage. But $P=I^2R$ can also be written as $P=V^2/R$ and hence a discrepancy would occur. Where am I going wrong?

$\endgroup$
0

3 Answers 3

2
$\begingroup$

Here are the possible ways of writing power

power

Look at how resistance exists in wires. The power is inversely proportional to resistance if one uses the voltage as a criterion, the more resistance in the way the smaller the power transferred for the same voltage. The smaller the current the less power is dissipated by a given resistance on the way and can be transferred where it is needed.

$\endgroup$
0
2
$\begingroup$

As we have $P = I\Delta V $, the same amount of power can be delivered either at high currents and low potential differences or at low currents and high potential differences. In either case, the same amount of internal energy is transferred to the surroundings by heat as you correctly pointed out, so the decision is made primarily based on economic reasons. From the definition of resistance, we have: $$ R = \frac{\Delta V}{I} $$ Hence, if low potential differences and high currents were used, the resistance of the materials would have to be lower in order to deliver the same power. Materials that have low resistances include rare metals like silver and gold that are too expensive to be used commercially. Hence the choice of low currents and high potential differences is made, so that inexpensive copper wires can be used that generally have a high resistance. This high potential can also be conviniently reduced or increased by a transformer.

Hope this helps.

$\endgroup$
1
  • $\begingroup$ Oh, alright. Thanks for clearing my doubt! $\endgroup$
    – CannedOrgi
    Commented May 29, 2021 at 13:24
1
$\begingroup$

When we talk about high voltages we do not mean a high potential difference between the two ends of the wire. In any case we want to minimise the potential difference across the wire as this obviously means a loss of potential in the wire( the thing that is needed to power things)

in the equation P=v^2/r

v is the potential difference across the wire, and R is the resistance along that stretch of wire. so by using this equation we can minimise the power losses by lowering the potential difference across the wires, Well how do we do this?

in P=IV the V is the potential at a point(measured against 0 potential)

so if the initial potential is increased at the start of the cable, for the same power, the current must be lower. Which, reduces losses in the wire as per I^2*r. Which means the potential difference ACROSS the cable is lower as less is being wasted as heat

You're confusing PD across the wire, vs the initial potential.

Which is exactly why "voltage" is such a bad word as it means different things

$\endgroup$

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