# Please clarify Voltage/Current relationship

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Thread Starter

#### levydee

Joined Oct 18, 2016
14
Good afternoon,

I am learning about basic electronics and attempting to establish a firm foundation of understanding on the basic principles involved. However I have run into a problem. Ohm's law is rather basic. V = IR. Great, no problem. But in the context of a transformer, how does that apply? From what I have read, if you step up the voltage, you lose current, if you step down the voltage, you increase the current. But according to Ohm's law, the only things that effect current is resistance and voltage.

Free power does not exist, so how do you determine voltage gain/Current loss and vice versa?

To clarify, if I have 10 volts on a circuit with 500 ohms resistance, my current is 20ma. If I step up the 10 volts to 50 volts(with 500 ohm resistance still), I magically get 100ma! Unfortunately that is not how it works.

Thank you.

#### BR-549

Joined Sep 22, 2013
4,938
Why does 100 ma. seem magical to you? Why was the 20 ma. NOT magical to you?

Ohm's law works for 50 volts just like it does for 10 volts.

#### Sinus23

Joined Sep 7, 2013
246
@BR-549 .OP is talking about transformers. Which is probably why he stated that 10V and 20mA stepped up to 50V at 100mA doesn't make sense. One is 200mW and the other is 5W. What I think is getting the thread starter confused is the 500 ohm resistance he used in his example in relations with transformers and Ohms law.

#### BR-549

Joined Sep 22, 2013
4,938
levydee................show us a schematic. Your question is not clear.

#### shteii01

Joined Feb 19, 2010
4,644
Consider a hill.

This hill has a gentle slope. You descent from the top of the hill to the bottom at a leisurely pace of 4 km/h.

What if the hill has a sharp slope. You are forced to descent fast in order to stay upright. You speed is 7 km/h.

What changed between the two hills? The slope. Is there magic?

Thread Starter

#### levydee

Joined Oct 18, 2016
14
Consider a hill.

This hill has a gentle slope. You descent from the top of the hill to the bottom at a leisurely pace of 4 km/h.

What if the hill has a sharp slope. You are forced to descent fast in order to stay upright. You speed is 7 km/h.

What changed between the two hills? The slope. Is there magic?
I guess I am missing a fundamental concept than. It is my understanding that you cannot create power from nothing. If I can theoretically step up(via transformer) an input voltage of 10v to 10000v, the power output must be the same correct? That means their has to be a drop in the current provided. So how do you measure the current when it is stepped up to 10000v as opposed to when it was only 10v?

#### shteii01

Joined Feb 19, 2010
4,644
I guess I am missing a fundamental concept than. It is my understanding that you cannot create power from nothing. If I can theoretically step up(via transformer) an input voltage of 10v to 10000v, the power output must be the same correct? That means their has to be a drop in the current provided. So how do you measure the current when it is stepped up to 10000v as opposed to when it was only 10v?
Um... Did you really had to bring up transformers? That goes into fields and fluxes and stuff! You really really REALLY don't want to go there.

Unless... you are trolling?

#### BR-549

Joined Sep 22, 2013
4,938
You are mis-understanding fundamentals. I still don't understand your question. Draw a picture.

#### hp1729

Joined Nov 23, 2015
2,304
Good afternoon,

I am learning about basic electronics and attempting to establish a firm foundation of understanding on the basic principles involved. However I have run into a problem. Ohm's law is rather basic. V = IR. Great, no problem. But in the context of a transformer, how does that apply? From what I have read, if you step up the voltage, you lose current, if you step down the voltage, you increase the current. But according to Ohm's law, the only things that effect current is resistance and voltage.

Free power does not exist, so how do you determine voltage gain/Current loss and vice versa?

To clarify, if I have 10 volts on a circuit with 500 ohms resistance, my current is 20ma. If I step up the 10 volts to 50 volts(with 500 ohm resistance still), I magically get 100ma! Unfortunately that is not how it works.

Thank you.
Voltage is electrical pressure. Current is a result. In the transformer example the current is a possible allowed current. At the lower secondary voltage we have a potential of a higher current. The current does not automatically increase. Current depends on resistance and voltage.

#### DGElder

Joined Apr 3, 2016
351
I guess I am missing a fundamental concept than. It is my understanding that you cannot create power from nothing. If I can theoretically step up(via transformer) an input voltage of 10v to 10000v, the power output must be the same correct? That means their has to be a drop in the current provided. So how do you measure the current when it is stepped up to 10000v as opposed to when it was only 10v?

Assume an ideal 1:10 step up transformer. On the primary side you have an ideal voltage source connected across the primary coil and across the secondary coil you have a resistor. Now assume you put 10V AC on the primary coil and you measure 100 ma of current in the primary. You know the secondary will be at 100V AC. What is the current in the secondary? It has to be 10mA. The power into the primary is 10V*0.10A = 1 Watt. So the secondary has to deliver 1 watt of power. Therefore the current in the secondary must be 1 Watt/100V = 10mA. I also know that the secondary circuit resistance must be 10K ohms (R=V^2/P)

But then you might say "what if the resistance in the secondary circuit is only 10 ohms? Given a 100V secondary voltage across 10 ohms the current would be 10A - so I am getting 100V*10A=1000 Watts! 1000 times the power I put into the transformer!"

