Let the following circuit :

so we have \( I_1 + I_2 + I_3 + \frac{V_x}{0.5 \Omega} = 0 \iff \frac{V_x}{-j \Omega} + \frac{V_x}{(1+j) \Omega} + \frac{V_1}{1 \Omega} + 2V_x = 0 \\\)
and we have : \( -V_1 + 12(0°) + V_x = 0 \iff V_x =V_1 - 12(0°) \\ \)
Finally \( V_1 \approx 8.653(3.1798°) \)

 

Remember what I said in another thread about the value is making sure that your set up equations were correct, because mistakes there generally can't be caught later since you are merely solving the same problem for a different circuit? The result is that you spend a lot of wasted time and effort working a different problem only to arrive at an answer that, except by sheer coincidence, is guaranteed to be wrong?

Well, first, you are being sloppy with your ground reference. The left-hand side is fine, because Vx is defined in the diagram as the voltage across the capacitor, which is also the voltage across the series combination of the resistor and inductor. But on the right hand side, you just say V1/jΩ. But V1 is NOT defined as the voltage across anything. It is defined as the voltage AT a particular node. This ONLY has meaning in the context of what that voltage is relative to. You are clearly assuming that the bottom node is your 0 V reference, which is fine. But you need to DEFINE that to be the case -- put a 0 V reference symbol on it. While this is commonly called the circuit "ground", that is not strictly correct, but it is sufficiently widespread that any reader will know what you mean -- but you need to define it.

So, let's say that the bottom node has been defined as your "ground".

Do you agree that the current flowing downward through the 1 Ω resistor is V1/jΩ?

You're also still being sloppy with your units -- and when the fact that they aren't working out can no longer be ignored, instead of stopping and figuring out what is going on, you simply drop them all together.

Your dependent source is a voltage-controlled current source. That means that the transfer coefficient (often referred to as the 'gain') can't be just a number, regardless of how sloppy the author might be (and many of them are VERY sloppy). The output is a current and the input is a voltage, so the transfer coefficient has to have units of current per voltage. With just a '2' there, it is incomplete as to whether it is 2 amperes per volt, or 2 milliampere per killovolt, or 2 megamperes per microvolt, or 2 something that is a measure of charge flow per something that is a measure of energy per unit charge. Most likely the author meant 2 amperes per volt, making the correct value inverse ohms, also known as 'mhos' or, more properly per the SI standard, siemens.

So the diagram should be labeled as (2 S)·Vx, or Vx/(0.5 Ω).

You compound this problem by dropping the units on the 12 V @ 0° -- BTW, just like you don't need to include a real or imaginary coefficient that happens to be zero, you don't need to include an angle that happens to be zero. It is understood.

Also, remember what I said about the correctness of the answer to most engineering problems can be established from the answer itself?

Let's see if that the case.

Your answer is that V1 = 12 V (again, the 0° angle is understood if omitted).

Well, if V1 is 12 V and V2 = V1 - 12 V, that makes V2 = 0.

That makes Vx = 0, which makes the current in the current source equal to zero.

But the voltage across the 1 Ω resistor on the left is NOT zero, it is V1, which you claim is 12 V. So there is 12 A of current flowing downward in it. It can't go through the current source, because it is outputting zero current. So it must be going over to the right hand side via the bottom node and up through the capacitor and/or the resistor/inductor branches. But, nope, that's not possible because the voltage across both of those branches is Vx, which is 0 V.

So, clearly, the answer that V1 = 12 V @ 0° is not correct.

It may or may not be correct for the problem you actually did the work for, but it's not correct for the problem you were given.

Finally, remember how I said that if you continued to be sloppy in your set up equations and in tracking your units that you would come to grief soon or later....

Engineering is both an art and a science. The art is learning from your own mistakes, and the science is learning from the mistakes of others. I am trying to help you practice the science of engineering, but you seem intent on practicing the art. That's fine -- practicing the art is actually the more effective approach to really learn things, albeit usually the slower and more painful one, too.

 

Remember what I said in another thread about the value is making sure that your set up equations were correct, because mistakes there generally can't be caught later since you are merely solving the same problem for a different circuit? The result is that you spend a lot of wasted time and effort working a different problem only to arrive at an answer that, except by sheer coincidence, is guaranteed to be wrong?

