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Newbie - Current, Voltage, Resistance, Power and Transformer theory

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hdjim69

Jan 1, 1970
0
I'm self-teaching myself electronics and the only place to ask
questions is in this forum so please excuse me if this has been asked
a million time already but after reading several books, I still have
questions on these topics. No real values, just theory in this
question.

I'm just starting the section on AC and the book is explaining why we
(homes and industry) use AC instead of DC and the use of transformers.
Now, the books says the reason we use AC is to minimize power loss.
That homes and industry need a lot of current and if we were using DC
we'd need to push a huge amount of current through the transmission
lines and the higher the current the more we'd lose in heat loss. OK
fine. But now let's see what happens in AC. Rather then pushing a
huge amount of current we have a very high voltage say 200,000 to
600,000 volts and a low amount of amps (current). But how can we have
this HUGE amount of "pressure" (the typical explanation of what
voltage is) and hardly any current ? I've been reading that voltage
and current are proportional - the more voltage the more current.
Ahh... but this isn't the case really since current is a variable
value. It depends on the amount of resistance. So getting back to the
transmission lines, if we have HIGH voltage and LOW current then
resistance MUST be high. E = I * R that is, if I is low R must be high
to get a high value of E. And resistance is what cause heat which
causes power loss. So how can we have low current + low resistance =
high voltage ?

In summary, if we have very high voltage and low current we must have
very high resistance which would eliminate just about all the current
so loss would be almost 100%.


TIA

J
 
hdjim69 said:
I'm self-teaching myself electronics and the only place to ask
questions is in this forum so please excuse me if this has been asked
a million time already but after reading several books, I still have
questions on these topics. No real values, just theory in this
question.

I'm just starting the section on AC and the book is explaining why we
(homes and industry) use AC instead of DC and the use of transformers.
Now, the books says the reason we use AC is to minimize power loss.
That homes and industry need a lot of current and if we were using DC
we'd need to push a huge amount of current through the transmission
lines and the higher the current the more we'd lose in heat loss. OK
fine. But now let's see what happens in AC. Rather then pushing a
huge amount of current we have a very high voltage say 200,000 to
600,000 volts and a low amount of amps (current). But how can we have
this HUGE amount of "pressure" (the typical explanation of what
voltage is) and hardly any current ? I've been reading that voltage
and current are proportional - the more voltage the more current.
Ahh... but this isn't the case really since current is a variable
value. It depends on the amount of resistance. So getting back to the
transmission lines, if we have HIGH voltage and LOW current then
resistance MUST be high. E = I * R that is, if I is low R must be high
to get a high value of E. And resistance is what cause heat which
causes power loss. So how can we have low current + low resistance =
high voltage ?

In summary, if we have very high voltage and low current we must have
very high resistance which would eliminate just about all the current
so loss would be almost 100%.


TIA

J

The power loss in a transmission line is calculated with the formula P
equals current squared times R:

P(loss) = I^2R

Therefore given any specific resistance the loss will be less if the
current is less.

Also, if you substitute for I in the above formula, using I = P/V, you
get:

P(loss) = P(used)^2R / V^2

Using this formula, we can see that the higher the voltage the smaller
the loss since the voltage is in the denominator. Note also that not
only is V in the denominator but is is squared which would make the
divisor even larger and P(loss) smaller.
 
P

Peter Bennett

Jan 1, 1970
0
I'm just starting the section on AC and the book is explaining why we
(homes and industry) use AC instead of DC and the use of transformers.
Now, the books says the reason we use AC is to minimize power loss.
That homes and industry need a lot of current and if we were using DC
we'd need to push a huge amount of current through the transmission
lines and the higher the current the more we'd lose in heat loss. OK
fine. But now let's see what happens in AC. Rather then pushing a
huge amount of current we have a very high voltage say 200,000 to
600,000 volts and a low amount of amps (current). But how can we have
this HUGE amount of "pressure" (the typical explanation of what
voltage is) and hardly any current ? I've been reading that voltage
and current are proportional - the more voltage the more current.
Ahh... but this isn't the case really since current is a variable
value. It depends on the amount of resistance. So getting back to the
transmission lines, if we have HIGH voltage and LOW current then
resistance MUST be high. E = I * R that is, if I is low R must be high
to get a high value of E. And resistance is what cause heat which
causes power loss. So how can we have low current + low resistance =
high voltage ?

The Power company is delivering Power to us, and power is voltage
times current. So, to deliver a given power, a higher voltage will
permit a lower current.

