Finding the constraint equation of a circuit with a dependent voltage

[johnsmith7565]

Homework Statement

Use the node voltage method to find v0 in the circuit shown.

Relevant Equations

(V0-v1)/R1 + (v1-v2)/R2+…= 0

What I’m stuck on is finding the constraint equation for i on the top question . I don’t know how to find it when it’s through a voltage source and not over any resistors. (I can’t use ohm‘s law) After I find i the problem should become easy to solve. I know that v1 = 10 and v2 = 20i . The KCL for v0 is (v0-10)/10 + v0/24 + (v0-20i)/20. I could write the KCL for v1 and v2 but that doesn’t address how to find the constraint which is what I need. And when I write them and solve for my answer is way off. Something tells me I need a new approach.

 

[BvU]

I could write the KCL for v1 and v2

Do us a favour and write them out. There also is a KCL for the bottom node; it should read something like $$i_\Delta + {V_0\over 40\;\Omega}-20\,i_\Delta = 0$$ if I am not mistaken (not an expert here, just searching for $n$ equations with $n$ unknowns)

The KCL for v0 is (v0-10)/10 + v0/24 + (v0-20i)/20

I suppose you mean $$
{10\; {\sf V}- V_0\over 10\;\Omega} - {V_0\over 40\;\Omega} + {V_2-V_0\over 20\;\Omega} = 0 \qquad ?
$$(considering the penciled arrows)

$\$

 

[johnsmith7565]

For v1: -i +(v1-v0)/10 + (v1 - v2)/30
For v2: I don’t know the current going through the dependent voltage source but I do know that current + (v2-v0)/20 + (v2-v1)/30.

 

[BvU]

Ah, my bad: it's not a dependent current source but a dependent voltage source .
Let me call $i_3$ the current going into the bottom node (named V3...) [edit]from the right
So thus far we have $$ \begin{align*}
{V_0- V_1\over 10\;\Omega} + {V_0\over 40\;\Omega} + {V_0-V_2\over 20\;\Omega} &= 0 \tag{V0}\\ \\
-i_\Delta +{V_1-V_0\over 10\;\Omega} + {V_1 - V_2\over 30\;\Omega}&=0\tag{V1}\\ \\
i_3 + {V_2-V_0\over 20\;\Omega} + {V_2 - V_1\over 30\;\Omega}&=0\tag{V2} \\ \\
i_\Delta + {V_0\over 40\;\Omega}-i_3 &= 0 \tag{V3} \\ \\
V_1 &= 10\; {\sf V} \\ \\
V_2 &= -20\;i_\Delta
\end{align*}$$and now I count 6 equations with only 5 unknowns -- so I must be doing something wrong, still ...

Oh, well, an opportunity to learn

By the way, the homework statement states 'use the voltage node method' and it looks as if we use the current node method so far... ?

$\$

 

[johnsmith7565]

Ah, my bad: it's not a dependent current source but a dependent voltage source .
Let me call $i_3$ the current going into the bottom node (named V3...)
So thus far we have $$ \begin{align*}
{V_0- V_1\over 10\;\Omega} + {V_0\over 40\;\Omega} + {V_0-V_2\over 20\;\Omega} &= 0 \tag{V0}\\ \\
-i_\Delta +{V_1-V_0\over 10\;\Omega} + {V_1 - V_2\over 30\;\Omega}&=0\tag{V1}\\ \\
i_3 + {V_2-V_0\over 20\;\Omega} + {V_2 - V_1\over 30\;\Omega}&=0\tag{V2} \\ \\
i_\Delta + {V_0\over 40\;\Omega}-i_3 &= 0 \tag{V3} \\ \\
V_1 &= 10\; {\sf V} \\ \\
V_2 &= -20\;i_\Delta
\end{align*}$$and now I count 6 equations with only 5 unknowns -- so I must be doing something wrong, still ...

Oh, well, an opportunity to learn

By the way, the homework statement states 'use the voltage node method' and it looks as if we use the current node method so far... ?

$\$

I'm in school currently, but when I have some down time I'll try setting the node labeled v2 as the ground. I think this might help out because the 60 V source would be in-between the current ground node and v1, which would make a supernode and give a constraint equation for idelta in terms of v1 and the current ground.

 

[DaveE]

By the way, the homework statement states 'use the voltage node method' and it looks as if we use the current node method so far... ?

OK, I had to look that up. I also wasn't familiar with the name. Khan Academy says "voltage node method" is what I call KCL.

 

[BvU]

I had a fun day learning the quirks of Qucs (quite universal circuit simulator) .
Almost converted me to Linux, but that curve I don't think I can handle, so I'll have to live with a buggy windows version.

