KCL Problem with Dependant sources

[bran_1]

Homework Statement

I'm trying to find the voltage drop V1 of the following circuit, assuming Vin=1v, and for the life of me can't seem to get the right answer.

Spoiler: Picture of Circuit

Homework Equations

KCL, Ohm's law

The Attempt at a Solution

I basically tried to use KCL at the top node,

Vin/2+2V1 = V1/3 + Vin/6, so 1/2=-2V1+V1/3+V1/6, (1/2)/(1.5)=V1=1/3, but apparently that's the incorrect answer

Thanks for any help

 

[gneill]

Hi bran_1, Welcome to Physics Forums.

I don't understand the terms of your node equation. For example, the first term of the LHS (left hand side) is Vin /2. How do you arrive at that? And on the RHS you have Vin/6, but Vin is not the node voltage. Can you explain your reasoning?

 

[bran_1]

Hi bran_1, Welcome to Physics Forums.

I don't understand the terms of your node equation. For example, the first term of the LHS (left hand side) is Vin /2. How do you arrive at that? And on the RHS you have Vin/6, but Vin is not the node voltage. Can you explain your reasoning?

Yes, Vin/2 is ohm's law, using input voltage (1v) and the resistance (2ohm) to find the Current through the resistor, and thus the current going into the node. As for the Vin over 6, since the 3ohm and 6ohm resistors are in parallel, they share the same voltage, do they not? This is just ohm's law again, except for the current leaving the node through this branch.Edit: I didn't get it correct though, so my logic is flawed somewhere...

 

[cnh1995]

Yes, Vin/2 is ohm's law, using input voltage (1v) and the resistance (2ohm) to find the Current through the resistor, and thus the current going into the node. As for the Vin over 6, since the 3ohm and 6ohm resistors are in parallel, they share the same voltage, do they not? This is just ohm's law again, except for the current leaving the node through this branch.Edit: I didn't get it correct though, so my logic is flawed somewhere...

You need not assume any value for Vin.
You should combine the 3 ohm and 6 ohm resistors in parallel and use their equivalent resistance (for simplicity). You'll still have the two nodes, and three currents instead of four.

Write the KCL (node voltage) equation again and simplify.

 

[bran_1]

You need not assume any value for Vin.
You should combine the 3 ohm and 6 ohm resistors in parallel and use their equivalent resistance (for simplicity). You'll still have the two nodes, and three currents instead of four.

Write the KCL (node voltage) equation again with and simplify.

I tried that also, it didn't seem to make a difference, as I got the same answer.
Vin/2 + 2V1 = V1/2
Vin = -3V1
V1=-(1/3)V, which is incorrect.

Is there some rule with dependent sources that I'm unaware of?

 

[gneill]

Yes, Vin/2 is ohm's law, using input voltage (1v) and the resistance (2ohm) to find the Current through the resistor, and thus the current going into the node.

Vin is not the potential difference between the two ends of the 2 Ω resistor. Only the left side of that resistor is at potential Vin. What is the potential at its other end? So then what is the potential difference?

 

[bran_1]

Wouldn't it be VA (of the node) on the other side? I'm not sure how to find the node voltage though

 

[gneill]

Wouldn't it be VA (of the node) on the other side? I'm not sure how to find the node voltage though

It's an unknown at this point. You'll solve for it after writing your node equation.

 

[bran_1]

It's an unknown at this point. You'll solve for it after writing your node equation.

Alright, so I'd have (Vin-Va)/2+2V1=V1/2, right? Could I call the bottom node node B, and say it's V=0? So then V1=Va-Vb? -> V1=Va?

 

[gneill]

Alright, so I'd have (Vin-Va)/2+2V1=V1/2, right? Could I call the bottom node node B, and say it's V=0? So then V1=Va-Vb? -> V1=Va?

Better.

Note that you can see that Va = V1 just by inspecting the circuit diagram. You don't need to introduce a different variable. Why not just use V1?

Yes, the bottom node makes an excellent choice for the reference node.

 

[bran_1]

Better.

Note that you can see that Va = V1 just by inspecting the circuit diagram. You don't need to introduce a different variable. Why not just use V1?

Yes, the bottom node makes an excellent choice for the reference node.

View attachment 112093

So using that, I have
Vin-V1=-3V1+V1 -> V1=-0.5V

 

[gneill]

Looks good.

 

[bran_1]

Looks good.

So then the power delivered by the dependent source would be:

P=VI , -0.5 * (2*-0.5) =1/2 watts

 

[gneill]

So then the power delivered by the dependent source would be:

P=VI , -0.5 * (2*-0.5) =1/2 watts

Yes, that looks right.

 

[bran_1]

Yes, that looks right.

Thanks for all your help!