How step-up transformers help in transmission of electrical energy over long distances?

i have read just read this blog:

http://www.blueraja.com/blog/194/do-transformers-obey-ohms-law

below blockquotes is taken from the blog above:

We connect a 1 volt AC generator to a 1 ohm resistor and measure the current. By Ohm’s Law, we should get 1 ampere of current.

Now imagine we stuck a 1:10 transformer in the circuit, splitting our one circuit into two electrically-separate circuits. The confusion arises from the following question: does the current through the resistor go up because the voltage went up, or down because the transformer needs to conserve power?

Treating the transformer as a 10-volt AC voltage source in the right circuit, we use Ohm’s law to see that the current through the resistor has gone up – it is now 10 amps. In order to preserve power, this means that in the left circuit our original AC power source is now drawing 100 amps of current, 100x what it was drawing before.

and, i have something in my textbook stating:

The large scale transmission and distribution of electrical energy over long distances is done with the use of transformers. The voltage output of generator is stepped-up (so that the current is reduced and consequently, the $$I^2R$$ loss is cut down). It is then transmitted over long distances to an area sub-station near the consumers. There the voltage is stepped-down. It is further stepped-down at the distributing sub-stations and utility pole before a power supply of $$240 V$$ reaches our homes.

well, as told in the blog, step-up transformers increases both voltage and current proportionally on the secondary side (right side), so $$I^2R$$ loss increases.. this is confusing me.. have i mistaken something or have a big misconception??

also, why the textbook refers to $$I^2R$$ loss, why not $$VI$$ loss or $$\frac{V^2}{R}$$ loss, which includes voltage, which definitely increases??

• @BlueRaja - Danny Pflughoeft kindly look at the question.. – PranshuKhandal Feb 10 at 17:15
• If transformers step up both voltage and current, where does the power come from? – The Photon Feb 10 at 17:22
• @The Photon because it draws even more power from the source, as 1:10 step-up transformer drew x100 power from source, its in the blog – PranshuKhandal Feb 10 at 17:26
• Duplicate – Farcher Feb 10 at 19:26

step-up transformers increases both voltage and current proportionally on the secondary side

This is incorrect.

Say you have a 1:100 turns-ratio transformer, with (for example) 100 V input and 10 kV output, and the load on the secondary draws 1 A (for example). Then the source on the primary side will have to provide 100 A, not 0.01 A.

The current is stepped down in the same proportion (for an ideal, lossless transformer) as the voltage is stepped up.

without transformer current was 1 amp. but with transformer, it became 10 amp.

The example assumes an ideal AC voltage source, with the same source voltage before and after inserting the transformer.

In this case, it's correct that the current delivered to the load increases by inserting the transformer, so 10 A is delivered to the resistive load. But also notice that the current drawn from the source increased even more --- to 100 A.

If your load was located far from the source, you'd rather put the transformer near the source and only send the 10 A current over the long transmission line, rather than put it near the load so that the transmission line has to carry 100 A.

Even better (as far as reducing $$I^2 R$$ losses in transmission) you could put a 1:100 transformer near the source, and a 10:1 transformer at the load. Then the source would still need to source 100 A. And the load would still get 10 A. But the transmission line would only need to carry 1 A.

• got it, thnx :) – PranshuKhandal Feb 10 at 18:02
• @PranshuKhandal, what example? Please edit your question to include the example you want to ask about. – The Photon Feb 10 at 18:02
• it was in the blog, i said, without transformer current was 1 amp. but with transformer, it became 10 amp. – PranshuKhandal Feb 10 at 18:05
• Please edit your question to quote the relevant part of the blog. We want the question to still be understandable for future readers, even if the blog post gets removed. – The Photon Feb 10 at 18:07

Sorry for my english !

I think you are making a confusion between the resistance of the transmission line with the equivalent load of the user.

Imagine that you have 10 houses that each consume 10 KW. You must therefore provide 100 KW. At the user level, the voltage must be 110 V and therefore the total current would be roughly 1000 A.

If you pass these 1000 A in a line of several hundred miles, Joule losses $${{R}_{\text{line}}}{{I}^{2}}$$ in the line will be huge.

The solution is to raise the voltage at the input of the line. For example, (not real...) if the transport voltage along the line is 10 KV then for 100 KW, the current in the line is only 10 A and therefore Joule losses $${{R}_{\text{line}}}{{I}^{2}}$$ in the line are much reduced.

At the user level, it is necessary to lower the voltage again to 110 V.

• thnx, for the help :) – PranshuKhandal Feb 10 at 18:08

In any discussion of this type of problem, it is vital to be clear as to exactly what voltage, current and resistance one is talking about.

So, consider a source of 240 V alternating current. That is, 240 V at the output terminals of the source. The source is capable of supplying energy at a high power level; many megawatts, without any drop in voltage.

We want to power a 2400 Watt electric heater from this source. So we plug one in a few feet away, the source supplies 10 Amperes of current at 240 V , and the heater delivers heat at a 2400 W rate. This also means that the heater must have a load resistance ($$R_{Load}$$) of 24 Ohms

Now, we decide to connect our source directly to a distant heater; so distant that the transmission lines have a significant resistance of their own, $$R_{Tran}$$.

Now the source sees a slightly higher resistance, and so delivers a slightly lower current. The source delivers a slightly lower power ($$V^2/R_{Total}$$) and slightly lower current, but now the energy is divided between the heater and the transmission wires.

(Most importantly, there may be thousands of heaters at the destination end of the line, all in parallel. This lowers the effective load resistance, but not the transmission line resistance. So a large fraction of the power coming out of the source winds up wasted to space, or overheating the wires.)

So now we take the 240 V current from the generator and run it through a step up transformer, to say 240,000 Volts, 1000 x as much. The transformer doesn't change the energy going through it, so the current on the output side is 1/1000 of the current on the input side.

This vastly reduced current flows through the same transmission resistance on its way to the load. By $$I^2R_{Tran}$$ the loss in the transmission line is $$1/1,000,000$$ of what it was previously.

It only remains to run the high voltage current through a step-down transformer close to the load to complete the almost loss-free transmission.

There will be some minor losses in the step up and step down transformers

• but why do we use I×I×R only and not V×I or V×V/R – PranshuKhandal Feb 10 at 18:18
• We use the quantities that we know. We don't know the voltage drop along the transmission line; we do know the current. Read the first sentence of my answer... – DJohnM Feb 10 at 18:21
• but we know voltage has increased – PranshuKhandal Feb 10 at 18:24
• and we derived p=i×i×r by p=v×i and v=i×r – PranshuKhandal Feb 10 at 18:26
• @PranshuKhandal, Be sure you understand what "Voltage" means. Voltage is only defined between two distinct points in a circuit. When we say that the power delivered by a generator is $V\times I$, the $V$ is the voltage between the generator's two output terminals. When we say that the power that is lost by a length of wire is $V^2/R$, the $V$ in that case is the voltage between the two ends of the wire, which is completely different from the voltage at the output of the generator or the voltage at the load or anywhere else in the circuit. – Solomon Slow Feb 10 at 19:18