Electrotechnic exercise (help)

[The Electrician]

I think the OP has the right numerical answer (although I'm having trouble following how he got it.) I did the simple voltage divider approach, but since it should not matter from which direction the voltage division is done, I did it both ways (Eqns 2 & 4) to ensure the answer was the same both ways. MAPLE worksheet attached.

There's my mistake. In the table of component values, where it says w = 50 rad/s, I just mentally changed that to 50 Hz, not 50 rad/second. This is in Europe right, where the power is 50 Hz?

With that change both my solutions will get Z = 2.

I also did a mesh solution which also gets Z=2.

 

[mandiatutti]

I did take an official exam at the university the 13/09/2018, i asked the professor if i can take the text of the exercise home to try solve it by myself since i did not solve it during the given time. This is not home exercise. I just wanted to try to solve it before the official release of the results. If you think I'm a fake, well, I'm just not. I was just damn frustrated because i couldn't find a Solution and wanted it badly asap.

 

[mandiatutti]

There's my mistake. In the table of component values, where it says w = 50 rad/s, I just mentally changed that to 50 Hz, not 50 rad/second. This is in Europe right, where the power is 50 Hz?

With that change both my solutions will get Z = 2.

I also did a mesh solution which also gets Z=2.

Thank you very much for your work! Appriciated it!

 

[Ratch]

Hello everyone,
I took an exam last week and there was an exercise (look at picture) which I have no idea how to solve it.
The text reports that the current generator is: j(t)=J_M sin(wt+alpha) and that the value reported by the rms voltmeter is 0. I have to find the value of the impedance Z.

I had an idea and i tried to solve it that way, but i got to nothing... I assumed that, becouse of the rms voltmeter, on C4 there is no current and then i applied the mesh current method...
Please help me i'm litterally desperate

View attachment 43039

Sorry I am late in helping you, but my wife and I were on vacation driving around Lake Superior last week.

This is a fairly straight forward problem Don't let that voltmeter baffle you. All it tells you is the ends of branch D-F are at the same voltage when it indicates zero. In other words, branch D-F is a virtual short when that condition occurs. This allows you to easily calculate the loop current of the southwest and southeast loop (i1 and i2). The equality of the D-F endpoint voltages means that i1 = i2, because if they were unequal, i1 and i2 would have different values. Putting it all together and assuming counter-clockwise loop currents we get 3 equations and 3 unknowns. the north loop is constant at 5 amps. The equations are shown below.

Why is there a phase given for the source current? What is the phase relative to? Whether it is a sin or cos source, the impedance of the components will be the same. As you can see, I get a Z impedance of 1.25 ohms resistance.

Ratch

 

[The Electrician]

Sorry I am late in helping you, but my wife and I were on vacation driving around Lake Superior last week.

This is a fairly straight forward problem Don't let that voltmeter baffle you. All it tells you is the ends of branch D-F are at the same voltage when it indicates zero. In other words, branch D-F is a virtual short when that condition occurs. This allows you to easily calculate the loop current of the southwest and southeast loop (i1 and i2). The equality of the D-F endpoint voltages means that i1 = i2, because if they were unequal, i1 and i2 would have different values. Putting it all together and assuming counter-clockwise loop currents we get 3 equations and 3 unknowns. the north loop is constant at 5 amps. The equations are shown below.
View attachment 43181
Why is there a phase given for the source current? What is the phase relative to? Whether it is a sin or cos source, the impedance of the components will be the same. As you can see, I get a Z impedance of 1.25 ohms resistance.

Ratch

You didn't read all the other responses to post #1, did you?

I made my careless mistake, and you failed to notice that the capacitors and inductor are not microfarads and microhenries, but rather millifarads and millihenries.

 

[Ratch]

You didn't read all the other responses to post #1, did you?

I made my careless mistake, and you failed to notice that the capacitors and inductor are not microfarads and microhenries, but rather millifarads and millihenries.

I was wondering about the mf in the problem. Years ago I remember seeing micro-micro farads sometimes being written as mmf and being called "mickey-mikes". See this link and read paragraph #3. https://www.westfloridacomponents.com/blog/is-mf-mfd-the-same-as-uf/ . I don't remember seeing any cap being marked as mF. Usually they will be marked thousands of uf such as 5000 uf. So there is a measure of confusion. The solution is z=2 if the "m" is interpreted as milli. However, I believe my solution method is shorter and more succinct.

Ratch

 

[The Electrician]

I was wondering about the mf in the problem. Years ago I remember seeing micro-micro farads sometimes being written as mmf and being called "mickey-mikes". See this link and read paragraph #3. https://www.westfloridacomponents.com/blog/is-mf-mfd-the-same-as-uf/ . I don't remember seeing any cap being marked as mF. Usually they will be marked thousands of uf such as 5000 uf. So there is a measure of confusion. The solution is z=2 if the "m" is interpreted as milli. However, I believe my solution method is shorter and more succinct.

Ratch

You used 3 equations (i1 and i2 are not asked for) and involved the current source.

For short and succinct, apply the voltage divider formula twice, obtaining expressions for the voltage across C3 and across Z, equate them and solve one equation. The current source needn't be involved::



This is what I essentially did in excruciating detail in post #14, and Laplace did explicitly in post #20 (he did voltage dividers in both directions, solving for Z twice using this method).

 

[Ratch]

You used 3 equations (i1 and i2 are not asked for) and involved the current source.

For short and succinct, apply the voltage divider formula twice, obtaining expressions for the voltage across C3 and across Z, equate them and solve one equation. The current source needn't be involved::

View attachment 43184

This is what I essentially did in excruciating detail in post #14, and Laplace did explicitly in post #20 (he did voltage dividers in both directions, solving for Z twice using this method).

The problem is essentially balancing a Wheatstone or impedance bridge. That is sometimes hard for a newby to understand. I wanted to use loop equations like the OP did. But, you are right, it can be solved using one equation and the current source does not affect the result.

Ratch