Relating Transformer Parameter I2 to V1 (Non-ideal) Updated

[ZenSerpent]

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Homework Statement

Relate secondary current to primary voltage in a single-phase transformer having resistive load for non-ideal situation.

Homework Equations

KVL, Voltage across resistor and inductor, ac voltage as a sine wave, reluctance, Faraday's law,other equations relevant to working principle of transformer.

The Attempt at a Solution

Good day everyone! I would like to verify whether my attempt for relating secondary current to primary voltage in a single-phase transformer having resistive load is correct or not. Are there other losses that I have not taken into account? If so, feel free to critique my work. This is actually a work of my colleagues and I for our research. So here's our go at it:Using KVL for the equivalent circuit in the primary side gives,v_1=v_{R_1}+v_{L_1}Substituting their corresponding values gives

v_{1max}{\mathrm{sin} \omega t\ }=i_1R_1+L_1\frac{di_1}{dt}Solving DE for i_1 gives

i_1=\frac{v_{1max}}{R^2_1+{{\omega }^2L}^2_1}\left(R_1{\mathrm{sin} \omega t\ }-\omega L_1{\mathrm{cos} \omega t\ }\right)\Combining the sine wave and cosine wave and simplifying gives

i_1=\frac{v_{1max}}{\sqrt{R^2_1+{\omega }^2L^2_1}}{\mathrm{sin} \left(\omega t-{\mathrm{arctan} \frac{\omega L_1}{R_1}\ }\right)\ }\Now since N_1i_1=\mathcal{R}\mathrm{\Phi }\, solving for {\Phi } gives

\mathrm{\Phi }=\frac{N_1i_1}{\mathcal{R}}\. This is the flux that flows to the secondary circuit after applying reluctance of the core.
Substituting i_1=\frac{v_{1max}}{\sqrt{R^2_1+{\omega }^2L^2_1}}{\mathrm{sin} \left(\omega t-{\mathrm{arctan} \frac{\omega L_1}{R_1}\ }\right)\ }\ to \mathrm{\Phi }=\frac{N_1i_1}{\mathcal{R}}\ gives\mathrm{\Phi }=\frac{N_1v_{1max}}{\mathcal{R}\sqrt{R^2_1+{\omega }^2L^2_1}}{\mathrm{sin} \left(\omega t-{\mathrm{arctan} \frac{\omega L_1}{R_1}\ }\right)\ }\

To determine the flux linkage in the secondary winding, we multiply N2 to both sides, givingN_2\mathrm{\Phi }=\frac{N_2N_1v_{1max}}{\mathcal{R}\sqrt{R^2_1+{\omega }^2L^2_1}}{\mathrm{sin} \left(\omega t-{\mathrm{arctan} \frac{\omega L_1}{R_1}\ }\right)\ }\Taking derivative wrt t gives the voltage across the secondary circuit. That is,v_2=N_2\frac{d\mathrm{\Phi }}{dt}=\frac{N_2N_1v_{1max}\omega }{\mathcal{R}\sqrt{R^2_1+{\omega }^2L^2_1}}{\mathrm{cos} \left(\omega t-{\mathrm{arctan} \frac{\omega L_1}{R_1}\ }\right)\ }\Using KVL on the secondary circuit, we getv_2=v_{R_2}+v_{R_L}+v_{L_2}\Substituting their corresponding values give,\frac{N_2N_1v_{1max}\omega }{\mathcal{R}\sqrt{R^2_1+{\omega }^2L^2_1}}{\mathrm{cos} \left(\omega t-{\mathrm{arctan} \frac{\omega L_1}{R_1}\ }\right)\ }=i_2R_2+i_2R_L+L_2\frac{di_2}{dt}\\frac{N_2N_1v_{1max}\omega }{\mathcal{R}\sqrt{R^2_1+{\omega }^2L^2_1}}{\mathrm{cos} \left(\omega t-{\mathrm{arctan} \frac{\omega L_1}{R_1}\ }\right)\ }=i_2(R_2+R_L)+L_2\frac{di_2}{dt}\Solving DE for i_2 gives,

