# Resistance in a circuit with transformers

1. Feb 20, 2011

### woodentsick

1. The problem statement, all variables and given/known data
Hey PF!
I had a problem that was bugging my mind. We know that in any circuit, the total resistance has to equal the total voltage divided by the total current, right? That is, R=V/I. And we also know that when you have a step up or step down transformer, power is conserved, right? (Let's assume a perfect transformer with the wires having almost no resistance). So assume this situation:

You have a circuit of 12V and 2A, making total power = 24W. You have a step-up transformer from 120 turns to 240 turns. Therefore, your second circuit will have 24V and 1A. Correct? Now let's imagine the second circuit has a light bulb in it with a resistance of 10 Ohms. However, that does not follow the R=V/I rule because 10 does not equal 24/1. So my question is, how does the resistance in a circuit affect the way the transformer works?

2. Relevant equations

R=V/I
Primary power = secondary power

3. The attempt at a solution

I have searched around a little, and the only solution I can find are that the current is somehow reduced, and power is lost. However I don't understand how this works. How can power be lost?

Woodentsick

2. Feb 20, 2011

### tiny-tim

Welcome to PF!

Hey Woodentsick! Welcome to PF!
No, you can specify the voltage, but the current will depend on the resistance R2 of the secondary circuit, which shows up as an apparent resistance across the primary coil of R2/T2, where T is the turn ratio.

eg if the primary circuit has a voltage V, with a resistance R1, and with voltage V1 across the primary coil, and if the secondary circuit has a resistance R2, and with voltage V2 across the secondary coil, then …

V2 = I2R2

V2 = TV1 and I2 = I1/T

So V = I1R1 + V1

= I1(R1 + V1/I1)

= I1(R1 + V2/I2T2)

= I1(R1 + R2/T2) …

ie the primary coil looks as if it has the resistance of the secondary circuit, divided by the square of the turn ratio.

(And I1V1 = I2V2.)

3. Feb 20, 2011

### woodentsick

Hello tiny-tim,

Thanks for your reply, but I am afraid I do not fully understand it :(

Firstly, how did you get

V = I1R1 + V1? Shouldn't V = I1R1 only?

And secondly, if you don't mind, could you also explain the next few steps?

Thanks so much,

Woodentsick

4. Feb 20, 2011

### Delta²

Tiny tim presents the truth from mathematical point of view. I would like to focus on the energy-power point of view.

When you say that we have a power of 24W at a voltage of 12V and current 2A the whole thing is abit not well presented. It would be more accurate to say that the maximum power is 24W, and since this power is given at 12V then the maximum current is 2A.

So while the whole system is within the maximum power limits (that is the power consumed does not exceed 24W) then the voltage on the primary would be 12V, the voltage on secondary 24V but the current on the primary can be anywhere between 0 and 2A and on secondary between 0 and 1A. How much exactly is the current depends on the resistance load we put on the secondary. In your example you set a load resistance of 10Ohm which means that the current will be 2.4A and the power will be 2.4^2*10=57.6W which is outside the power limits, so we dont know exactly how the system will behave in this case (most likely there would be a voltage drop at both the primary and the secondary so the current will be lower also such that for the new current I it will be I^2*10=24 that is I=1,549 and the new voltage will be not 24V but 15,49V).

However if you set a load resistance of 100Ohm then the current will be 0,24A (which is between 0 and 1A) and the power 5,76W which is below 24W. In this case the system works as intended and with the tiny-tim equations the primary "would feel" a resistance 100/4=25Ohm (we set R1=0 , R2=100, T=2) though the 100Ohm resistance is connected to the secondary.

This resistance the primary feels is a consequence of the conservation of energy. Since energy is conserved the power on primary and secondary should be equal, hence

$$V_1I_1=V_2I_2$$ hence $$I_1=\frac{V_2}{V_1}I_2$$. For I_2=0,24A we get I1=0,48A. So the primary is like it feels a resistance $$R_2^{'}=\frac{V_1}{I_1}=\frac{12}{0.48}=25Ohm$$.

