Nodal Analysis with Dependent Voltage Source

[marsupial]

Homework Statement

Here is a diagram of the circuit and my redrawing with currents drawn:

We are to find the power delivered by the dependent voltage source. I found a solution online, but I am not sure why they had the KCL line as:

$$ \frac{v_1 - 160} {10} + \frac{v_1} {100} + \frac{v_1 - 150i_\sigma} {50} = 0 $$

I understand all of it except for the way they handled the dependent voltage source. My attempt is below.

Homework Equations

KCL

The Attempt at a Solution

$$ i_1 + i_\sigma = i_2 $$

$$ \frac{160-v_1} {10} + \frac{v_1} {10} = \frac{v_1 - 150i_\sigma} {50} $$

So, I either have the current drawn the wrong way (although in the answer published online it is in the same direction, or I am reading/interpreting the way the current is represented wrong. Any help would be appreciated.

 

[berkeman]

I understand all of it except for the way they handled the dependent voltage source.

The right-hand dependent source looks to be a current source, no?

 

[lewando]

The right-hand dependent source looks to be a current source, no?

Looks like a current-controlled voltage source.

 

[berkeman]

Ah, thanks. Then their KCL equation makes sense.

 

[marsupial]

Can someone help me understand why what I did was incorrect and the reasoning behind the correct approach to the dependent voltage source? If I've drawn the currents the correct way, it appears they have $\frac{150i\sigma - v_1} {50}$

 

[berkeman]

Can someone help me understand why what I did was incorrect and the reasoning behind the correct approach to the dependent voltage source? If I've drawn the currents the correct way, it appears they have $\frac{150i\sigma - v_1} {50}$

It just looks like simple sign errors. I find it easiest to write the sum of all currents leaving a node equals zero. For me, it's too hard to try to label a circuit diagram with my best guesses at which way the currents go before writing the KCL equations. I just write the sum out = zero.

Do you see how the KCL equation they wrote reflects that paradigm? Do you see the sign error in your equation?

 

[marsupial]

I do see how it reflects what you are saying, but we were taught to do it this way and the answer in the pdf I found online also drew the current as exiting the v_1 node. I would just like to understand how it works that way too. Also, in an example we had in our lecture notes, where there was also a dependent voltage source, current controlled, but not between 2 resistors, they did the dependent source minus the V node, which is what they appear to have done in this instance. I usually do what is on the left minus what is on the right, so I was just wondering if with dependent sources I should do it the other way around, or if there is reasoning that would lead to that, say due to the voltage drop or rise.

Also, just looking at what I did compared to what they did, they don't seem to have written all the currents coming out, as they have $i_1$ as $\frac {v_1 - 160} {10}$ and I have it as $\frac {160 - v_1} {10}$

 

[cnh1995]

If I've drawn the currents the correct way,

There is no one correct way.
But you have to write the KCL equation accordingly, consistent with the assumed current directions. From their equation, it looks like they have assumed all three currents to be outgoing currents at the upper node (V1).

In your assumption, you have assumed i1 and iσ to be incoming currents and i2 to be outgoing current.

Check the voltage polarity you've marked for the 100 ohm resistor and your assumed direction of iσ. Are they consistent with each other?

 

[marsupial]

$i_\sigma$ was given as an incoming current in the diagram they supplied (the first diagram - I only added currents to the second). I see I wrote the polarity wrong for the 100 ohm resistor. But even with that change I didn't get the correct answer (using the KCL I gave).

If the polarities are reversed to how they should be for the 100 ohm resistor, then $i_\sigma$ would be written as $\frac{-v1} {100}$ wouldn't it?

I can see from that, that their equation holds as $i_\sigma = i_1 + i_2$

But I was taught that you just put in the currents as you think they should be, and the results would be a negative if they were in the reverse order. But this comes out to a completely different value. I am not sure why they chose $i_1$ as going out rather than going in, except perhaps taking cue from the direction of $i_\sigma$ ??

Also, did I approach the writing of $i_2$ correctly - that it is $v_1 - 150i_\sigma$ rather than the other way around?

Thanks for your help!

 

[cnh1995]

I can see from that, that their equation holds as iσ=i1+i2iσ=i1+i2i_\sigma = i_1 + i_2

Right. They equated the sum of three outgoing currents to zero, so at least one of them has to be negative and you don't know which one(s) right now. It will be clear once you solve the equation using actual values.