# Difficulties in KVL signs (RL circuit)

If we start from bottom left and rotate clockwise, then first we visit resistor negative terminal so I expect -Ri but the book says the below equation:

So why isn't the Ohm's law for the resistor negative? (I follow this famous convention in KVL, whenever you encounter the positive terminal of an element you mark that expression positive else you mark it negative)

• What is the source here? There is not an external voltage source... so we can guess the inductor is the source (if pre-charged). But then the current will flow in an opposite direction. It starts from +VL... enters the upper end of the resistor thus creating +VR... leaves its lower end thus creating -VR... and returns where it started - -VL. – Circuit fantasist Nov 24 '19 at 22:45
• @Circuitfantasist actually the book is assuming that a source has been present at some point. – Mahdi Nov 24 '19 at 22:57
• So this is the situation - a pre-charged inductor discharges through a resistor as I described above. It behaves as a current source that creates the voltage drop R.i across the resistor (and itself). The inductor is pre-charged by connecting a voltage source in the circuit so that to inject charging current into the lower terminal of the inductor. Then the source is removed and the inductor continues "moving" the current in the same direction; hence the present polarity of Vl... – Circuit fantasist Nov 24 '19 at 23:26
• i(t) can be shown in either direction. The sum is still =0 – Tony Stewart Sunnyskyguy EE75 Nov 24 '19 at 23:37
• @TonyStewartSunnyskyguyEE75 what I don't understand is that why Ri is positive. I think as current enter resistor from its (conventional) negative terminal then we should have -Ri + L di/dt. – Mahdi Nov 24 '19 at 23:52

With the current flowing as shown, $$v_L = L\frac{di}{dt}$$ and $$v_R = -iR$$ with $$\v_R\$$ being negative as expected.
Using $$\v_R\$$ and $$\v_L\$$ with KVL yields $$v_R=v_L$$ or $$-v_R+v_L=0$$ Note there is a negative sign as well.
Substituting the first 2 equations yields $$-(-iR) + L\frac{di}{dt} = 0$$ $$iR + L\frac{di}{dt} = 0$$