i_2=\frac{N_2N_1v_{1max}\omega }{\mathcal{R}\sqrt{R^2_1+{\omega }^2L^2_1}\left[{\left(R_2+R_L\right)}^2+{\omega }^2L^2_2\right]}\left[\left(R_2+R_L\right){\mathrm{cos} \left(\omega t-{\mathrm{arctan} \frac{\omega L_1}{R_1}\ }\right)\ }+\omega L_2{\mathrm{sin} \left(\omega t-{\mathrm{arctan} \frac{\omega L_1}{R_1}\ }\right)\ }\right]\Combining the sine and cosine waves gives and simplifying gives,

i_2=\frac{N_2N_1v_{1max}\omega }{\mathcal{R}\sqrt{R^2_1+{\omega }^2L^2_1}\sqrt{{\left(R_2+R_L\right)}^2+{\omega }^2L^2_2}}{\mathrm{sin} \left\{\omega t+{\mathrm{arctan} \left[\frac{R_1\left(R_2+R_L\right)-{\omega }^2L_1L_2}{R_1\omega L_2+\left(R_2+R_L\right)\omega L_1}\right]\ }\right\}\ }\
Critiquing this will be of great help for our group's thesis. Thank you so much. :)

 

[rude man]

≪ mod note: thread moved from technical forum so template is missing ≫

Homework Statement

[/B]
Relate secondary current to primary voltage in a single-phase transformer having resistive load for non-ideal situation.
v_1=v_{R_1}+v_{L_1}

If L1 is the primary inductance then you're off to a bad start. Or you need to show a diagram of your non-ideal transformer, identifying these parameters.

I would start with basic equations for your transformer:

v1 = L₁ di₁/dt + M di₂/dt
v2 = M di₁/dt + L₂ di₂/dt

L₁, L₂ are primary & secondary inductances
M = mutual inductance (have you considered this?)
v₁, v₂ = pri & sec voltages
i₁, i₂ = pri & secondary currents

and work from there. These equations accommodate loading, leakage inductances, winding resistance, etc.

If you want to include consideration of core fluxes see my Insight article at https://www.physicsforums.com/insights/misconceiving-mutual-inductance-coefficients/

 

[ZenSerpent]

Thanks Sir rude man. Basing from your comment,

v1 = L1 di1/dt + M di2/dt
v2 = M di1/dt + L2 di2/dt

This implies that this is a system of DE and I need to solve for the corresponding functions of i1 and i2 as functions of time, right? As for the v1 and v2, shall I express them explicitly in the form v = v_max sin(ωt + θ)? I think this would be impossible for i2 since I don't know its phase angle yet.

M = mutual inductance (have you considered this?)
Well Sir, you can see in the last equation the following term N1*N2/R which is equal to μ*N1*N2*A/l. Is this not equal to \sqrt{L_1L_2}? I may be wrong. I don't know.

"These equations accommodate loading, leakage inductances, winding resistance, etc."
How are these inductances related to resistance? Is there an explicit formula expressing the relationship among these variables?

P.S. I'm an Applied Mathematics student, not an Engineering one. But I'm studying this field for our research. Thanks in advance. :)

 

[rude man]

Thanks Sir rude man. Basing from your comment,
v1 = L1 di1/dt + M di2/dt
v2 = M di1/dt + L2 di2/dt
0
This implies that this is a system of DE and I need to solve for the corresponding functions of i1 and i2 as functions of time, right? As for the v1 and v2, shall I express them explicitly in the form v = v_max sin(ωt + θ)? I think this would be impossible for i2 since I don't know its phase angle yet.

Let's assume an input voltage V = V₀sin(wt), then a series resistor R1 connected to the transformer primary v1, then a load resistor R2 across the secondary v2. Then the equations would be
(V - v1)/R1 = i1
v1 = L1 di1/dt + M di2/dt
v2 = M di1/dt + L2 di2/dt
v2 = -i2. (I hope this last answers your question about i2.)

