# In which orientation should the dummy voltage source be connected when determining the i-v characteristics?

Suppose, I have a circuit like this:

Now, in order to find the i-v characteristics, I have to connect a variable voltage source across the A-B terminals. Now, this can be done in one of two ways. The two ways are given:

This is the first way. Let us figure out the equation for this case. In the closed loop PQRSP, we can apply KVL,

2I-V2+5=0

This can alternatively be written as 2y-x+5=0

We can also find out the equation without using KVL. The voltage drop across the resistor R1 is V2-V1 (assuming that the current I flows from V2 to V1). Then I=(V2-V1)/2. We will eventually get 2y-x+5=0 from here.

This is the second way. We can determine the equation using KVL. In the closed loop TMNOT,

-2I+V2+5=0

or, -2y+x+5=0

Also, we can determine the equation in an alternative way. The voltage drop across R1 is V1+V2 or 5+V2. Then I=(V1+V2)/2. Then, from that we can derive -2y+x+5=0.

Here V2 is the variable voltage source. Now, I have found that the graph becomes different when I connect the variable voltage in the opposite orientation. The red line is for the first orientation, and the purple line is for the second orientation. I used KVL to find the equations. I initially expected both the graphs to be the same. I was unpleasantly surprised to find that that was not the case.

Now, say I was given the above circuit in my lab exam. How should I connect V2 to the circuit? In what orientation? I will get two different equations/graphs/i-v characteristics for the same circuit by connecting in two different orientations. How do I know which is the right one?

• A current pointing out of the positive end is negative. In both your cases a current following your arrows will be negative since both your arrows point out of the positive end. Since I'd likely be interested in the blue arrow direction I'd use the orientation found in the green arrow case. This would not look like your red line, though, if the x axis is the voltage of V2. I don't know how you got your equation. V2=0 gives 2.5 A and V2=5 gives 0 A in your first schematic. Commented Oct 25, 2023 at 11:40
• @periblepsis I edited my question to show how I got the equations. Should've done that the first time. My bad. Commented Oct 25, 2023 at 13:21
• Here's the I/V curve. Simple. Does your equation reproduce this? Commented Oct 25, 2023 at 13:38
• I've provided the complete schematic for LTspice that you can use. It provides the proper resulting load line (2 Ohm.) If you have any remaining questions, let me know. Commented Oct 26, 2023 at 4:07
• @periblepsis Thank you so much!! I am just a bit run-down; that's why I couldn't get back to you earlier. The thing is that I was able to derive your equations too by taking current (I) out of the negative terminal (i.e. by considering the current coming out of the negative terminal to be positive, but I find that convention a bit strange). I mean it is fine, but I thought that the well-recognized convention was that the current coming out of the positive terminal is positive. Commented Oct 26, 2023 at 14:21

I think your math isn't quite right

To be clear, you're trying to produce a graph that shows the current in the circuit, when an external voltage source is applied, where both V1 and V2 are assumed to be perfect sources, with no internal resistance, capable of sourcing or sinking infinite current.
I might gently note that this seems a bit odd to me, perhaps this is a specific task you've been set.

As you understand, to get the current in this circuit, you must find the voltage drop across the resistor R1, which is (V1 - V2). Where the resistance is 2Ω,
I = (V1-V2)/2, given that V1 is fixed at 5V,
I = (5 - V2)/2
You could rearrange this to 2I + V2 - 5 = 0, and substitute I=Y and V2=X to get this into the format
2Y + X - 5 = 0
If we then reverse the polarity of V2, as in your second circuit, we then change the polarity of V2 in the math, giving
I = (V1 + V2)/2 or
2Y - X - 5 = 0

For both of these, we're getting a value for the current flowing where a positive value is the current flowing clockwise (or out of the positive terminal of V1), and a positive anti-clockwise.
I see that in your first circuit, the arrow suggests a positive value for the current flowing anticlockwise. Where current flows in a circuit are absolute, but whether this is considered positive or negative depends on your perspective/math/measurement orientation. If we flip this, and consider an anticlockwise current positive for the first circuit we get
I = (V2-V1)/2 or
2Y - X + 5 = 0

Assuming the change in polarity of the current between the two circuits is deliberate the two graphs should be
2Y - X - 5 = 0, where X is the value of V2, and Y represents the value of anticlockwise current.
Then for the second circuit
2Y - X + 5 = 0, where X is the value of V2, and Y represents the value of clockwise current. Because both the current and V2 change orientation together, there isn't a polarity shift in the two graphs.

[after significant edit to question]

2I - V2 + 5 = 0
This is wrong, 5V is generated by V1, this is opposed by V2 (so this is -V2), but the voltage droped accross R1 is also a negative, so your equation should be
-2I - V2 +5 = 0
I've rearranged this to hopefully make this clear. If it helps, consider if V2 wasn't there, then the equation would have to be
5 - 2I = 0
Of course, you can flip the polarity to go from
-2I - V2 + 5 = 0 to
2I + V2 - 5 = 0. These are the same equation, in this case, you have an incorrect -2I

For the second circuit, now that you have flipped the polarity of V2, the equation becomes
-2I + V2 + 5 = 0 or
2I - V2 - 5 = 0
Which is how you have it.

I still don't quite understand the exercise. If the point is to generate a graph of voltage and current out, a graph that says for a given voltage across QR, then this current must be drawn, then I'd go with the first. Technically they both produce the same relationship, it's just that the second reverses the polarity - ie this considers a voltage RQ, rather than QR. It's still the same value.

• My second equation is exactly the same as your second equation. Commented Oct 25, 2023 at 13:05
• I edited my question to show how I got the equations. Should've done that the first time. My bad. Commented Oct 25, 2023 at 13:21