Another thing to consider is that you may, in fact, not have enough information to solve for ALL of the currents and voltages in the circuit. But you may have enough information to solve for the one(s) you are interested in.

 

Hi again,

Wbahn:
Oh ok that looks good.

cdummie:
Did you think about what happens when an impedance is in series with a constant current source yet?
That helps a lot too with this circuit.

When impedance is in series with current source, it means that current through that impedance is the current of that current source.

 

It just circulates current in that upper-left loop and has no interaction with the rest of the circuit. So it can be removed.

So that means that i can remove Ig1 from this circuit? If that's true, i don't know how it can help me since i knwo the value of Ig1 and removing it would not mean that i got rid of unknown value.

 

When impedance is in series with current source, it means that current through that impedance is the current of that current source.

Hi again,

Yes, very good, i am happy to see you understand this

Now what does that mean with regard to the only capacitor in this circuit? Answering that question will help simplify the circuit even more, and so little by little the circuit gets more and more simplified. You probably want to draw the circuit again after considering this little fact.

 

So that means that i can remove Ig1 from this circuit? If that's true, i don't know how it can help me since i knwo the value of Ig1 and removing it would not mean that i got rid of unknown value.

It doesn't help directly with getting rid of the unknown variable -- but isn't simplifying the circuit and having one less mesh to deal with worth something?

 

When impedance is in series with current source, it means that current through that impedance is the current of that current source.

And, just like a voltage source that is in series with a current source, what does this suggest that you can do with that impedance?

 

Hi again,

Yes, very good, i am happy to see you understand this

Now what does that mean with regard to the only capacitor in this circuit? Answering that question will help simplify the circuit even more, and so little by little the circuit gets more and more simplified. You probably want to draw the circuit again after considering this little fact.

Let's see, you said constant current source, but in this case, all sources are AC sources, if they were constant i could remove branch with capacitor, because of it's infinite voltage, but i think that i can't do it if sources are AC, and in this case, they are.

 

A sinusoidal source with constant magnitude, frequency, and phase is considered a constant source in AC steady state. Consider that you model it as a phasor whose value is a constant.

 

It doesn't help directly with getting rid of the unknown variable -- but isn't simplifying the circuit and having one less mesh to deal with worth something?

Well, that's right, smaller circuit means less equations.

 

And, just like a voltage source that is in series with a current source, what does this suggest that you can do with that impedance?

I am not really sure what are you trying to say.

 

A sinusoidal source with constant magnitude, frequency, and phase is considered a constant source in AC steady state. Consider that you model it as a phasor whose value is a constant.

Does that means that i could remove the branch with capacitor? If that is true, why i can't do that when i have resitor in series with current source and capacitor, and what would happen if there was voltage source instead of current source in this case?

 

You can't remove the branch, but you can remove the impedance that is in series with the current source since, like the case of the series element being a voltage source, the current source will produce whatever voltage is necessary to cancel out the voltage developed across the impedance. A current source masks the voltage of components that are in series with it.

If it's a voltage source instead of a current source, this doesn't apply. But a voltage source masks the currents of components that are in parallel with it.

 

Hi again,

Does that means that i could remove the branch with capacitor? If that is true, why i can't do that when i have resitor in series with current source and capacitor, and what would happen if there was voltage source instead of current source in this case?

When there is an impedance in series with a current source the value does not matter. For example, if we have a 1 ohm resistor in series with a 2 amp current source the current though the resistor is 2 amps, and if we have a 100 ohm resistor in series with a 2 amp current source then the current through the resistor is still 2 amps. The current affects the rest of the circuit, but only affects the voltage at the point between the current source and the resistor. Since only the current matters and not the voltage at the center node, might as well make that resistance the most convenient value that helps to solve the REST of the circuit (not that one node though), and that value is zero. So we short out the impedance.
Note we can not solve for the center node anymore (the one that connects the resistor to the current source) but we can solve for the rest of the circuit, and that helps a lot. Also, i said constant current source because that's what you have, although it is a constant AC source.
Note that when the current peak is at 1 amp the current entering the node beyond the resistor is +1 amp, and when the current peak is at -1 amp the current entering the node beyond the resistor is -1 amp, and the resistor, again, never changes any of this.

We are looking at the cap right now, but if you have a resistor this works the same way. Short out the resistor. The only way you can do this though is if that is the only connection to the current source. If there are other connections to the current source at that common node you have to look at those too.
If you care to prove any of this, do a nodal analysis and see that C drops out of the solution for the output voltage (factors out) so you wont find C in the equation even though it is in the circuit. This is also true for other elements that factor out...they will not be present in the output equation.

 

You can't remove the branch, but you can remove the impedance that is in series with the current source since, like the case of the series element being a voltage source, the current source will produce whatever voltage is necessary to cancel out the voltage developed across the impedance. A current source masks the voltage of components that are in series with it.

