How are amps and volts related?

[Coolmanjack]

If amperage = voltage/resistance, and power lines carry a cheap ton of volts, why aren't the amps proportional?

Lest say the line carries 10,000 volts and has 500 ohms of resistance ( I'm pulling these numbers from my rectum) wouldn't the line have 20 amps of current?

I'm assuming it has to do with AC vs DC. It seems that AC does not follow ohms law and you can trade amperage and voltage, am I correct? If so what is the law for AC and does trading volts and amps work on DC?

 

[MarkW]

Read this short page...

 

[unksol]

Ohms law still applies. For a purely resistive loads at peak voltage it looks very similar to DC. But there are other things to consider like the phase angle, average voltage etc. None of this really matters for your question. Or at least what I think your question is.

The way the power grid works (simplified)is to transfer power from the station at very high voltage at very low current to minimize losses. When it get to substations or at the pole to your house a transformer lowers the voltage to 120V. In your "example" 10,000V*20A = 200,000W of power. 200,000W /120V = 1666A. Obviously there are losses in the transformer that affect the max but you get the idea. Your PSU does the same thing. 120V AC in at 10A is 1200W. It's switched to DC and dropped to 12V @ 100A. That would be at full load and 100% efficient. Normally it's close to 85-95%.

DC you would normally just use a regulator to lower 12V to say 5V. Inductors/transformers require changes in an electric field to cause changes in the magnetic field and induce a voltage on the other side. DC is constant voltage so to do something like this you build a circuit to switch the voltage on and off rapidly to generate an alternating waveform to feed the transformer. This is how you step up voltage from say a 9V to 10,000 volts for a stun gun

 

[guanyu210379]

V = R . I is only the simplified formula for schools and this should be relatively close for simple DC circuits, only to simple DC circuits.
If you are running AC or voltage with frequency, if you have capacitors, inductors, diodes, transistors (FET, MOS, etc.), etc. if you include the losses, etc. The calculation will get really nasty.
AC calculation still follows Ohm's law but the complete calculation, not the simplified calculation.

 

[MU_Engineer]

I think the best way to answer the OP's question is that there are two reasons that very high voltages used. Unksol mentioned them but I will expand upon it.

1. The amount of current you can deliver in a wire of a given composition is determined by its size. The more current, the larger the wire has to be.

2. Power = current * voltage. You can deliver a bunch more power over the same wire by increasing the voltage.

3. Voltage drops (resistive losses) in AC systems is actually V=I^2 * R. R is a fixed quantity for a particular wire, so your voltage losses quadruple when you double the amperage. Going back up to the second equation, if you double the voltage, you halve the current, and thus reduce your voltage losses by 75%.

The power company saves a lot of money by using high-voltage distribution systems as it allows them to lose less power to resistive losses in the lines and also spend less money by using smaller, less-expensive wires. Going from distribution voltages of 600-2300 volts around 1900 to 7200+ volts by the 1930s also allowed for rural areas to be electrified. The lower voltages simply had too much voltage drop over long lines (resistance is measured in ohms per foot) to feed customers more than a few miles beyond the generation plant.

Your typical pole or pad-mount residential transformer takes 7200 volts and turns it into 125 to 200 amps of 240 volt service. Drawing the 200 amps at 240 V from the transformer causes the transformer to draw less than seven amps at 7200 volts. Other common voltages to feed customer transformers are 13,200 volts and 19,920 volts, particularly to larger customers that draw thousands of amps of 480 volt service and to very isolated rural areas. The power company will use voltages like 34 kV, 69 kV, 115 kV and 161 kV to transmit power between generation plants and substations, and voltages from 230 kV up to 765 kV is used to send power several states away.

 

[Someone Somewhere]

99% of the time V=IR works just fine, even on AC systems.

To go back to the OP's original example, a power line might have a resistance (from one end to another) of maybe one ohm - 500Ω is going to be pretty excessive, even on very long lines.

However, the main thing that determines the current flow is the resistance of the load connected to the end of it - you need to sum up the load resistance along with both legs of the power line (I'm assuming this is a single-phase or DC line for simplicity).

That could be 1Ω (supply conductor) + 1000Ω (load) + 1Ω (return) = 1002Ω. I = V/R so I~=9.98A.

Your estimate would only apply if you short-circuited the power line at the far end, so the power line's resistance was the only thing that mattered.

 

[Reynod]

Enjoying this .. .takes me way way back ... I can even smell resin cored solder.

Trying to resist the urge to post further.

AC vs DC ... E=IR ... however the frequency of the alternating waweform modifies R into Impedance Z.

P = EI or I2R ... which is power ... or ability to do real work.

Essentially AC is much easier to deal with in terms of changing the voltage (high voltage for transmission means lower losses over distance) and using a transformer (on the pole or substation to bring it down from KV to 240 or 120) enables you to bring it easily down for household appliances.

There were many arguments to use DC (direct current) for transmission back at the turn of the last century but these were largely ignored. he main reason for using AC on the grid was to enable transformation to higher voltages in order to drop line losses (P=IE=I2R) and if the conductor size remains the same, when E is increased in the equation E=IR then I must necessarily decrease, in turn decreasing losses as the square of I). But now we have the ability to transform AC (at all thermal, hydro and wind generators) and DC (at solar generators) to any level of DC we desire and transmit, usually to residential or commercial loads which tend to use DC anyway. If need be it can be converted back to AC at industrial loads (motors usually).

 

[Kewlx25]

If "amperage = voltage/resistance" was always true, then a 35k volt static shock would blow my finger to kingdom-come.

I think it is meant to describe an upper-bound, but I'm not sure.

 

[Someone Somewhere]

Technically, it's only true where there's no reactance (capacitance or inductance).

Because of the very short time frame, the inductance limits how fast the current can change (rise). And then the charge starts to run out.

The more correct formula is V=IZ, where Z is impedance (the vector sum of resistance and reactance).