You would indeed get 1000W, but you can not get more power out than you put in. If the secondary resistance were 10 ohms then the primary resistance (because the secondary impedance is reflected back into the primary) would be only 0.1 ohms and therefore the primary current would be 100A. Primary power = 10V*100A = 1000W = Secondary power.

Since we are talking about an ideal transformer the primary coil has no resistance. The primary impedance is a^2 times the impedance that the secondary coil sees - where a is the turns ratio: primary turns/secondary turns. In this case a =1/10, so the primary impedance is 1/100 of the resistance in the secondary circuit.

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#### KeepItSimpleStupid

Joined Mar 4, 2014
5,090
But in the context of a transformer, how does that apply? From what I have read, if you step up the voltage, you lose current, if you step down the voltage, you increase the current. But according to Ohm's law, the only things that effect current is resistance and voltage.
In a transformer Power is conserved: e.g. Pp=Ps or Vp*Ip = Vs*Is

Ac voltage is an entirely different animal to work with. Ohms law works with resistive loads. Ohms law for AC requires that the voltage measured is the RMS voltage and the load is RESISTIVE. Most meters ASSUME a sine wave, Measure a precision rectified sine wave and average it and multiply it by a "fudge factor". So, the meter has a particular frequency response AND the input must be a sine wave.

Fundamental concepts (Capacitor): The Voltage cannot change instantaneously.
(Inductor): The current through an inductor cannot change instantaneously.
In a resistor, voltage and current can change instantaneously.

Any component and even wires may have capacitance and inductance, but it is usually ignored.

#### MrChips

Joined Oct 2, 2009
24,641
Thread Starter is mixing up how Ohm's Law applies to a resistor and power conservation through a transformer.

A resistance of 500Ω will demand 20mA when 10V is applied across the resistor. Ohm's Law I = V/R applies.
Power consumed by the resistor = I x V = 0.02A x 10V = 0.2W

If we increase the voltage across the resistor to 50V, current I = V/R = 50/500 = 100mA
Power consumed by the resistor = I x V = 0.1A x 50V = 5W

The transformer will deliver the power demanded by the load.
Conservation of power across the transformer means that the power drawn from the primary winding equals the power delivered to the secondary winding (for a 100% efficient transformer, i.e. no power loss).

On the primary side, the power supplied, I x V must still match the power on the secondary side. By changing the turns ratio of the transformer you have created a different transformer. Hence you have a new situation that is different from the first.

There is no free lunch. There is no inconsistency here.

#### Blockhead

Joined Oct 23, 2016
9
Good afternoon,

I am learning about basic electronics and attempting to establish a firm foundation of understanding on the basic principles involved. However I have run into a problem. Ohm's law is rather basic. V = IR. Great, no problem. But in the context of a transformer, how does that apply? From what I have read, if you step up the voltage, you lose current, if you step down the voltage, you increase the current. But according to Ohm's law, the only things that effect current is resistance and voltage.

Free power does not exist, so how do you determine voltage gain/Current loss and vice versa?

To clarify, if I have 10 volts on a circuit with 500 ohms resistance, my current is 20ma. If I step up the 10 volts to 50 volts(with 500 ohm resistance still), I magically get 100ma! Unfortunately that is not how it works.

Thank you.
You are making a simple mistake in your assumption that both the Primary and Secondary windings will have the same resistance. This will only be the case when the voltage and current you apply to the primary winding is the same as the voltage and current you get from the secondary winding.
If you are increasing the voltage at the secondary winding then to do that you will have more turns on the secondary. If you have more turns then the wire will have be thinner. This means the resistance of the secondary will be higher than the primary winding.
If you use your sample voltage and current of 10 volts and 20mA the total power consumed will be 200mW (Power = Voltage x Current). You will get virtually same output power available from the secondary. (A well designed transformer is 98% or more efficient.)
If you have 200mW available at the secondary then you can calculate the current available from the secondary. Power = Voltage x Current
200mW = 50V x Current, so secondary Current = 200mW/50V = 4mA. You could then work out the resistance of the secondary winding
Resistance = Voltage/Current, R = 50/4mA = 12,500 Ohm.
Hope that helps

Thread Starter

#### levydee

Joined Oct 18, 2016
14
Awesome, thank you for the valuable information from everybody. That definitely clears things up for me. Cheers!

#### EM Fields

Joined Jun 8, 2016
583
Good afternoon,

I am learning about basic electronics and attempting to establish a firm foundation of understanding on the basic principles involved. However I have run into a problem. Ohm's law is rather basic. V = IR. Great, no problem. But in the context of a transformer, how does that apply? From what I have read, if you step up the voltage, you lose current, if you step down the voltage, you increase the current. But according to Ohm's law, the only things that effect current is resistance and voltage.

