fundamental voltage misunderstanding

[Designer69]

Per ohms law if you increase voltage and keep resistance the same, current goes up.

Then how come we use step-up transformers (increase voltage) to keep current low and keep cable size small.
Where is the disconnect here, what am I missing?

 

[luckylerado]

Per ohms law if you increase voltage and keep resistance the same, current goes up....

Uhh. No.

:ashamed1:

 

[Smart $]

It's quite simple if you think in terms of power.

P=E_LI_L=E_HI_H

"L" is low voltage
"H" is high voltage​

As you can see, resistance doesn't even enter the general equation.

 

[dkidd]

Uhh. No.

:ashamed1:

Uhh, yes. :?:happysad:

 

[Carultch]

Per ohms law if you increase voltage and keep resistance the same, current goes up.

Then how come we use step-up transformers (increase voltage) to keep current low and keep cable size small.
Where is the disconnect here, what am I missing?

Ohm's law specifically refers to voltage DROP across a resistive component. Not the voltage of the full circuit.

When you use transformers to trade voltage and current (similar to how gears trade torque and rotation speed), your equation is that current*voltage on both sides of the transformer is approximately the same (neglecting internal lossses). Reducing the current in the high voltage line means that you reduce the voltage drop across that same resistance. The voltage drop across the transmission line becomes a much smaller fraction of the total voltage, and therefore more power can be delivered to the load.

 

[gar]

160322-1124 EDT

Designer69:

The disconnect is that you have not had an adequate education on basic electrical circuit theory.

You described the relation of voltage and current with respect to a resistive load. You did not discuss power in relation to voltage and current. By the way, power equations are not part of Ohm's law. The power and energy relationships were developed by Joule many years after Ohm show the relationship of voltage, current, and conductance. Do you know how Ohm created a variable voltage source in the early 1800s to perform his experiments? Do you have any idea of the nature of electrical instruments that early pioneers in the electrical field had with which to determine these basic concepts?

An equation for power is P = V * I . So we can distribute a given amount of power with less current and less wire by just raising the distribution voltage. That is what the transformer does, and why it is used.

.

 

[gadfly56]

Per ohms law if you increase voltage and keep resistance the same, current goes up.

Then how come we use step-up transformers (increase voltage) to keep current low and keep cable size small.
Where is the disconnect here, what am I missing?

Don't forget that the resistance you're talking about has two (2) components. One is the resistance of the wire and the other is the resistance of the load. You want the resistance of the wire to be insignificant compared to the resistance of the load. For very long runs, the voltage drop over the run can be significant and you may not get enough voltage to operate the load properly. Try calculating the VD for a 10 amp load using 5, 50, 100, 250, and 500 volts at 500 feet starting with #14. Assume that the base case is 100 volts (to establish the resistance of the load). Say you don't want more than a 3 volt drop, and see how large you have to make the conductor in each case to meet that limit.

 

[K8MHZ]

Uhh. No.

:ashamed1:

Uh, yes x2.

 

[K8MHZ]

Per ohms law if you increase voltage and keep resistance the same, current goes up.

Then how come we use step-up transformers (increase voltage) to keep current low and keep cable size small.
Where is the disconnect here, what am I missing?

Well, for one thing, we don't 'keep the resistance the same' when using transformers.

The design resistance for a desired power output is determined by voltage. The higher the voltage, the higher the resistance needed for a desired power output.

When we step up voltage, the amount of current needed to supply a certain amount of power decreases. That's why we can use smaller conductors (but larger insulators).

 

[Designer69]

160322-1124 EDT

Designer69:

The disconnect is that you have not had an adequate education on basic electrical circuit theory.


.

yes you are correct Gar, I partied and slept thru most of college, paying the price now.

I see what my mistake was though, cannot combine power and resistance like I did. thanks

 

[Bugman1400]

yes you are correct Gar, I partied and slept thru most of college, paying the price now.

I see what my mistake was though, cannot combine power and resistance like I did. thanks

That is not true either since P=V*V/R but, that is only because I=V/R (or V = I*R).

 

[Ingenieur]

Per ohms law if you increase voltage and keep resistance the same, current goes up.

