Resistance in a circuit with transformersby woodentsick Tags: circuit, electricity, magnetism, resistance, transformer, transformers 

#1
Feb2011, 05:57 AM

P: 6

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 stepup 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? Thanks in advance for your help, Woodentsick 



#2
Feb2011, 06:53 AM

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Hey Woodentsick! Welcome to PF!
eg if the primary circuit has a voltage V, with a resistance R_{1}, and with voltage V_{1} across the primary coil, and if the secondary circuit has a resistance R_{2}, and with voltage V_{2} across the secondary coil, then … V_{2} = I_{2}R_{2} V_{2} = TV_{1} and I_{2} = I_{1}/T So V = I_{1}R_{1} + V_{1} = I_{1}(R_{1} + V_{1}/I_{1}) = I_{1}(R_{1} + V_{2}/I_{2}T^{2}) = I_{1}(R_{1} + R_{2}/T^{2}) … ie the primary coil looks as if it has the resistance of the secondary circuit, divided by the square of the turn ratio. (And I_{1}V_{1} = I_{2}V_{2}.) See also http://en.wikipedia.org/wiki/Transfo...sic_principles. 



#3
Feb2011, 07:19 AM

P: 6

Hello tinytim,
Thanks for your reply, but I am afraid I do not fully understand it :( Firstly, how did you get V = I_{1}R_{1} + V_{1}? Shouldn't V = I_{1}R_{1} only? And secondly, if you don't mind, could you also explain the next few steps? Thanks so much, Woodentsick 



#4
Feb2011, 08:01 AM

P: 256

Resistance in a circuit with transformers
Tiny tim presents the truth from mathematical point of view. I would like to focus on the energypower 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 tinytim 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 [tex]V_1I_1=V_2I_2[/tex] hence [tex] I_1=\frac{V_2}{V_1}I_2[/tex]. For I_2=0,24A we get I1=0,48A. So the primary is like it feels a resistance [tex]R_2^{'}=\frac{V_1}{I_1}=\frac{12}{0.48}=25Ohm[/tex]. 



#5
Feb2011, 10:12 AM

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Hello Woodentsick!
So the total voltage V equals the sum of the voltage drops across the two components I_{1}R_{1} across the resistor, and V_{1} across the coil. 



#6
Feb2211, 02:56 AM

P: 6

Hello tinytim,
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 tinytime 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 R_{1} = R_{2}/T^{2} ? (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 



#7
Feb2211, 04:00 AM

P: 256

If we use impendance we ll get that the total impendance on the primary is [tex]Z_1+\frac{Z_2}{T^2}[/tex] with [tex]Z_i=\sqrt{R_i^2+(L_if\frac{1}{C_if})^2}[/tex] where [tex] R_i [/tex] is ohmic resistance [tex] L_i [/tex] is inductance [tex] C_i [/tex] is capacitance [tex] i=1,2[/tex] f is the frequency of the alternating voltage on the primary. 



#8
Feb2211, 06:03 AM

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Hello Woodentsick!
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
Feb2211, 06:40 AM

P: 6

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 [tex]Z_1+\frac{Z_2}{T^2}[/tex], is Z1 the impedance originally on the primary, and Z2 the impedance on the secondary? 



#10
Feb2311, 12:17 AM

P: 256

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




#11
Feb2311, 01:08 AM

P: 4,513

[tex]Z = (Z_L + Z_R)Z_C[/tex] where Z_R = R = winding resistance, and Z_C is the interwinding capacitance. 



#12
Feb2311, 03:28 AM

P: 256

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




#13
Feb2311, 04:26 AM

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(I think only doom is measured in impendance! ) 



#14
Feb2411, 02:58 AM

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