Transformers: high voltage, low current AND low resistance?

[Laura555]

From my understanding, transformers allow electricity to travel long distances with minimal energy loss because they initially use a step-up transformer which reduces the current (which reduces the heat loss and hence energy loss). However, I do not understand how a step-up transformer manages to create a high voltage and low current AND also has a low resistance (to minimise heat loss). From Ohm's law, the resistance of the primary circuit (in picture above) should be:
R=V/I= 6/0.1 = 60 Ohms
The resistance of the secondary circuit, however, should be:
R=V/I = 48/0.0125 = 3840 Ohms. [I know this must be incorrect, because it would make no sense as it would lead to massive heat loss!]
Obviously, the power is the same in both circuits, P=IV=0.6W.
My question is:
1. What is the resistance of the secondary (middle) circuit and why does Ohm's law not work here?
2. Also, if the type of wire used in all the circuits was the same, they should have identical resistances (if there are no appliances are in the circuits). So how could 48V (48J of energy per coulomb) produce such a low current - if the energy isn't moving the electrons to produce a high current or being wasted as heat to the surroundings, where is it going?

Many thanks for any help offered!

 

[BvU]

Hello Laura,

Not the way it works. In the figure, the ohmic resistance in the primary circuit on the left is exactly zero (very unrealistic): there is no power lost. Same for the 48 V circuit. And the resistance (in the transformer coil) in the circuit on the right is also zero, since it manages to deliver 0.1 A at 6 V.

In reality they use the thickest possible wire for the coil turns that's still economical.

So:
1. Zero. Ohm's law doesn't work because it concerns alternating current here.
2. In a more realistic case the wire has resistance, say 0.1 $\Omega$/m. For a consumer 500 m away that is 100 Ohm (50 plus 50 back). To deliver 1 kW at 110 V to this customer, so 9 Ampere, the power line connection dissipates 8100 Watt (100 Ohm at 9 A is 900 Volt, 900 Volt x 9 Ampere = 8100 Watt). So 90% of the power supplied is lost.

Same power line, now with transformers: up by a factor 100, so 11000 volt. Now the current in the high-voltage line is only 0.09 A, so the voltage loss over the power lines is 9 V. 9 V x 0.09 A = 0.81 W, an improvement by a factor of 10000 ! (simply because the power loss is $I^2R$). So its economical to accept a bit of loss in the transformers (we didn't take that into account yet) to reduce the power loss. Actual high-voltage lines work at even higher voltages: 100000 V or more.

And 48 volt is indeed 48 Joule/Coulomb, but it is also 48 (Joule/s)/(Coulomb/s), i.e. Watt/Ampere

--

 

[davenn]

Not the way it works. In the figure, the ohmic resistance in the primary circuit on the left is exactly zero (very unrealistic): there is no power lost. Same for the 48 V circuit. And the resistance (in the transformer coil) in the circuit on the right is also zero, since it manages to deliver 0.1 A at 6 V.

and to add to what BvU has stated
when dealing with AC, Ohms Law is not applicable. Ohms Law works well for basic DC circuits
We don't have plain resistance of wires or coils of wires any more. instead we now have impedance
Impedance deals with how an AC supply is affected by a coil of wire, and transmission lines. We have to look at things like inductive reactance and
capacitive reactance.

this comment by BvU is also very important for you to grasp ...

So its economical to accept a bit of loss in the transformers (we didn't take that into account yet) to reduce the power loss

with your circuit, you are not likely to get the 0.0125A @ 48V out of the first transformer because of the losses in that transformer
The same will happen in the second transformer with its losses.
Lets say, for example sake (without getting bogged down in the maths), out of the first transformer you can get a max of 0.01A instead of your 0.0125A. See the little bit of loss there?, that means you now have only 0.01A plus any losses in the wires between the transformers, say 0.009A
Now you are down-converting again to 6V. and that second transformers loss is added. There may only be 0.005A available @ 6V on the output

No system is perfect, You are NEVER going to get out the same power you put in. There will ALWAYS be losses

Dave

 

[BvU]

Well, more treats:
To add to what dave added (also to correct myself and to nitpck a little ): when dealing with AC, Ohms Law is quite applicable for pure resistance. But - in contrast with direct current - with AC we have a phenomenon that voltage and current can be out of phase (the sines don't go though the maximum at the precise same moment due to reactance and inductance). To be sure, dave explains that quite well.
My estimate was that in this stage of your curriculum you haven't been introduced to these aspects yet. So my answer 1. was formulated a bit crudely - which may have triggered dave.