4 independent equations, 4 unknowns. Solve classically or, preferably, using phasors which I guess you haven't covered.

M = mutual inductance (have you considered this?)
Well Sir, you can see in the last equation the following term N1*N2/R which is equal to μ*N1*N2*A/l. Is this not equal to \sqrt{L_1L_2}? I may be wrong. I don't know.

. M = k√(L1 L2) and I don't see the "k" which is the coupling coefficient and affects mutual inductance. I believe you assumed unity coupling (all primary flux coupling into the secondary). This is close to true for a power transformer with its high-permeability core, but for (for example) an IF radio transformer it's far from true.

I will try to find time to pay more attention to your rather extensive (and impressive!) post.

"These equations accommodate loading, leakage inductances, winding resistance, etc."
How are these inductances related to resistance? Is there an explicit formula expressing the relationship among these variables?

Inductances L1 and L2 are the inductances of the primary & secondary windings respectively and have no direct relationship to the respective winding resistances.

P.S. I'm an Applied Mathematics student, not an Engineering one. But I'm studying this field for our research. Thanks in advance. :)

You can certainly use classical calculus to solve these equations, though phasors are preferred since they change the diff. eqs. to algebraic ones. But you do need at least basic circuit analysis to do anything with the equations I provided.

Have you looked at my Insight paper? You need not be concerned with the main topic but I do relate electrical parameters to magnetic ones which may be of interest to you.

 

[ZenSerpent]

For this set of equations,

(V - v1)/R1 = i1
v1 = L1 di1/dt + M di2/dt

Are you trying to say that
(V0sin(wt) - L1 di1/dt + M di2/dt)/R1 = i1, right? So V and v1 are actually different? If V is the input voltage, then what is v1?

Also, I said this statement:

I think this would be impossible for i2 since I don't know its phase angle yet.
I'm actually referring to v2 instead of i2. I apologize for that.

And for this equation:
v2 = -i2
Why is this so? I don't understand. I mean, they don't even have the same units. Are you referring to their respective magnitudes only?

M = k√(L1 L2) and I don't see the "k" which is the coupling coefficient and affects mutual inductance. I believe you assumed unity coupling (all primary flux coupling into the secondary). This is close to true for a power transformer with its high-permeability core, but for (for example) an IF radio transformer it's far from true.
Well, for this Sir, the result excluding k just surfaced from my derivations. Probably it is due to my assumption above that after reluctance is divided to the mmf, the resulting flux will link to the secondary side.

Have you looked at my Insight paper? You need not be concerned with the main topic but I do relate electrical parameters to magnetic ones which may be of interest to you.
I did, and you came up with the equation M = √(k1k2L1L2) which is pretty interesting. How do I know the values for k's?

 

[rude man]

For this set of equations,

(V - v1)/R1 = i1
v1 = L1 di1/dt + M di2/dt

Are you trying to say that
(V0sin(wt) - L1 di1/dt + M di2/dt)/R1 = i1, right? So V and v1 are actually different? If V is the input voltage, then what is v1?

you're a bit sloppy with a sign here, but otherwise yes. V is the input voltage to resistor R1 in series with the transformer primary. v1 is the transformer primary voltage (the low end of both primary & secondary assumed grounded). Sorry I can't draw a diagram right now so go with what I described earlier.

Also, I said this statement:
I think this would be impossible for i2 since I don't know its phase angle yet.
I'm actually referring to v2 instead of i2. I apologize for that.

still don't see what your problem is . 4 equations, 4 unknowns. You are solving for all four of them and v2 is one of them.

And for this equation:
v2 = -i2
Why is this so? I don't understand. I mean, they don't even have the same units. Are you referring to their respective magnitudes only?

That was a big old blooper! I meant to write v2 = -R2 i2. R2 is a resistor across the secondary winding (the transformer load).