Free power does not exist, so how do you determine voltage gain/Current loss and vice versa?

To clarify, if I have 10 volts on a circuit with 500 ohms resistance, my current is 20ma. If I step up the 10 volts to 50 volts(with 500 ohm resistance still), I magically get 100ma! Unfortunately that is not how it works.

Thank you.
In an ideal transformer, the power delivered into a load, by the secondary, will be equal to the power put into the primary, by the source.

For example, if the secondary is delivering 12 volts into a 12 ohm resistor, then the current through the resistor will be: I = E/R = 12V/12Ω = 1 ampere, and the power dissipated by the resistor will be:
P = EI = 12V × 1A = 12 watts.

Now, assuming that the source is 120 volt mains, it must put 12 watts into the transformer's primary in order to get 12 watts into the load. That means that since the transformer's working with 120 volt mains, then to get 12 watts from the mains it must draw I = P/E = 12W/120V = 0.1 ampere and the mains will think it's supplying 12 watts into R = E/I = 120V/0.1A = 1200 ohms.

#### WBahn

Joined Mar 31, 2012
26,398
Good afternoon,

I am learning about basic electronics and attempting to establish a firm foundation of understanding on the basic principles involved. However I have run into a problem. Ohm's law is rather basic. V = IR. Great, no problem. But in the context of a transformer, how does that apply?
Basic answer: It doesn't apply.

Ohm's Law is a relationship that applies to some materials and components and not to others. Specifically, it applies to materials that are "ohmic". What is an "ohmic" material or device, you might ask. Somewhat circularly, it is a material that obeys (at least to a good approximation) Ohm's Law. A transformer is not an ohmic device, so Ohm's Law simply doesn't apply to it. What does apply to a transformer, to a reasonable first-order approximation, is the notion of power conservation between input and output.

#### EM Fields

Joined Jun 8, 2016
583
Basic answer: It doesn't apply.

Ohm's Law is a relationship that applies to some materials and components and not to others. Specifically, it applies to materials that are "ohmic". What is an "ohmic" material or device, you might ask. Somewhat circularly, it is a material that obeys (at least to a good approximation) Ohm's Law. A transformer is not an ohmic device, so Ohm's Law simply doesn't apply to it. What does apply to a transformer, to a reasonable first-order approximation, is the notion of power conservation between input and output.
I don't believe there's anything about Ohm's law which states that E/I must be a constant in order for R = E/I to be valid at the instant of measurement, and there are no materials or components which are truly ohmic in the sense which you propose so, with respect, I must state that I view your premise, and conclusion, as flawed.

#### WBahn

Joined Mar 31, 2012
26,398
I don't believe there's anything about Ohm's law which states that E/I must be a constant in order for R = E/I to be valid at the instant of measurement, and there are no materials or components which are truly ohmic in the sense which you propose so, with respect, I must state that I view your premise, and conclusion, as flawed.
Perhaps you should learn more about Ohm's Law, which merely states that the current between two points in a conductor is directly proportional to the voltage across those two points. Then go discover what "proportional to" means.

And trying to nitpic on "truly ohmic in the sense" I propose is not going to get you anything since, if nothing else, I specifically covered that base in bounding it as a good approximation (and for most ohmic materials it is a damn good approximation over a very wide range of conditions).

Beyond that, you can view it however you like. Then, when you are ready, you can go and educate yourself of what Ohm's Law is and what an ohmic material is.

#### KeepItSimpleStupid

Joined Mar 4, 2014
5,090
I think the deal here is, "If you have a hammer", you can't see everything as a nail".
To solve problems, you generally have to make assumptions that you believe are valid for the conditions.

It gets weird when 0 isn't an attainable number sometimes and 1e18 and 4e18 can be considered the same "value" because your interested in "order of magnitude" types of changes.

A bulk material property is Resistivity, but that is a function of temperature.
R = pL/A is the resistance of some length of homogeneous material with A cross-sectional area A, that has the material property p (Rho) of resistivity.

#### EM Fields

Joined Jun 8, 2016
583
Perhaps you should learn more about Ohm's Law, which merely states that the current between two points in a conductor is directly proportional to the voltage across those two points. Then go discover what "proportional to" means.

And trying to nitpic on "truly ohmic in the sense" I propose is not going to get you anything since, if nothing else, I specifically covered that base in bounding it as a good approximation (and for most ohmic materials it is a damn good approximation over a very wide range of conditions).

Beyond that, you can view it however you like. Then, when you are ready, you can go and educate yourself of what Ohm's Law is and what an ohmic material is.
Hmm...
You change your avatar but, apparently, you can't change your hostile demeanor when your point of view is questioned.

The point is that regardless of the properties of the material, be it an insulator, a semiconductor, or a conductor, if it isn't forced into superconductivity, Ohm's law can always be used to determine the voltage across, the current through, or the resistance of a chunk of material at any instant in time if two of the three quantities are known. And, for the purpose of responding to the OP's query, it's perfectly fine to consider the transformer as a perfect transducer where power out equals power in.

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