Power is lost in the transmission lines due to resistance in the wires
- that power is sometimes called "I squared R loss" since power may be
calculated from I*I/R.
 
J

Jamie

Jan 1, 1970
0
hdjim69 said:
I'm self-teaching myself electronics and the only place to ask
questions is in this forum so please excuse me if this has been asked
a million time already but after reading several books, I still have
questions on these topics. No real values, just theory in this
question.

I'm just starting the section on AC and the book is explaining why we
(homes and industry) use AC instead of DC and the use of transformers.
Now, the books says the reason we use AC is to minimize power loss.
That homes and industry need a lot of current and if we were using DC
we'd need to push a huge amount of current through the transmission
lines and the higher the current the more we'd lose in heat loss. OK
fine. But now let's see what happens in AC. Rather then pushing a
huge amount of current we have a very high voltage say 200,000 to
600,000 volts and a low amount of amps (current). But how can we have
this HUGE amount of "pressure" (the typical explanation of what
voltage is) and hardly any current ? I've been reading that voltage
and current are proportional - the more voltage the more current.
Ahh... but this isn't the case really since current is a variable
value. It depends on the amount of resistance. So getting back to the
transmission lines, if we have HIGH voltage and LOW current then
resistance MUST be high. E = I * R that is, if I is low R must be high
to get a high value of E. And resistance is what cause heat which
causes power loss. So how can we have low current + low resistance =
high voltage ?

In summary, if we have very high voltage and low current we must have
very high resistance which would eliminate just about all the current
so loss would be almost 100%.


TIA

J
simple.
AC is simpler to transport because transformers are used to set the
final voltage you need..
the transport of voltage over long halls is done in much higher
voltage than what you're going to use at the destination.
Higher voltages requires smaller wire, less weight and cost.
and being that it's AC, its easy to transfer the voltage to the levels
you need at the destination via a transformer.
as the voltage gets lower, the size of the wire requirement gets
larger.
now, you can transport DC at high voltages also over long runs how
ever, the cost of dropping the voltage down to the level you need and
the size of the converter is not practical.
so, ok, lets say we transport it at the voltage level we are going
to use at the destination, the wire gauge would have to be so large over
any great distance, that wouldn't be practical for a pole. it would
have to be placed in the ground, and even then, still not practical.

simply speaking, the use of a transformer to convert the energy is
far more sensible than any other means. and transformers require AC.
 
P

PeteS

Jan 1, 1970
0
Peter said:
The Power company is delivering Power to us, and power is voltage
times current. So, to deliver a given power, a higher voltage will
permit a lower current.

Power is lost in the transmission lines due to resistance in the wires
- that power is sometimes called "I squared R loss" since power may be
calculated from I*I/R.
I think the OP is hung up on this:

<It depends on the amount of resistance. So getting back to the
transmission lines, if we have HIGH voltage and LOW current then
resistance MUST be high>

Not at all.

We provide a high voltage so the current will be *relatively* low. The
actual resistance is the parallel addition of the loads (ignoring
reactive effects for now) plus the resistance of the cable.

So the current becomes I = V/R (or more properly in this case I = E/R
but that's another discussion....) where V is the source EMF and R is
the effective load.

As others have noted, for a given load, a higher voltage means a lower
current which leads to lower power losses (copper loss).


As the load gets heavier (higher current) then we get larger losses, of
course. By making the source EMF *very high* the current is still low
enough *at this point in the power delivery system* to keep power losses
at a reasonable level.

Before we get to the plant/house/trailer/bar the voltage is transformed
to <some national standard> of 110V - 240V (RMS). That lower voltage is
suitable for consumer (and industrial) equipment and the power loss
within a house (for example) at 110V (again, for example) is negligible
[unless you are running lots of lamps for some interesting recreational
drugs, of course].

Anyway, your hangup point - 'there's high voltage for low current so it
must have been high resistance' is mistaken; we know there is a certain
power load, so we raise the voltage *in the transmission system* such
that the energy can actually be successfully transferred, whatever the
load provided it is within the spec range.

Cheers

PeteS
 
D

Dorian McIntire

Jan 1, 1970
0
hdjim69 said:
I'm self-teaching myself electronics and the only place to ask
questions is in this forum so please excuse me if this has been asked
a million time already but after reading several books, I still have
questions on these topics. No real values, just theory in this
question.