M = k√(L1 L2) and I don't see the "k" which is the coupling coefficient and affects mutual inductance. I believe you assumed unity coupling (all primary flux coupling into the secondary). This is close to true for a power transformer with its high-permeability core, but for (for example) an IF radio transformer it's far from true.
Well, for this Sir, the result excluding k just surfaced from my derivations. Probably it is due to my assumption above that after reluctance is divided to the mmf, the resulting flux will link to the secondary side.

yes I think that was the assumption which is fine for a power transformer and certain others having relatively high-permeability cores so that the flux is well confined to the intended magnetic path.

Have you looked at my Insight paper? You need not be concerned with the main topic but I do relate electrical parameters to magnetic ones which may be of interest to you.
I did, and you came up with the equation M = √(k1k2L1L2) which is pretty interesting. How do I know the values for k's?

That depends on how the transformer is manufactured. For most transformers k1 and k2 are the same or closly the same so there is just one coupling coefficient and then M = k√(L1 L2). For power transformers, even small ones, k ~ 1.0.

 

[The Electrician]

Since you apparently want a steady state solution, there's no need to solve differential equations. Using the Laplace variable s, a mesh solution can be simply obtained:

 

[rude man]

Since you apparently want a steady state solution, there's no need to solve differential equations. Using the Laplace variable s, a mesh solution can be simply obtained:

View attachment 112324

Knowledge or application of the Laplace transform is not necessary if the steady-state, sinusoidal solution is sought. Phasors suffice. But the OP is probably not familiar with phasors either, he/she stating that he/she is not an engineering but an applied math student. So classical calculus is probably his/her only way to go.

 

[ZenSerpent]

Sirs, I already figured it out. But the equations really looked very hideous. I may be able to post it later.

 

[ZenSerpent]

The Electrician, may I know what the initial equations looked like before they were set up in the matrix and before their Laplace transforms were taken? I also have a somehow considerable understanding of Laplace transforms and transfer functions btw, Sirs, but not that deep enough.

 

[The Electrician]

There's no need to perform any Laplace transforms to perform steady state AC circuit analysis.

See this page just beyond half way down, under the section heading "Generalized s-plane impedance" at this page: https://en.wikipedia.org/wiki/Electrical_impedance. The variable s can be taken to be simply jω for AC circuit analysis.

Then read this page: https://en.wikipedia.org/wiki/Mesh_analysis for a short overview of the technique of phasor mesh analysis.

The two mesh equations are:

1.) (s*L1+R1)*I1 + (-s*M)*I2 = V1
2.) (-s*M)*I1 + (s*L2+R2+RL)*I2 = 0

The minus sign in the term "-s*M" is there because of the dotted ends of the transformer windings. If one of the dots were on the other end of the winding, the sign would be changed.

In post #8, I showed the equations in matrix form, which allows the use of a matrix solver to obtain the solution. But any technique for solving a pair of simultaneous linear equations can be used.

 

[The Electrician]

You might find this helpful: http://jazapka.people.ysu.edu/ECEN 2632 Chapter 9.pdf

Especially look at page 6.

 

[ZenSerpent]

The Electrician, I now understand what you're trying to say and I tried so solve them manually. Luckily, I got the same answer for the magnitude and angle.
If we translate this back in the t-domain, will i(t) = L^(-1){V1(s)}*(magnitude derived)*sin(wt + (angle derived)) or i(t) = V1max*magnitude derived* sin(wt + (angle derived)) or should the phase angle be negative? Thanks.

 

[The Electrician]

The Electrician, I now understand what you're trying to say and I tried so solve them manually. Luckily, I got the same answer for the magnitude and angle.
If we translate this back in the t-domain, will i(t) = L^(-1){V1(s)}*(magnitude derived)*sin(wt + (angle derived)) or i(t) = V1max*magnitude derived* sin(wt + (angle derived)) or should the phase angle be negative? Thanks.

The phase of the current with respect to the applied voltage is delayed, as you would expect in an inductor, but the current in the secondary of your transformer will experience a possible additional 180° phase shift depending on where the windings are dotted: https://en.wikipedia.org/wiki/Polarity_(mutual_inductance)

 

[ZenSerpent]

How about if it is dotted the same way as you did in formulating the initial equations?