Well, first, you are being sloppy with your ground reference. The left-hand side is fine, because Vx is defined in the diagram as the voltage across the capacitor, which is also the voltage across the series combination of the resistor and inductor. But on the right hand side, you just say V1/jΩ. But V1 is NOT defined as the voltage across anything. It is defined as the voltage AT a particular node. This ONLY has meaning in the context of what that voltage is relative to. You are clearly assuming that the bottom node is your 0 V reference, which is fine. But you need to DEFINE that to be the case -- put a 0 V reference symbol on it. While this is commonly called the circuit "ground", that is not strictly correct, but it is sufficiently widespread that any reader will know what you mean -- but you need to define it.

So, let's say that the bottom node has been defined as your "ground".

Do you agree that the current flowing downward through the 1 Ω resistor is V1/jΩ?

You're also still being sloppy with your units -- and when the fact that they aren't working out can no longer be ignored, instead of stopping and figuring out what is going on, you simply drop them all together.

Your dependent source is a voltage-controlled current source. That means that the transfer coefficient (often referred to as the 'gain') can't be just a number, regardless of how sloppy the author might be (and many of them are VERY sloppy). The output is a current and the input is a voltage, so the transfer coefficient has to have units of current per voltage. With just a '2' there, it is incomplete as to whether it is 2 amperes per volt, or 2 milliampere per killovolt, or 2 megamperes per microvolt, or 2 something that is a measure of charge flow per something that is a measure of energy per unit charge. Most likely the author meant 2 amperes per volt, making the correct value inverse ohms, also known as 'mhos' or, more properly per the SI standard, siemens.

So the diagram should be labeled as (2 S)·Vx, or Vx/(0.5 Ω).

You compound this problem by dropping the units on the 12 V @ 0° -- BTW, just like you don't need to include a real or imaginary coefficient that happens to be zero, you don't need to include an angle that happens to be zero. It is understood.

Also, remember what I said about the correctness of the answer to most engineering problems can be established from the answer itself?

Let's see if that the case.

Your answer is that V1 = 12 V (again, the 0° angle is understood if omitted).

Well, if V1 is 12 V and V2 = V1 - 12 V, that makes V2 = 0.

That makes Vx = 0, which makes the current in the current source equal to zero.

But the voltage across the 1 Ω resistor on the left is NOT zero, it is V1, which you claim is 12 V. So there is 12 A of current flowing downward in it. It can't go through the current source, because it is outputting zero current. So it must be going over to the right hand side via the bottom node and up through the capacitor and/or the resistor/inductor branches. But, nope, that's not possible because the voltage across both of those branches is Vx, which is 0 V.

So, clearly, the answer that V1 = 12 V @ 0° is not correct.

It may or may not be correct for the problem you actually did the work for, but it's not correct for the problem you were given.

Finally, remember how I said that if you continued to be sloppy in your set up equations and in tracking your units that you would come to grief soon or later....

Engineering is both an art and a science. The art is learning from your own mistakes, and the science is learning from the mistakes of others. I am trying to help you practice the science of engineering, but you seem intent on practicing the art. That's fine -- practicing the art is actually the more effective approach to really learn things, albeit usually the slower and more painful one, too.

Hey WBahn tysm for ur detailled answer, i rechecked again on my equations, i've corrected my mistake i typed the "correct" answer in my paper, and when rewriting it here on latex i didnt want to check on my calculation so i did the math way but nvm, the equations i wrote of the circuit are valid dont you think so? also about the two factor for the VCCS, the professor's assistant never mentioned it in the course nor even pays attention to units while writing his answers, so since the first time i wrote a question in this helpful forum i started putting effort into the units,

@MrAi, i tried to use multisim and simulate the circuit here :
but somehow i dont get my answers in the analysis!, i want you pleeaaaaase to explain to me whats the difference between single frequency analysis and transient analysis, also i want to know how do you control the \( \omega \) to get the desired impedances over the circuit compenents , also what is the frequency of the VCCS is it the same as the independent source? also if we have two independent sources each one with a frequency -possibly same frequency- how do you calculate the \( \omega \) since \( X_L = j L \omega, X_c = \frac{-j}{C \omega} , \omega = ?\)


infinite thanks WBahn and MrAi for your efforts!!