I'm just starting the section on AC and the book is explaining why we
(homes and industry) use AC instead of DC and the use of transformers.
Now, the books says the reason we use AC is to minimize power loss.
That homes and industry need a lot of current and if we were using DC
we'd need to push a huge amount of current through the transmission
lines and the higher the current the more we'd lose in heat loss. OK
fine. But now let's see what happens in AC. Rather then pushing a
huge amount of current we have a very high voltage say 200,000 to
600,000 volts and a low amount of amps (current). But how can we have
this HUGE amount of "pressure" (the typical explanation of what
voltage is) and hardly any current ? I've been reading that voltage
and current are proportional - the more voltage the more current.
Ahh... but this isn't the case really since current is a variable
value. It depends on the amount of resistance. So getting back to the
transmission lines, if we have HIGH voltage and LOW current then
resistance MUST be high. E = I * R that is, if I is low R must be high
to get a high value of E. And resistance is what cause heat which
causes power loss. So how can we have low current + low resistance =
high voltage ?

In summary, if we have very high voltage and low current we must have
very high resistance which would eliminate just about all the current
so loss would be almost 100%.


TIA

J

Image you wanted to transmit 1Kw (kilowatt) of power from point A to point
B. This 1Kw of power could represent 1V (volt) at 1000A (amps) or it could
represent 1000V at 1A or an infinite number of other possibilities. The
product of volts and amps in all cases is 1000 watts. If your transmission
line has 1 ohm of resistance there are two scenarios:



A - You push 1000A of current of through the transmission line and
experience a power loss of 1000A * 1000A * 1Ohm (I squared * R) = 1,000,000W
(watts) or 1000Kw of power dissipated as heat in the transmission line. You
will measure a 1000V voltage drop across the transmission line and will
require 1001V to get your 1V at 1Kw to the load on the other end. You lose
1,000,000 watts of power in the process of transmitting 1000 watts.



B - You push 1 A of current through the transmission line and experience a
power loss of 1A * 1A * 1Ohm = 1 Watt of power dissipated as heat in the
transmission line. You will experience a 1V voltage across the power line
and so will need 1001 volts to get your 1000V at 1Kw to the load on the
other end. You lose 1W of power in the process of transmitting 1000 watts.



You will need a transformer on the other side of the transmission line to
convert your transmitted power to the voltage required by your load.
Transformers make converting power from high current - low voltage to high
voltage - low current and back again trivial in AC systems and with very
little power loss in the transformer itself.



The down side of high-voltage transmission is the possibility of insulation
failure, arc-over and, of course, accidental electrocution of personnel.



The same power transmission problems exist in other systems such as
hydraulic and mechanical systems. If you were transmitting a large amount of
hydraulic power over a long distance you might convert it to high
pressure-low flow using an appropriate hydraulic converter. When
transmitting a large amount of mechanical power a long distance you might
use the gear-train (mechanical equivalent of a transformer) to transmit the
power in the form of a high torque - low rpm. The power robber in all these
systems is friction due to movement of some medium whether its fluid,
rotating parts or moving charges.



Hope this helps.



Dorian
 
D

Dorian McIntire

Jan 1, 1970
0
Sorry, the first word is "Imagine" and not "Image".

I always screw that word up the same way and my work processor never helps.

Dorian
 
K

kell

Jan 1, 1970
0
hdjim69 said:
how can we have
this HUGE amount of "pressure" (the typical explanation of what
voltage is) and hardly any current ?

No current will flow through the power line coming over the mountain
until the people in your town start turning on lamps and air
conditioners. After all that stuff gets turned on, let's say the town
consumes a megawatt.
Using round numbers:
At 100,000 volts the town's megawatt consumption comes to ten amps.
Now, let's make the math easy and say the power line coming over the
mountain has one ohm of resistance, from end to end. You have ten amps
going through a one ohm resistor (the power line). Power dissipated in
a resistor is equal to current squared times resistance for 100 watts
of heat produced in the power line. Losing 100 watts out of a million
isn't too bad.
Now imagine they leave the same power line up but decide to put 120
volts through it so everybody can plug in directly and the power
company doesn't have to put up all those transformers on the telephone
poles... Your town will draw more than eight thousand amps through a
power line with a resistance of one ohm. You can figure out what
happens next.
 
S

Stanislaw Flatto

Jan 1, 1970
0
hdjim69 said:
I'm self-teaching myself electronics and the only place to ask
questions is in this forum so please excuse me if this has been asked
a million time already but after reading several books, I still have
questions on these topics. No real values, just theory in this
question.

