# Can inductance or resistance be negative or “infinite”? Since the potential different or voltage drop is positive (am I allow to say voltage drop or potential difference in the presence of a changing current?), the current flows from right to left since $\Delta V$ is positive. That is, $V_b$ is at a higher potential than $V_a$.

Case (1), Increasing current.

So traversing from b to a (in the direction of the current) I get a

$-IR - LI' = 9 \implies -2R - 0.5L = 9$

I get $-IR$ because the voltage drops from a higher to a lower potential as it crosses the resistor. I get $-LI'$ because since the current is increasing, the inductor opposes this increase by "becoming" a battery with - on the "a" side and + on the "b" side and I should get a voltage drop across that.

Case (2), Decreasing current.

So traversing from b to a (in the direction of the current) I get a

$-IR + LI' = 5 \implies -2R - 0.5L = 5$

I get $-IR$ because the voltage drops from a higher to a lower potential as it crosses the resistor. I get $LI'$ because since the current is decreasing, the inductor opposes this decrease by "becoming" a battery with + on the "a" side and - on the "b" side and I should get +V by gaining potential.

Solving

http://www.wolframalpha.com/input/?i=RowReduce{{-0.5%2C-2%2C9}%2C{-0.5%2C-2%2C5}}

I get some absurd answer. What is wrong with my argument?

$V_{AB} = V_A - V_B$ so node A is more positive than node B side when $V_{AB}>0$

According to the Passive Reference Convention, current enters the positive terminal of a resistor so the current is from node A to node B.

We have

$V_{AB} = 2A \cdot R + 0.5A/s \cdot L = 9V$

and

$V_{AB} = 2A \cdot R - 0.5A/s \cdot L = 5V$

which yields

$R = 3.5 \Omega$

$L = 4 H$

[EDIT]: Given the numerous questions jak has asked in the comments, I've added a drawing to aid in "seeing" how to get the equations above.

Note that regardless of the direction one "travels" around the loop for KVL, you get the same equation:

$V_{ab} = V_R + V_L = IR + L\frac{dI}{dt} = 2A \cdot R + 0.5A/s \cdot L = 9V$

Also note that I take $V_{ab}$ to be the voltage across nodes a & b with the polarity shown as per the double-subscript convention. And further, that I chose the current direction and passive component voltage polarities according to the passive reference (or sign) convention. • What? Doesn't $V_{AB}$ mean the potential difference from $V_a$ to $V_b$? – Hawk Aug 10 '12 at 21:18
• $V_{AB} > 0$ when node A is more positive than node B. Colloquially, $V_{AB}$ is the voltage read by a voltmeter when the read lead is place on node A and the black lead is placed on node B. – Alfred Centauri Aug 10 '12 at 21:28
• @jak, see this – Alfred Centauri Aug 10 '12 at 21:46
• Because it isn't the other way around; the way I have it is the notational convention: $V_{AB}$ is the voltage drop from node A to node B. – Alfred Centauri Aug 10 '12 at 22:04
• @jak, see the added diagram and info to my answer. Hope it helps. – Alfred Centauri Aug 11 '12 at 15:40

Your formulas are false. In both case you have : RI+LI'=U. That gives you : 2R+0.5L=9 and 2R-0.5L=5 So R=3.5 Ohm; L=4 Henry

Notice: The arrow in potential indicates the highest potential Va. Current flows from the highest potential to the lowest as would do a stream down a mountain.

• Why do you have $+IR$? $\Delta V_{ab} = V_b - V_a > 0 \iff V_b > V_a$ – Hawk Aug 10 '12 at 21:20
• With method, it is quite simple: 1) Find which of Va or Vb is the greatest 2) Current flows from the greatest potential to the lowest. 3) Write the potential for each component. In a resistror and in an inductance, the current flows from the highest potential to the lowest: U=RI; U=Ldi/dt – Shaktyai Aug 11 '12 at 8:18