And what will the equation in the time domain will look?

 

[The Electrician]

How about if it is dotted the same way as you did in formulating the initial equations?

Substitute some values for L1, L2, M, R1, R2, RL and ω into the expression I obtained for I2/V1:

and find the angle of the result. If the dot on one winding of the transformer is reversed, then substitute M = -M and re-evaluate the expression.

And what will the equation in the time domain will look?

This is your homework, not mine. What do you get for the time domain equation? I see a zero and a couple of poles in the expression, so I would expect a moderately complicated time domain expression.

 

[ZenSerpent]

I got this answer. Is this acceptable?

 

[ZenSerpent]

Sir, also noticed that the answer is not in Amperes.

 

[ZenSerpent]

Oops. I forgot the convolution theorem.

 

[The Electrician]

View attachment 112408

I got this answer. Is this acceptable?

This is not the steady state result. You will note that you have a decaying exponential factor. What you have is the result of suddenly applying a sine wave at t=0. Get rid of the exponential factor and the remaining part should be the steady state result.

But you already have the steady state result in phasor format. Why have you bothered to invoke Laplace transforms?

 

[ZenSerpent]

This is not the steady state result. You will note that you have a decaying exponential factor. What you have is the result of suddenly applying a sine wave at t=0. Get rid of the exponential factor and the remaining part should be the steady state result.

But you already have the steady state result in phasor format. Why have you bothered to invoke Laplace transforms?

So the expression outside the brackets is the answer?

I applied LT since I need it to be in the time domain. Is there something wrong in doing it?

 

[The Electrician]

You're making me do a lot of work here. Let me explain why using phasor analysis saves a lot of work, and is the right way to do things when you want a steady state result. I'm going to use Mathematica to do a lot of the work involved here. You may have to do it by hand, or with some other method.

First, let's suppose we have a very simple frequency dependent circuit consisting of a LR voltage divider, with a sine wave source driving the input:

Using phasor analysis, we could obtain an expression for the output (I should have labeled the output; it's the right two terminals) using the voltage divider rule. We get Vout/Vin = R/(R+sL)

Let's use Laplace transforms to get an expression for the time domain output with a simple sine wave voltage of angular frequency ω applied as Vin.

First get an expression for the transform of a simple sine wave:

Now let's get the inverse Laplace transform of our LR circuit output with a sine wave input. A very important point here is that using this method assumes that Vin is zero until time t=0; at t=0, the sine wave is suddenly applied. In the Laplace domain we multiply the transform of the input sine wave times the expression for the output of the circuit and derive the Inverse Laplace transform to get the time domain voltage at Vout:

This consists of a transient part (in red), a decaying exponential, and a steady state part (in black), which is the voltage waveform after the exponential decays to zero.

Let's plot this waveform. I'll arbitrarily pick some values for R, L and ω, and plot the output waveform (in red) starting at t=0 for several cycles of the waveform. The plot also shows the waveform of Vin (in blue) scaled by 1/5 to make it fit the plot nicely:

You can see the effect of the initial transient (the decaying exponential) at the beginning. Now let's eliminate the exponential part of the time domain expression and plot it again:

Now we have only the steady state part of the output waveform. This shows that the output is just a sine wave phase shifted with respect to the input.

In the next post, I'll show how the phasor method of analysis gets the same result.

 

[The Electrician]

Now let's use the phasor method to get the output voltage waveform of the circuit.

The output voltage is given by Vout = Vin \frac {R}{R+sL}

To get a time waveform we substitute s = jω. Mathematica uses a special symbol for the imaginary unit j; it should be obvious in what follows.