Physics is interesting: there's always a next layer of complexity to explore !

 

[CWatters]

In case you are confused... The currents shown on your diagram are not due to the resistance of the windings. It's the load (aka light bulb) and the transformer winding ratios that dictate the current.

First the voltages (work from left to right):

Starting at the left with the 6V supply. The winding ratio of the first transformer is 1:6 so the voltage is stepped upto 48V. The next transformer has a winding ratio of 6:1 so that steps down the voltage to 6V.

Now for the currents (work right to left)

Start on the right hand side of the drawing with the load/bulb. The 0.1A current is due to the resistance of the bulb. The resistance of the bulb is not specified but it must be around 60 Ohms. The current drawn from the transformer is therefore I = 6/60 = 0.1A.

The turns ratio of the right hand transformer is 6:1. So the current on the transmission line is 0.1/6 = 0.0167A. Aside: The drawing says 0.0125A which I believe is a minor error.

The transformer on the left has a winding ratio of 1:6 so the current drawn from the 6V AC source is increased to 0.0167 * 6 = 0.1A

 

[jbriggs444]

48:6 = 8:1 winding ratio. The 0.0125 A is correct for an 8:1 ratio.

 

[davenn]

48:6 = 8:1 winding ratio. The 0.0125 A is correct for an 8:1 ratio.

assuming no losses ( but we know there are )

 

[CWatters]

Slaps head. Never was good at maths

 

[sophiecentaur]

In case you are confused... The currents shown on your diagram are not due to the resistance of the windings. It's the load (aka light bulb) and the transformer winding ratios that dictate the current.

First the voltages (work from left to right):

Starting at the left with the 6V supply. The winding ratio of the first transformer is 1:6 so the voltage is stepped upto 48V. The next transformer has a winding ratio of 6:1 so that steps down the voltage to 6V.

Now for the currents (work right to left)

Start on the right hand side of the drawing with the load/bulb. The 0.1A current is due to the resistance of the bulb. The resistance of the bulb is not specified but it must be around 60 Ohms. The current drawn from the transformer is therefore I = 6/60 = 0.1A.

The turns ratio of the right hand transformer is 6:1. So the current on the transmission line is 0.1/6 = 0.0167A. Aside: The drawing says 0.0125A which I believe is a minor error.

The transformer on the left has a winding ratio of 1:6 so the current drawn from the 6V AC source is increased to 0.0167 * 6 = 0.1A

That's a really good way to think about it. You have to stick, strictly to that and not try to second guess what's going on when you are half way through the process, though. You need to get the 'causes and effects' in the right order or you can find apparent paradoxical behaviour.
@Laura: have you ever seen the school / college demonstration, using wires strung about the lab and 'safe' voltages, throughout? It shows a dim bulb when you use no transformers and a bright bulb when you introduce the transformers.

 

[anorlunda]

when dealing with AC, Ohms Law is not applicable. Ohms Law works well for basic DC circuits

Just to nitpick some more, Ohms Law works just fine for AC circuits using impedance rather than resistance. Impedance can be inductive, capacitive, resistive, or any combination of those. In AC analysis, Voltage, Current, and Impedance are all complex.

 

[sophiecentaur]

Just to nitpick some more, Ohms Law works just fine for AC circuits using impedance rather than resistance. Impedance can be inductive, capacitive, resistive, or any combination of those. In AC analysis, Voltage, Current, and Impedance are all complex.

And to nitpick further, the term "Ohm's Law" implies a linear resistor (a metal at constant temperature, actually). 'Resistance' doesn't have to be constant and there are many components that are definitely Resistive (i.e. no inductance of capacitance) but which are non-ohmic. Bottom line is that the term Resistance does not imply there's Ohm's Law operating. Glib use of the term Ohm's Law can and will introduce confusion somewhere along the line. Resistance is just the ratio between Volts and Current and is in Ohms but I can't understand why so many people don't feel that it's important to distinguish. This is not an irrelevant point. Merely re-wording a post could take care of the problem easily and avoid confusion.

 

[BvU]

My estimate is and remains that for Laura all this "complexity" is undesirable in this stage of her introduction to electricity.

 

[sophiecentaur]

I agree. If CWatters had got his sums right (let him that is without sin etc. etc.) the thread could have terminated on Post #3. We started bickering amongst ourselves.