Then how come we use step-up transformers (increase voltage) to keep current low and keep cable size small.
Where is the disconnect here, what am I missing?

losses = i^2 R
so for constant R the lower the current the lower the losses
and it's a square ratio
decrease R by 10% and losses drop by 10
decrease i by 10 and losses by 1.21 or 21%

also
v drop = i x R
the lower the i the lower the drop

as was pointed out the power required by a load is assumed constant
so as v increases i decreases and vice versa

 

[Carultch]

Per ohms law if you increase voltage and keep resistance the same, current goes up.

Then how come we use step-up transformers (increase voltage) to keep current low and keep cable size small.
Where is the disconnect here, what am I missing?

See my attached image to explain this concept.

Circuit 1 is the circuit that contains the "high voltage" source with a voltage of V0, the resistance of the transmission line (RT), and the primary side of the transformer. The current through circuit 1 is I1. The voltage delivered to the transformer primary is V1.

Circuit 2 is the circuit that contains the secondary side of the transformer and the load to which we intend to deliver power, modeled as a simple resistor of resistance RL. The transformer secondary is the voltage source of this circuit, which has a voltage of V2. The current in this circuit is I2.

It's true that for each individual resistance in this circuit, Ohm's law applies. For the load, this is V2 = I2*RL. For the inevitable resistance in the transmission line, the voltage across this resistance is (V0 - V1), and Ohm's law is (V0 - V1) = I1*RT.

 

[Carultch]

Previous post continued

For the transformer, when we assume ideal 100% efficiency, power in must equal power out. Therefore, V1*I1 = V2*I2.

So our set of equations is:
V1*I1 = V2*I2
V0 - V1 = I1*RT
V2 = I2*RL

Given values would likely be RL, V2, RT and V0. The unknowns are I1, I2, and V1. I2 is trivial, but the others are not. This is a challenging system of equations to solve, as you'll get a quadratic formula. I'm using software to assist, to show the results.

Let's apply a couple of sets of numbers.

Common to all:
V2 = 120V
RL = 144 Ohms
RT = 2000 Ohms

Consider two cases for source voltage. V0 = 1000 V and V0 = 10000V.

When V0=1000V:
V1 = 723.6 V
I1 = 0.1382A
I2 = 0.8333A
Power delivered to load RL = 100 Watts
Power loss in transmission line RT = 38.2 Watts
Total power produced at source = 138.2 Watts
Efficiency: 72.36%

When V0 = 10000V:
V1 = 9980 V
I1 = 0.01002A
I2 = 0.8333A
Power delivered to load RL = 100 Watts
Power loss in transmission line RT = 0.2008 Watts
Total power produced at source = 100.2 Watts
Efficiency: 99.8%

Now consider the case that the transformer isn't even present, but the transmission line resistance is just as high. The voltage still needs to be 120V at the load, to deliver 100 Watts. This means that the current both in the source and the load is 0.8333A, and in order to get through the transmission line resistance of 2000 Ohms, the voltage drop is 1667 Volts. Therefore, a total of 1787 Volts must be present at the source. In this example, a total of 1388 Watts are lost to the transmission line, so the efficiency of the whole system is 6.7%.

You can see that the lower you can make the current in the transmission line, the less voltage will drop across it, and the less power will be lost within it. To reduce the primary current, the voltage output of the primary circuit will have to increase, and the transformer will be there to step it down. Without the transformer (or a modern equivalent component taking its place), you can see that there isn't a way to reduce the current through the transmission line, as the full current would have to make it from source to load. The higher the primary voltage, the lower the primary current, and the less power is lost within the transmission line.

 

[GoldDigger]

You can avoid the need for quadratic equations by simply replacing the transformer in the primary circuit with the resistance RL reflected across the turn count of the transformer.
The circuit is then a simple voltage applied to pair of series resistors, and the voltage across the transformer is one leg of the voltage divider.

 

[mivey]

Good gracious!

 

[Smart $]

Good gracious!

A lot of hoopla for such a basic concept, eh?

 

[kwired]

yes you are correct Gar, I partied and slept thru most of college, paying the price now.

I see what my mistake was though, cannot combine power and resistance like I did. thanks

The information provided so far hopefully makes it more clear why POCO's use higher voltages to transmit power over longer distances.

A drop of 1 volt @ 115,000 volts is nothing compared to what a 1 volt drop is on a 12 volt system.

 

[Tony S]

To add proof to the voltage/current discussion. OK it’s a UK transformer but as far as I know they work the same both sides of the atlantic.

 

[mivey]

A lot of hoopla for such a basic concept, eh?

Kinda like re-sodding the whole green for one pitch mark.