The phasor result is a sine wave if the input is a sine wave, which is the basic assumption of phasor analysis. The magnitude of the output sine wave is given by the magnitude of our (complex arithmetic) expression for the output fraction: \frac {R}{R+j ω L}

The magnitude is given by the square root of the sum of the real part squared and the imaginary part squared, and the angle is the Arctan of the imaginary part divided by the real part. So we would need to separate the real and imaginary parts. Mathematica has a function to do this, and it's relatively simple for this simple circuit. For your transformer it will be much more complicated. Herewith the separation, with the imaginary part in red:

Now we calculate the magnitude and angle of our output sine wave:

We can plot the scaled input sine wave, and the output sine wave using those values:

Here for comparison purposes is the plot of the steady state time waveform from the previous post:

It's identical. Notice how that even for this very simple circuit, the time domain expression derived with a Laplace transform was moderately complicated.

The expression for your transformer is going to be immensely complicated if you derive it via Laplace transform. Using the magnitude and angle expression derived from the phasor analysis I gave in post #8 will be much easier and less complicated looking.

The result you got in post #1 seems fairly simple, but you have a variable, (script) R, involving flux in the core and flux linkages between primary and secondary. This hides a lot of complication; you don't need to involve flux anyway. Using mutual inductance (M) is the way to go.

But, if you must use the Laplace transform, remember that the exponential part should be eliminated if you want the steady state result.

The result you have in post #18 doesn't look right. The exponential part and steady state part should be added together, not multiplied.

 

[ZenSerpent]

Correct me if my understanding of this is wrong.

So using this circuit, you determined the output to input ratio as R/(R + sL). Now since, this is Vo(s)/Vi(s), you wanted to obtain the function for vo(t), given that vi(t) = sin(wt). So what you basically did was you took the LT of vi(t) so that we will have an explicit function for Vi(s) then you arranged the expression for the ratio of the two s.t. Vo(s) = Vi(s) R/(R + sL)
Then you took the inverse laplace for this, which sounds like convoluting sin(wt) * L^(-1){R/(R + sL)}. The result is the sum of a decaying exp function and another nondecaying function. As t tends to infinity, obviously the decaying exp function approaches 0 and the nondecaying part is the vo(t) in steady state.

If that's the case, then my intuition tells me that my previous solution, excluding vo sinwt, needs to be convoluted with vo sinwt, right? Since in the first place, I took the ILT of the right hand side of I2(s)/V1(s) which is the expression on the paper excluding vo sin wt. And I just mistakenly multiplied it by vo sin wt since I have no prior idea on convolution theorem. But just a few hours ago, I was able to learn it.

 

[The Electrician]

Correct me if my understanding of this is wrong.

So using this circuit, you determined the output to input ratio as R/(R + sL). Now since, this is Vo(s)/Vi(s), you wanted to obtain the function for vo(t), given that vi(t) = sin(wt). So what you basically did was you took the LT of vi(t) so that we will have an explicit function for Vi(s) then you arranged the expression for the ratio of the two s.t. Vo(s) = Vi(s) R/(R + sL)
Then you took the inverse laplace for this, which sounds like convoluting sin(wt) * L^(-1){R/(R + sL)}. The result is the sum of a decaying exp function and another nondecaying function. As t tends to infinity, obviously the decaying exp function approaches 0 and the nondecaying part is the vo(t) in steady state.

If that's the case, then my intuition tells me that my previous solution, excluding vo sinwt, needs to be convoluted with vo sinwt, right?

Don't say "convoluted"; say "convolved".

Since I don't how you derived your previous solution, which I assume involved finding Inverse Laplace transforms, I can't answer your question.

You correctly described what I did. You need to do that with the phasor expression for I2/V1 I gave in post #8.

 

[The Electrician]

As you obviously know, in the time domain what you need to do is convolve the function sin[ωt] with the Inverse Laplace transform of the ratio I2/V1. But it's much easier to do in the Laplace domain where multiplication replaces convolution. So your big problem is to find the Inverse Laplace transform of the product of the transform of the input sine wave and the transform of the ratio I2/V1. In other words, you need to find this:

I wouldn't even think of trying to do this without the aid of modern mathematical software; but then I'm spoiled.

Just to give a taste of what your result is going to look like, here's what Mathematica gives for the Inverse transform of just the ratio I2/V1.

