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What i am seeking is to explain why the relationship here is not V2/R2 instead it's given as (V1-V2)/R2

enter image description here

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    \$\begingroup\$ V2 is the voltage across both resistors. \$\endgroup\$ – Ignacio Vazquez-Abrams Oct 23 '16 at 14:46
  • \$\begingroup\$ yeah,but why? i mean R1 doesn't seem to have any relationship with V2.Am i examinating it with the wrong scope? @IgnacioVazquez-Abrams \$\endgroup\$ – Konstantinos T. Pantelis Oct 23 '16 at 14:52
  • \$\begingroup\$ You have noticed that there is a 2H inductor in that circuit I hope. \$\endgroup\$ – JIm Dearden Oct 23 '16 at 14:55
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    \$\begingroup\$ The voltage across R2 is V2 - V1 where both V1 and V2 are referenced to ground. \$\endgroup\$ – Peter Smith Oct 23 '16 at 15:07
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    \$\begingroup\$ You have a constant current source as well by the way \$\endgroup\$ – crowie Oct 23 '16 at 16:25
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Because ohm's law is actually:

$$\Delta V=RI <=> Va-Vb=RI_{ab}$$

Where Va is the voltage at one terminal of the resistor, and Vb is the voltage at the other terminal. I just added the little index to the current so that you know its direction.

The delta is often omitted for simplicity, and this usually leads to a lot of confusion from beginners.

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  • \$\begingroup\$ Sometimes Ohm's Law is used like $$ ESR= $$ $$ \Delta V \over \Delta I $$ \$\endgroup\$ – Sunnyskyguy EE75 Oct 23 '16 at 21:26
  • \$\begingroup\$ In the context of my answer, ΔV is the difference in potential across the resistor, which is the same as the difference of the potential differences in relation to ground (voltages Va and Vb) at the terminals of the resistor. When calculating ESR you are actually taking the difference between voltages for different currents, like when trying to calculate resistance from a diode's I-V curve, so in that case ΔV means a voltage difference (a differance in potential difference in relation to ground) and not strictly a potential difference. \$\endgroup\$ – Chi Oct 24 '16 at 1:06
  • \$\begingroup\$ Of course, understading that a difference of potential differences is the same as a potential difference is very confusing for beginners and heck, even for some professionals, but that's what we get for not having an absolute 0 potential reference. I'm not saying that we should have one though, as that would be as stupid as defining a universal "0 height" for calculating gravitational potentials in different planets. \$\endgroup\$ – Chi Oct 24 '16 at 1:12
  • \$\begingroup\$ when you get 20 to 40 yrs experience , you will know the difference very well. I was actually referring to the linear region of a nonlinear component, circuit or network using simply, Ohm's Law. try to keep it real. \$\endgroup\$ – Sunnyskyguy EE75 Oct 24 '16 at 1:43
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I don't understand the difficulty, at all. With currents spilling outward on the left and currents spilling inward on the right, I get:

$$\begin{align*} \frac{V_1}{R_1} + \frac{V_1}{R_2} &= \frac{0\:\textrm{V}}{R_1} + \frac{V_2}{R_2} + i_s \\ \\ \frac{V_1}{R_1} + \frac{V_1}{R_2} &= \frac{V_2}{R_2} + i_s \\ \\ \frac{V_1}{R_1} + \frac{V_1}{R_2} - \frac{V_2}{R_2} &= i_s \\ \\ \frac{V_1}{R_1} + \frac{V_1-V_2}{R_2} &= i_s \end{align*}$$

I'm really flummoxed why you can't get there. But you haven't really exposed your thinking much, either.

As an aside, another way of writing the second node is:

$$\begin{align*} \frac{V_2}{R_2} + \frac{1}{L} \int V_2\:\textrm{d}t &= \frac{V_1}{R_2} \\ \\ \frac{1}{L} \int V_2\:\textrm{d}t &= \frac{V_1-V_2}{R_2} \end{align*}$$

But, assuming \$i_s\$ is a constant current source, your circuit can also be reduced to a voltage source with a series resistance into an inductor. So a DC voltage and an R+L load.

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