If it's a voltage source instead of a current source, this doesn't apply. But a voltage source masks the currents of components that are in parallel with it.

Ok, i tried nodal analysis since you said that i don't have enough info about components of the circuit but i do have enough info to find voltage U. So i tried to find it this way:



Ok, so this is the original picture of the circuit where i labeled the nodes, and the way i did that, i have to find U23=U20-U30 and that is actually the voltage U that i am looking for.

Since there are four nodes in this circuit, it means that i have to form 3 equations where one of them would be U10=E1 so i have two more equations to make.

G21*U10 + G22*U20 + G23*U30 = I2

and

G31*U10 + G32*U20 + G33*U30=I3

Firstly, i will determine all admittances in the circuit:

G11=0
G22=1/R2 + 1/R5 + 1/R6
G33=jωC + 1/R2 + 1/R6

G21=G12=0
G31=G13=0
G32=G23=1/R2 + 1/R6

Now, currents:

I1=Ig1 + Ig3

I2=Ig2 + E2/R2

I3=Ig4 - Ig3 - E2/R2

when i put this in equations:

E1=U10
0*U10 + G22*U20 + G23*U30 = I2
G31*U10 + G32*U20 + G33*U30=I3


i have:

E1=U10
0*U10 + (1/R2 + 1/R5 + 1/R6)*U20 + (1/R2 + 1/R6)*U30 = Ig2 + E2/R2
0*U10 + (1/R2 + 1/R6)*U20 + (jωC + 1/R2 + 1/R6)*U30=Ig4 - Ig3 - E2/R2


But, obviously, there are four unknowns in the last two equations, so i don't know how to solve it (first equation is irrelevant here since term with U10 in both equations is zero). What i am supposed to do now? I haven't noticed any mistakes here, so i would like to know what is wrong here, since i can't get the solution.

 

Have you thought of transforming E2,R2 into an equivalent current source and parallel resistance?

 

Re-read post #21

You can solve your equations (after some minor corrections) for U20 and U30, form the expression U20-U30 (which is equal to U) and I think you'll see that Ig2, E1 and C don't appear in it. In other words, there will be no unknowns in the expression for U.

Your equations have a couple of sign errors shown in red below; also, get rid of the jωC term (per post #32) shown in blue. Solve the two equations and you should get your (symbolic) answer; substitute numerical values for the variables to get the final numeric value for U.

0*U10 + (1/R2 + 1/R5 + 1/R6)*U20 - (1/R2 + 1/R6)*U30 = Ig2 + E2/R2
0*U10 - (1/R2 + 1/R6)*U20 + (jωC + 1/R2 + 1/R6)*U30=Ig4 - Ig3 - E2/R2

 

Have you thought of transforming E2,R2 into an equivalent current source and parallel resistance?

But, how that can simplify the solving procedure here?

 

Hi again,



When there is an impedance in series with a current source the value does not matter. For example, if we have a 1 ohm resistor in series with a 2 amp current source the current though the resistor is 2 amps, and if we have a 100 ohm resistor in series with a 2 amp current source then the current through the resistor is still 2 amps. The current affects the rest of the circuit, but only affects the voltage at the point between the current source and the resistor. Since only the current matters and not the voltage at the center node, might as well make that resistance the most convenient value that helps to solve the REST of the circuit (not that one node though), and that value is zero. So we short out the impedance.

When you say that we cannot solve that one node, you mean the voltage of that one node not the current, since we can't find correct voltage of that node since we changed the value of impedance, but everything else will remain the same as in the original circuit.


Ok, i have another idea now, as i stated in my first post, if i am looking for thevenin generator, i need to know it's voltage and it's inner resistance, Thevenin resistance is R2 and now i have an idea for Thevenin voltage, and here it is(note that i removed the branch with R6 since i am looking for Thevenin generator):



which means that the voltage of the branch with E2 and R2 (which is the voltage that i am looking for) is Et=E2-R2(Ig4-Ig3) = 1 - 2*10^3(0.5-j)10^(-3) = 1-1+j = jV

so now i have the following circuit:



So now I=Et/(Rt+R6) = j/6 10^(-3) A which means that the voltage across the R6 is U=I*R6 = 2j/3 V

Is this the correct way to solve this and is this the correct solution?

 

Why are you ignoring post #36? I showed your two equations with minor errors corrected and told you that if you solve those two simultaneous equations you will get the correct answer.

The answer you have obtained in post #38 is not correct.

 

Why are you ignoring post #36? I showed your two equations with minor errors corrected and told you that if you solve those two simultaneous equations you will get the correct answer.

The answer you have obtained in post #38 is not correct.

I haven't ignored it, i tried it and i still had Ig2 in final result, so i thought that i could try different approach, anyway, what is wrong with approach in my previous post?