KCL Problem with Dependant sources

In summary, the conversation was about finding the voltage drop V1 in a circuit using KCL and Ohm's law. The attempt at a solution involved using Vin/2 and Vin/6, but the logic was flawed somewhere. The expert advised to combine the resistors in parallel and write the KCL equation again. The correct solution was found by using the bottom node as the reference node and solving for V1, which was found to be -0.5V. The power delivered by the dependent source was calculated to be 1/2 watts.
  • #1
bran_1
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0

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.
QWnesTY.jpg

Homework Equations


KCL, Ohm's law

The Attempt at a Solution


I basically tried to use KCL at the top node,
Code: 
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
 
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  • #2
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?
 
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  • #3
gneill said:
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...
 
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  • #4
bran_1 said:
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.
 
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  • #5
cnh1995 said:
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?
 
  • #6
bran_1 said:
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?
 
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  • #7
Wouldn't it be VA (of the node) on the other side? I'm not sure how to find the node voltage though
 
  • #8
bran_1 said:
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.
 
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  • #9
gneill said:
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?
 
  • #10
bran_1 said:
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.

upload_2017-1-25_9-13-34.png
 
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  • #11
gneill said:
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
 
  • #12
Looks good.
 
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  • #13
gneill said:
Looks good.

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

P=VI , -0.5 * (2*-0.5) =1/2 watts
 
  • #14
bran_1 said:
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.
 
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  • #15
gneill said:
Yes, that looks right.
Thanks for all your help!
 

1. What is KCL (Kirchhoff's Current Law) Problem with Dependent Sources?

KCL Problem with Dependent Sources refers to a situation in an electric circuit where Kirchhoff's Current Law cannot be directly applied due to the presence of dependent sources, such as voltage-controlled or current-controlled sources. This can make the circuit analysis more complex and may require the use of additional techniques, such as nodal analysis or superposition, to solve the circuit.

2. How do I identify a KCL Problem with Dependent Sources?

A KCL Problem with Dependent Sources can be identified by examining the circuit and determining if it contains any dependent sources, such as voltage-controlled or current-controlled sources. If so, KCL may not be directly applicable and additional techniques may be needed to solve the circuit.

3. Can KCL still be used in a circuit with dependent sources?

Yes, KCL can still be used in a circuit with dependent sources, but it may require the use of additional techniques to properly apply it. For example, in a circuit with a voltage-controlled source, the controlling voltage can be replaced with a variable resistance and KCL can be applied as usual.

4. What are some common techniques for solving a KCL Problem with Dependent Sources?

Some common techniques for solving a KCL Problem with Dependent Sources include nodal analysis, superposition, and Thevenin's or Norton's theorem. These techniques allow for the conversion of dependent sources into independent sources, making it easier to apply KCL and solve the circuit.

5. How can I simplify a circuit with dependent sources to make it easier to apply KCL?

To simplify a circuit with dependent sources, you can use techniques such as source transformations, where dependent sources are converted to independent sources, or Thevenin's or Norton's theorem, which can reduce a complex circuit into a simpler equivalent circuit. These techniques can make it easier to apply KCL and solve the circuit.

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