 

[The Electrician]

I assume you are a university student. Find out if your institution has network access to a modern mathematical software such as Mathematica, Maple, Matlab, etc. Such a software should enable you to derive inverse Laplace transforms.

 

[rude man]

Sir, also noticed that the answer is not in Amperes.

You should avoid the use of the laplace transform altogether. That also precludes convolution. Instead, either use phasors as I described earlier, or, since you're a math major not an engineering one, solve the equations classically.

The powerful laplace transform enables transient as well as steady-state solutions but this is of no interest here.

 

[ZenSerpent]

Oh my! So the best option is phasor analysis. Using that on z = I2(s)/V1(s) the time domain function is i2(t) = |z| sin(wt - arg(z)) right?

 

[The Electrician]

Oh my! So the best option is phasor analysis. Using that on z = I2(s)/V1(s) the time domain function is i2(t) = |z| sin(wt - arg(z)) right?

I explained this in post #23

 

[The Electrician]

You should avoid the use of the laplace transform altogether. That also precludes convolution. Instead, either use phasors as I described earlier, or, since you're a math major not an engineering one, solve the equations classically.

The powerful laplace transform enables transient as well as steady-state solutions but this is of no interest here.

Did you not read the posts following #19?

 

[ZenSerpent]

Sir rude man, yup, failure to perform convolution is a serious mistake. But if I were to perform convolution, I'll get an expression as hairy as the one posted by The Electrician in the post above. So, I think phasor analysis is the way to go. And I now gained some good insight just now. And I noticed that the intuition is much similar to the concepts discussed in complex number analysis. Now everything appears to be clear to me.

 

[ZenSerpent]

The Electrician, so the expression I mentioned in #29 is correct right? One thing I noticed is that all the units cancel out. Why is this?

 

[The Electrician]

The Electrician, so the expression I mentioned in #29 is correct right? One thing I noticed is that all the units cancel out. Why is this?

sin(ω t) has units of volts. That's the units of your final answer.

 

[rude man]

I explained this in post #23

Yes, but don't say "I(s), V(s) etc. Avoid "s" completely. Say I(jw), V(jw) etc. which are phasors.

 

[The Electrician]

Yes, but don't say "I(s), V(s) etc. Avoid "s" completely. Say I(jw), V(jw) etc. which are phasors.

On this page: https://en.wikipedia.org/wiki/Electrical_impedance, a little over halfway down under the heading "Generalized s-plane impedance" is the justification for using s to stand for jω. It's common usage for "circuits that are driven with a steady-state AC signal".

You even do it yourself, for example here: https://www.physicsforums.com/threads/transfer-function-for-rc-and-rl-circuits.738708/#post-4663732

and here: https://www.physicsforums.com/threa...cy-response-of-a-circuit.726774/#post-4593798

 

[ZenSerpent]

sin(ω t) has units of volts. That's the units of your final answer.

Well this sounds counter-intuitive to me as, v1max is in volts and sin wt is dimensionless.

So the amplitude of the input sine wave, which is v1 max doesn't really matter at all in the i2(t)?

 

[The Electrician]

Well this sounds counter-intuitive to me as, v1max is in volts and sin wt is dimensionless.

So the amplitude of the input sine wave, which is v1 max doesn't really matter at all in the i2(t)?

The expression V1max*sin(ω t) derives its units from the amplitude variable V1max. Plain old sin(ϖ t) is really 1*sin(ϖ t), and derives its units from the amplitude 1--that is, 1 volt. Does that make sense to you?

 

[ZenSerpent]

So instead of

i2(t) = |z| sin(wt - arg(z))

It needs to be

i2(t) = V1max*|z| sin(wt - arg(z))

Or I'll just have to consider
i2(t) = |z| sin(wt - arg(z)) with sin(wt -arg(z)) having a unit of volts?

 

[ZenSerpent]

What if the initial amplitude is v1 max =/= 1. Will I not be able to see its value in the expression in i2(t). I apologize as this still confuses me up to now.