I'm just starting the section on AC and the book is explaining why we
(homes and industry) use AC instead of DC and the use of transformers.
Now, the books says the reason we use AC is to minimize power loss.
That homes and industry need a lot of current and if we were using DC
we'd need to push a huge amount of current through the transmission
lines and the higher the current the more we'd lose in heat loss. OK
fine. But now let's see what happens in AC. Rather then pushing a
huge amount of current we have a very high voltage say 200,000 to
600,000 volts and a low amount of amps (current). But how can we have
this HUGE amount of "pressure" (the typical explanation of what
voltage is) and hardly any current ? I've been reading that voltage
and current are proportional - the more voltage the more current.
Ahh... but this isn't the case really since current is a variable
value. It depends on the amount of resistance. So getting back to the
transmission lines, if we have HIGH voltage and LOW current then
resistance MUST be high. E = I * R that is, if I is low R must be high
to get a high value of E. And resistance is what cause heat which
causes power loss. So how can we have low current + low resistance =
high voltage ?
*************************************************************************
In summary, if we have very high voltage and low current we must have
very high resistance which would eliminate just about all the current
so loss would be almost 100%.

I read most of the replies, and would suggest to open again the chapter
on transformers. In there should be mentioned the fact that a
transformer reflects the load connected to one of its windings by a
ratio of its windings, so:

Nin/Nout = Zin/Zout (Zout is the final load)

So Zin represents to the source a much higher value and can be supplied
with higher voltages and smaller currents for given power consumption.

And in this way you reduce the ohmic losses on the intermediate
connecting lines.

HTH
************************************************************************


Stanislaw
Slack user from Ulladulla.
 
D

Dorian McIntire

Jan 1, 1970
0
Stanislaw Flatto said:
I read most of the replies, and would suggest to open again the chapter on
transformers. In there should be mentioned the fact that a transformer
reflects the load connected to one of its windings by a ratio of its
windings, so:

Nin/Nout = Zin/Zout (Zout is the final load)

So Zin represents to the source a much higher value and can be supplied
with higher voltages and smaller currents for given power consumption.

Actually the formula for impedance as it relates to turns ratio is:

(Nin/Nout)Squared = Zin/Zout or Nin/Nout = Sqrt(Zin/Zout)


Dorian
 
H

Homer J Simpson

Jan 1, 1970
0
That homes and industry need a lot of current and if we were using DC
we'd need to push a huge amount of current through the transmission
lines and the higher the current the more we'd lose in heat loss. OK
fine.

There is no DC transformer. Edison's system was never practical.
But now let's see what happens in AC. Rather then pushing a
huge amount of current we have a very high voltage say 200,000 to
600,000 volts and a low amount of amps (current). But how can we have
this HUGE amount of "pressure" (the typical explanation of what
voltage is) and hardly any current ?

It's like gearing. Very fast but less force instead of very slow but high
force.
 
J

jasen

Jan 1, 1970
0
I'm self-teaching myself electronics and the only place to ask
questions is in this forum so please excuse me if this has been asked
a million time already but after reading several books, I still have
questions on these topics. No real values, just theory in this
question.
Rather then pushing a
huge amount of current we have a very high voltage say 200,000 to
600,000 volts and a low amount of amps (current). But how can we have
this HUGE amount of "pressure" (the typical explanation of what
voltage is) and hardly any current ?

It's like the difference between a fire hydrant (high flow low pressure)
and and a water blaster (high pressure low flow),

the 330kV distribution lines have isulators over 1m in length, they need to
be that size to resist the pressure of all that voltage.
I've been reading that voltage
and current are proportional - the more voltage the more current.

Correct, that is Ohm's law...
they are only proportional when the resistance remains the same.
Ahh... but this isn't the case really since current is a variable
value. It depends on the amount of resistance. So getting back to the
transmission lines, if we have HIGH voltage and LOW current then
resistance MUST be high.

when you're dealing with AC impedance is the term that's used and yes the
impedance is high. You can think of impedance being like resistance until
you encouter the part in the book that explains the difference.

The high impedance seen on is the high voltage side of the step-down transformer.
On the low voltage side the real load is located.
the transformer by tansforming the high voltage into low voltage is also
transforming the low impedance load into a high impedance one (when measured
from the high voltage side).

Bye.
Jasen
 
S

Stanislaw Flatto

Jan 1, 1970
0
Dorian said:
Actually the formula for impedance as it relates to turns ratio is:

(Nin/Nout)Squared = Zin/Zout or Nin/Nout = Sqrt(Zin/Zout)


Dorian

Thanks for turning the clock back about 45 years, the times when "hi
fidelity" was using transformers, to connect spakers to push-pull 807
tubes. Then those calcullations were in front, were displaced by others
during time and my technical books are gathering dust.

Have fun

Stanislaw
Slack user from Ulladulla.
 
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