[HW] Decide v as a function of Vs

[Teszla]

Decide v as a function of Vs.

**broken link removed**

I'm trying to solve this. But the way it is drawn it looks to me like v = Vs? They look like they are the same? Could someone explain to me why this is not the case?

 

[prog_doctor]

Because v = Vs + V(3R-resistor) There is certainly a current flowing through the 3R resistor. A QUCS simulation of your circuit and the associated solution of v vs Vs may help you with the solution (if I succeeded in attaching the image, that is).

 

[Teszla]

Aren't v and Vs part of the same node? If the same node always have the same voltage, how can they be different in this case?

 

[prog_doctor]

v is the voltage from that middle node to ground (includes the voltage across the 3R resistor), wheres Vs is the voltage across the voltage source (does not include the voltag eacross the 3R resistor)

 

[Teszla]

Aha, I understand. But does that mean that v and Vs are different nodes? (As I understood the definition of a node is that it has the same voltage). If so, where exactly do the nodes for v and Vs meet?

 

[adi0803199]

In nodal analysis,any node's voltage is always referenced with a reference node.From the above given schematic,the ground is at the bottom of the circuit. Whereas Vs is the potential difference generated by the voltage source and not a nodal point. So v and Vs are not at the same potential.

 

[Teszla]

Ok. I take it that they should be considered as two different nodes, even though Vs is not technically a node. Is that definition ok?

Now, if I want to do a KCL for the node, how would the current that goes through Vs be defined?

 

[adi0803199]

Now from the above schematic,the left of the node v has a potential difference of v Volts between that point and the reference terminal(ground), (i.e) current through this part of the loop(say loop 1 -left side) is v/3R. Therefore v1=v/3. Now consider the node v,the other point that the 3R resistor is connected to (4/3v) volts. The higher potential for the 3R resistor is 4v1 = (4/3) v volts.So the current flows into the voltage source Vs from the second loop( right hand side).Then the current in the Voltage source branch is the current difference between the current in the loop 1(1/3 *(v/R)) and the current in loop 2(1/9*(v/R)).
Now that the current through has been found out,the value of v is (Vs+(current through source branch)*3R) Volts
I hope it makes some sense.

 

[Teszla]

Now from the above schematic,the left of the node v has a potential difference of v Volts between that point and the reference terminal(ground), (i.e) current through this part of the loop(say loop 1 -left side) is v/3R.

Yep, so far I'm following.

Therefore v1=v/3.

How do you draw this conclusion?

Now consider the node v,the other point that the 3R resistor is connected to (4/3v) volts. The higher potential for the 3R resistor is 4v1 = (4/3) v volts.So the current flows into the voltage source Vs from the second loop( right hand side).Then the current in the Voltage source branch is the current difference between the current in the loop 1(1/3 *(v/R)) and the current in loop 2(1/9*(v/R)).
Now that the current through has been found out,the value of v is (Vs+(current through source branch)*3R) Volts
I hope it makes some sense.

I'm not following here. I would just say that the current from v to the right side is (v-4V1)/3R. I don't understand what you are doing otherwise, maybe you could make some illustrations to explain it?

I still wonder how to express the current that goes from v through the voltage source?

 

[adi0803199]

1)Current through this part of the loop is v/3R,so the value of v1=(current through loop 1 )*(Resistance R) = (v/3R)*R Volts = v/3 Volts.
2) Now substitute the value of v1 to the voltage dependent voltage source on loop 2.Upon substitution, you get 4v1 = 4*(v/3) Volts = (4/3)*v Volts.
Now follow the previous thread and it will make better sense.

 

[Teszla]

1)Current through this part of the loop is v/3R,so the value of v1=(current through loop 1 )*(Resistance R) = (v/3R)*R Volts = v/3 Volts.
2) Now substitute the value of v1 to the voltage dependent voltage source on loop 2.Upon substitution, you get 4v1 = 4*(v/3) Volts = (4/3)*v Volts.

Ok. I understand.

Now follow the previous thread and it will make better sense.

What loops are you refering to? Which is loop 1 and which is loop 2? Where do they start and in what direction do they go (clockwise or counter-clockwise). I also wonder what method are you using to analyse the circuit? KVL?

 

[adi0803199]

The left side of the circuit (to the left of the Voltage Source -Vx) is loop1,here current flows in the anticlockwise direction.The right of the node v(right of the voltage source -Vx) is loop2. The current direction is again anticlockwise direction as 4V1 is at higher potential compared to voltage source Vx. Now see that thread again and I hope it makes some sense.

 

[Teszla]

Could you just write the calculations for how you determine the answer? It feels like this has gotten stuck and since you don't answer some of my questions (like what method of analaysis you use) it just takes forever to solve.

I would also welcome such calculations from anyone else who feels up for it.

 

[KerimF]

The sum of currents out of node "v" equals zero.
v/3R + (v-vs)/3R + (v-4*v1)/3R = 0
v + (v-vs) + (v-4*v1) = 0
3v - vs - 4*v1 = 0

But from the left branch (seen as a voltage divider):
v1 = v*R/(2R+R) = v/3

3v - vs - 4*v/3 = 0
9v - 3*vs - 4*v =0
5v - 3*vs = 0
v = 3*vs/5

 

[Teszla]

The sum of currents out of node "v" equals zero.
v/3R + (v-vs)/3R + (v-4*v1)/3R = 0
v + (v-vs) + (v-4*v1) = 0
3v - vs - 4*v1 = 0

But from the left branch (seen as a voltage divider):
v1 = v*R/(2R+R) = v/3

3v - vs - 4*v/3 = 0
9v - 3*vs - 4*v =0
5v - 3*vs = 0
v = 3*vs/5

Thanks. I just wonder how you motivate taking (v-vs)/3R? I'm sure it's correct, but I thought when doing this, that the resistance has to be between the voltages? I mean, it would be correct if the circuit looked like this:

**broken link removed**
Here I would say the current from v to ground would be (v-vs)/3R.

Does that mean there is no difference between this circuit and the first one? If so, why? Shouldn't the different locations of the voltage source have some impact on the circuit?

 

[KerimF]

Let us simplify the circuit and replace the left and right branches with a voltage source of "v" only.
If you get my point, we have now one loop that has 3 elements in series vs, 3R and v. I hope you can see that as long the polarity of the voltage sources are not changed, the current of the loop won't be affected if we place 3R anywhere on it.

 

[adi0803199]

The current won't change and hence both the circuit will yield the same result in this case.