# Help with opamp question!

Discussion in 'Homework Help' started by koshi, Apr 29, 2015.

1. ### koshi Thread Starter New Member

Sep 25, 2014
27
0

Hi,

I need help in solving this circuit. I have the answer sheet but I don't really understand the way they solve this. The method they use is Vx-V1/R + V2-V1/R = V1/R, then solve for K. I thought Vx-V1/R will give me Ix which is the total current flowing in the circuit? So why did the answer sheet use Ix to add something else? And what is this V1/R value means? Isn't the two R center between V1 parallel to each other?

Any help will be appreciated. Thanks.

2. ### WBahn Moderator

Mar 31, 2012
20,075
5,666
You need to learn order of operations.

Vx-V1/R is a voltage, Vx, minus a current, V1/R, since this expression is identical to (Vx-V1)/R

You are correct that Ix = (Vx-V1)/R, but what do you mean by this being the total current flowing in the circuit? Is it the current flowing in the dependent source? Is it the current flowing in the bottom resistor? And since you don't know Ix it doesn't help you find V1, does it?

The node V1 has three branches coming into it. You have an expression for the currents in one of them. What are the similar expressions for the currents in the other two? Then what constraint is placed on the relationship of these three currents by Kirchhoff's Current Law?

I don't know what "the two R center between V1 parallel to each other" means. None of those resistors are in parallel with either of the others. What is required for two components to be in parallel?

3. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,923
601
It seems to me they are applying Node-Voltage Method. Have you done any work with Node-Voltage Method?

4. ### koshi Thread Starter New Member

Sep 25, 2014
27
0
Hi,

(Vx-V1)/R + (V2-V1)/R = V1/R, they are in bracklets in anyone didn't notice. Anyways, I am not too familiar with node-voltage. I roughly understand the part where (Vx-V1)/R is the current flowing through the first resistor or the line linked to the Vx and (V2-V1)/R will give me the second current flowing on the most right resistor, R. But what about the resistor that is going to ground in the middle?

5. ### koshi Thread Starter New Member

Sep 25, 2014
27
0
Actually, I just thought of something, correct me if I am wrong. Because the current flows from lower to higher potential, from the source it will travel from - to +. So in the case of Vx, current will travel along the right side, whereas, V2 will travel along the left side. So, this means the resulting route for the current to join up and flow to is only through V1 to ground. Since it doesn't make sense if the current of opposite route to cross each other so it must go through the path to ground. But going through ground, does it make sense?

Last edited: Apr 29, 2015
6. ### WBahn Moderator

Mar 31, 2012
20,075
5,666
Apparently you are using "electron flow" in which you are taught that current is the flow of electrons and not the flow of electrical charge. That's unfortunate, but we can deal with it.

The current in any resistor is the difference between the voltages on each side of the resistor divided by the resistance. The voltage of the ground node is simply 0V.