 

[The Electrician]

So instead of

i2(t) = |z| sin(wt - arg(z))

It needs to be

i2(t) = V1max*|z| sin(wt - arg(z))

Or I'll just have to consider
i2(t) = |z| sin(wt - arg(z)) with sin(wt -arg(z)) having a unit of volts?

If V1max equals 1 (and therefore isn't explicitly shown), it still has units of volts, so your final result will have units of volts. In other words, your final result has units of volts whatever the value of V1max.

 

[ZenSerpent]

So the bottom line of all these is that we cannot see V1max in the function of i2(t) = |z| sin(wt - arg(z)) but the unit of sin(wt - arg(z)) is in volts since there will be a hidden 1 volt in there regardless of what the value of V1max is, right? (Even if V1max is not equal to 1.)

 

[The Electrician]

You should be able to see V1max in i2(t) = |z| sin(wt - arg(z)) because to show exactly where it is, we should write i2(t) = |z|*V1max*sin(wt - arg(z)). The only reason I didn't show it is because I assumed the input excitation voltage was 1 volt, just for simplicity. Whenever a sine wave voltage function is used, if there isn't an explicit multiplier for the peak voltage, assume it is unity.

You know that the standard mathematical sine function has a peak value of 1, even though the multiplier of 1 is not explicitly shown. If you want to show a sine function with a peak value of 3, you will have to write out the 3, as 3*sin(ω t), but its a convention to not show the multiplier if it's 1.

We don't need to assume that there's a hidden 1 if V1max is not 1 because then the numerical value of V1max will be explicitly present and the V1max expression is what carries the units. It's only when V1max is 1 that it won't (by convention) be shown as a multiplier of 1*.

I suppose I should have shown it in post #23, but I assumed it would be clear that I was using a multiplier of 1 for simplicity in my work, and that it carried units of volts.

 

[ZenSerpent]

That's what I'm trying to say several posts ago. Woah! Thank you so much The Electrician and rude man.

 

[ZenSerpent]

Sir Electrician. There is still a problem. I applied what you did in Mathematica using the expressions for the transformer. Attached are photos for the phasor one and the other one for the Laplace. I scaled both 10000x. It turns out that they don't look identical.

 

[The Electrician]

One big problem is that your expression for M, namely M = 3*SQRT(L1 L2) is not possible. The 3 you have there is the coupling coefficient, and the maximum physically possible value is 1. I wouldn't even choose it to be very close to 1 or you might have numerical problems; I wouldn't set to greater than .95 I set it to .5 in what follows.

I can't tell if you got the right Inverse Laplace tranform, but you shouldn't keep all those exponentials when you plot the result. They are likely to cause numerical problems. Here's what I got; I only show the first two parts with the steady state part first in red, and the first exponential. Only keep the steady state part. There's a minus sign in front of the whole thing due to the particular choice of dots on the transformer windings; if one dot is moved to the other side, it will change the sign of the result:

Using the expressions for magnitude and phase derived from the phasor solution here's what I get for 1/magnitude and phase using your numerical values. I calculated the reciprocal of the magnitude because that's what you need to multiply the transform expression by to make the red trace fit the plot:

Here's a plot of the input sine wave including a phase angle of -1.09383 radians. I left its magnitude as 1 in this plot, and multiplied the Laplace expression by 1/magnitude. You can see that both blue and red plots are on top of each other. So the result from the phasor solution is identical to the result from the Inverse Laplace transform:

 

[ZenSerpent]

Got it. But I now have this question. Notice that arg(z) is already negative. That means the general form should be i2(t) = V1max*|z|*sin(wt + arg(z))? Not i2(t) = V1max*|z|*sin(wt - arg(z)) for those circuits that involve inductance?

 

[ZenSerpent]

Alright, I get it. When we expanded the complex numbers into the real and imaginary part, we disregarded the negative sign in computing for its arg( ) but we promised to put the negative sign in sin(wt - arg(z)). So generally speaking, it's just |z| sin(wt + arg(z)) if we included the negative sign in the computation of arg(z).