5. Feb 20, 2011

### tiny-tim

Hello Woodentsick!
I assumed there was a resistor R1 in the primary circuit (it really doesn't make any sense to have a cirucit with no resistance in it! ), as well as the transformer.

So the total voltage V equals the sum of the voltage drops across the two components … I1R1 across the resistor, and V1 across the coil.
I don't get what you don't get.

6. Feb 22, 2011

### woodentsick

Hello tiny-tim,

Now I get what you mean, but I just want to clarify that I meant all the resistance in the primary circuit was due to the transformer. Because the primary circuit is 12V; 2A the resistance of the transformer will be 12/2 = 6 Ohms.

Well, because of the responses from tiny-time and Delta^2, this problem seemed almost solved in my eyes (thanks again guys!) However I talked to my physics teacher about it who said that that it was in fact much more complex than that and required an understanding of inductance, capacitance and impedance (I'm in high school and have no idea what those words mean).

But I'm confused, because isn't it as simple as R1 = R2/T2 ? (which I've managed to derive on my own so I don't feel bad about it :P)

Once again, thanks so much PF!

Woodentsick

Last edited: Feb 22, 2011
7. Feb 22, 2011

### Delta²

Your teacher is correct, it involves impendance (which is a term that combines inductance, capacitance and resistance). But at the high school level we neglect inductance and capacitance so that impedance equals ohmic resistance.

If we use impendance we ll get that the total impendance on the primary is $$Z_1+\frac{Z_2}{T^2}$$

with $$Z_i=\sqrt{R_i^2+(L_if-\frac{1}{C_if})^2}$$ where

$$R_i$$ is ohmic resistance
$$L_i$$ is inductance
$$C_i$$ is capacitance
$$i=1,2$$

f is the frequency of the alternating voltage on the primary.

Last edited: Feb 22, 2011
8. Feb 22, 2011

### tiny-tim

Hello Woodentsick!
Yup, I assumed you wouldn't know about impedance, so I didn't confuse the issue by mentioning it.

For the record (by which I mean, file it away for a few years! ), if you replace voltage by complex voltage, current by complex current, and resistance by impedance, then the equations in my previous post are all still correct.

(loosely speaking, resistors have real resistance R, coupled to the current I, but inductors and capacitors in an AC circuit have imaginary resistance X, called reactance, which depends on frequency, coupled to dI/dt instead of I, and you can add them all to get a general complex number called impedance Z = R + iX)

9. Feb 22, 2011

### woodentsick

Ohhh so to fully understand this problem, I'll need to learn about impedance? Well, now I know what's gonna be eating up my lunch break! :D Thanks guys, you've been REALLY helpful, and I'll report back once I learn more about electricity!

Till then, goodbye!

Woodentsick

P.S. Delta^2, when you say total impedance on the primary is $$Z_1+\frac{Z_2}{T^2}$$, is Z1 the impedance originally on the primary, and Z2 the impedance on the secondary?

10. Feb 23, 2011

### Delta²

Yes. Have in mind that Tiny-tim is more accurate than me on the definition of impendance.

11. Feb 23, 2011

### Phrak

Where did you get this equation? It looks wrong. The presents of a square root looks wrong. Impedance in series add like resistance in series, and impedance in parallel combine as resistance in parallel. It's that simple. And we never see frequency without a 2pi attached to it. For example, Z_L = 2pi i f L = i omega L, Z_C = 1/(2pi ifC) = 1/( i omega C).

$$Z = (Z_L + Z_R)||Z_C$$

where Z_R = R = winding resistance, and Z_C is the interwinding capacitance.

Last edited: Feb 23, 2011
12. Feb 23, 2011

### Delta²

I hate to say it but you are right phrak :)

13. Feb 23, 2011

### tiny-tim

dooom!

Well, the spelling anyway!

(I think only doom is measured in impendance! )

(the SI unit is the d'ohm, d'Ω)​

14. Feb 24, 2011

### Phrak

Yeah. Sorry about that. I was in a fowl and beflumexed mood--in part due to some past and partial enlightenment by his very self Tiny, by the way.

Last edited: Feb 24, 2011