(You wouldn't believe where you have to go to turn off the grid points in the circuit diagram )

Can't make it come up with a sensible answer when I use $\ 20\ i_\Delta\$ for the dependent voltage (gets 15.2 V but claims V2 = 0 at the same time ).

However, when I let step through a range of voltages, a solution emerges. Suppose I can't spoil it under PF rules but at least I can use it to see if the equations in the list are satisfied. Tomorrow (1 AM here )...

Hint: do the exercise with $\infty$ for the 40 $\Omega$ resistor ... piece of cake !

$\$

 

[DaveE]

I had a fun day learning the quirks of Qucs (quite universal circuit simulator) .

Have you tried LTSpice? What does Qucs in comparison have that makes it worth my time to investigate?

 

[BvU]

Didn't know either this morning, so just tried one. ( see post #2 )

 

[DaveE]

Didn't know either this morning, so just tried one. ( see post #2 )

You might try LTSpice then. It's the most popular free simulator. Of course it will be annoying learning a new system. It always is.

 

[Baluncore]

Here is an LTspice .asc file

 

[BvU]

Here is an LTspice .asc file

Very nice. Yet another wrench to get used to in my toolkit . First impression is better than Qucs (but I am slightly prejudiced - Analog Devices does that for me).

However, my intuition doesn't help me here: how is this

a dependent voltage source delivering $20 * i_\Delta$ Volts ?@johnsmith7565 : what do you find with my suggestive hint in #7 at the end ?

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[Baluncore]

... a dependent voltage source delivering 20 * iΔ Volts ?

Look up "F" in the help and you will see that it is a primitive current controlled current source.
The current flowing through voltage source V1 is multiplied by 20.

 

[BvU]

Can't find a simple current controlled voltage source, but a fixed voltage source with the value as found in #7 does the trick and gets the exact same result. Thanks!

Now on to doing things by hand and analytically .. counting and checking out eqns in #4

$\$

 

[Baluncore]

Can't find a simple current controlled voltage source, ...

E. Voltage Dependent Voltage Source.
F. Current Dependent Current Source.
G. Voltage Dependent Current Source.
H. Current Dependent Voltage Source.

There is now a more flexible general element.
B. Arbitrary Behavioral Voltage or Current Sources.

 

[BvU]

Boy, the stupidities one can come up with ! in #4
$$i_\Delta + {V_0\over 40\;\Omega}-i_3 = 0 \tag{V3} $$should of course be
$$i_\Delta - {V_0\over 40\;\Omega}-i_3 = 0 \tag{V3} $$ (all currents going into bottom node V3).

And then the equations are not independent ($V0+V1+V2+V3\Rightarrow 0=0\$) so we are left with 5 eqns with 5 unks. I will be able to sleep again .

Oops, forgot to also solve the set ...

----

E. Voltage Dependent Voltage Source.
...

Excavated that F2 allows inserting some things, but never guessed what h stands for.
Now embark on an expedition to enforce H1.value = 20 * V1.I ? (in my case -20, because flipping a symbol is too expert for me as well..

How come every different program is as if learning a different language on a different planet ?

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[Baluncore]

... because flipping a symbol is too expert for me ...

Control R will Rotate selected symbol clockwise.
Control E will flip selected symbol horizontal.

 

[BvU]

You are too indulgent. I'm just moping: why can't the program tell met that ? Searching help for flip or rotate yields no help whatsoever. I suppose being patient and working through yet another 'getting started' should do the trick

 

[Baluncore]

You are too indulgent. I'm just moping: why can't the program tell met that ?

The LTspice learning curve is exciting, as it has some overhangs.
There may still be a hint if you look in the edit pull down menu.
If you cannot see a function, ask here so others can benefit.

Help - Contents - Schematic capture - Keyboard shortcuts - - -
Help - Search - “rotate” - List topics - Schematic editing - - -

 

[johnsmith7565]

Very nice. Yet another wrench to get used to in my toolkit . First impression is better than Qucs (but I am slightly prejudiced - Analog Devices does that for me).

However, my intuition doesn't help me here: how is this

View attachment 295494​

a dependent voltage source delivering $20 * i_\Delta$ Volts ?@johnsmith7565 : what do you find with my suggestive hint in #7 at the end ?

$\$

Sorry for not responding earlier.

Trying to make the other node a ground node went no where.

To be honest I am a little bit confused... why would I plug in infinity for the 40 ohm resistor? How do I even do that? This problem seems very difficult for a basic text on node voltages! It seems like there is some simple trick to make this problem easier to solve...

 

[DaveE]

You are too indulgent. I'm just moping: why can't the program tell met that ? Searching help for flip or rotate yields no help whatsoever. I suppose being patient and working through yet another 'getting started' should do the trick

Yes, it's a pain. I've never liked the schematic capture part of LTSpice, but it does work well.
There is some good stuff online about how to use it. This one looked good to me. http://www.simonbramble.co.uk/lt_spice/ltspice_lt_spice.htm

 

[johnsmith7565]

I taught myself the loop current method today, and I've just solved this circuit using the loop-current method. IMO mush easier and straightforward with the circuit using the loop-current method.

 

[BvU]

Trying to make the other node a ground node went no where.

Not surprised.

To be honest I am a little bit confused... why would I plug in infinity for the 40 ohm resistor? How do I even do that?

Because that makes it a very simple circuit. And the results are useful when estimating which way the actual solution should go. Cryptic ?

Try to solve this one (so without the 40 Ohm) for V₀

This problem seems very difficult for a basic text on node voltages! It seems like there is some simple trick to make this problem easier to solve...

I agree with you. But I would be surprised if there was a trick. I do suppose the set of equations can be solved, but it's work and I am thoroughly lazy.

 

[The Electrician]

I taught myself the loop current method today, and I've just solved this circuit using the loop-current method. IMO mush easier and straightforward with the circuit using the loop-current method.

What result did you get?

 

[johnsmith7565]

What result did you get?

24 V for v0.

 

[The Electrician]

24 V for v0.

Do you know that to be correct? Was the answer given with the problem?

It's good that you learned another method to solve circuits.

The two methods alluded to in this thread are commonly called mesh analysis and nodal analysis in the USA.

I find that it's a good procedure to solve every circuit problem with both methods. This provides a check for the answer and is good practice for network solving.

Do you still need to do the nodal analysis?

 

[BvU]

24 V for v0.

Well done .

 

[johnsmith7565]

Do you know that to be correct? Was the answer given with the problem?

It's good that you learned another method to solve circuits.

The two methods alluded to in this thread are commonly called mesh analysis and nodal analysis in the USA.

I find that it's a good procedure to solve every circuit problem with both methods. This provides a check for the answer and is good practice for network solving.

Do you still need to do the nodal analysis?

Yes, the answer is in the back of the textbook. I don’t really need to use nodal analysis since I’m just doing this for my own pleasure. I’m just trying to learn the basics of circuits so I just care about arriving at the correct answer, not really the method. Initially, I thought nodal analysis would be the best way since the textbook told me to solve the question that way, but that turned out to be untrue.

 

[The Electrician]

Yes, the answer is in the back of the textbook. I don’t really need to use nodal analysis since I’m just doing this for my own pleasure. I’m just trying to learn the basics of circuits so I just care about arriving at the correct answer, not really the method. Initially, I thought nodal analysis would be the best way since the textbook told me to solve the question that way, but that turned out to be untrue.

I would think that it's better to care more about the method, because if you know the method(s) well, the correct answer is easy to get. I'm sorry I didn't see this post earlier because I would have shown you how to use nodal analysis to solve this problem. Here's how to do it.

First, designate the nodes from left to right as v1, v2 and v3; it's easy to make the connection of Vo to v2 at the end.

You then have 3 unknowns and need 3 equations, one for each node.

The equation at v1 is not a KCL equation, it's a constraint equation; it is just this: v1 = 10

The second (v2) equation is a genuine KCL equation and can be seen in the image which follows.

The third (v3) equation is another constraint equation: v3 = -20*IΔ, where IΔ is just the sum of the currents in the 10 ohm and 30 ohm resistors.

Solving the 3 equations gives the result shown:

 

[Baluncore]

I question the polarity of the current controlled voltage source.
It is shown inverted in the original diagram.

 

[The Electrician]

I question the polarity of the current controlled voltage source.
It is shown inverted in the original diagram.

Exactly what are you getting at when you say "I question the polarity of the current controlled voltage source."? That source is indeed shown inverted in the diagram as you assert, so what is your point?

 

[Baluncore]

It is OK. The 10 V source is sinking conventional current, so Id is negative.
Multiply by 20 and invert the CCVS to get +64 V, which sources the current to be sunk by the 10 V.

 

[BvU]

I question the polarity of the current controlled voltage source.
It is shown inverted in the original diagram.

Only means a negative answer instead of a positive

 

[Baluncore]

Only means a negative answer instead of a positive

Not quite.
I think if you reverse the CCVS polarity it produces 6.80851 V, and Vo becomes +7.65957 V.
The 10.0 V becomes a positive current source, so the CCVS is also positive.

 

[BvU]

My turn to be confused: if $V_2=+64$ V, the CCVS delivers $-$64 V and pumps 3.8 A towards point V2. The $-$64 is 20 times the $-$3.2 A $i_\Delta$.

If the CCVS sets $V_2=+ 6.80851$ V, the dial must read $-6.80851$ V which is minus 20 times the then $-$0.34 A $i_\Delta$.