# How to decide the voltage sign of inductor in a circuit?

In this simple circuit, as the current increase, the inductor will induce an emf and a current point up from B to A. The voltage across the inductor should be -L*dI/dt, then the battery voltage plus this emf equal to RI, voltage across the resistor. This is different from the equation because of the sign of the voltage across the inductor. I know this is not right, but I don’t know where is wrong. I feel the argument kind of like Kirchhoff voltage law, but I’m not sure it’s applicable here cause there is a changing flux in this loop due to the inductor.

• When you say "the voltage across the inductor", do you mean $V_A-V_B$ or $V_B-V_A$? Jul 12 at 16:38
• "The voltage across the inductor should be -L*dI/dt, then the battery voltage plus this emf equal to RI" But this is exactly the same as equation 7.65 in the text that you quote. Jul 12 at 16:48
• I mean VA-VB if the given current direction is positive. The induced emf by the increasing current is point form B to A which is negative, same as the sign of battery voltage. Then the voltage cross the resistor is positive from B to C. Jul 13 at 3:16
• To understand KVL and circuits, it is best to stop using the word "voltage" for a while, and understand the concepts of potential difference and various kinds of emfs first. There is the emf $\mathscr{E}_0$ due to battery, induced emf of inductor $-LdI/dt$ ((always with minus). The Kirchhoff second circuital law implies $\mathscr{E}_0 - LdI/dt = RI$. KVL is a restatement of this in terms of potential drops: $RI + LdI/dt - \mathscr{E}_0 = 0$ (potential drop on perfect inductor is $LdI/dt$ with plus, potential drop on battery is $\mathscr{E_0}$). Jul 14 at 23:36

the inductor will induce an emf and a current point up from B to A

This is not entirely correct. If the current increases, dI/dt > 0 necessarily, therefore the current will still be in the conventional direction ("downwards" toward the resistor to the ground).

The voltage across the inductor should be -L*dI/dt, then the battery voltage plus this emf equal to RI, voltage across the resistor.

This can be simply analyzed by Kirchoff's Law, where the battery (positive, "active" voltage) is equal to the resistor and inductor voltages (the "passive" voltages), as both are net current sinks.

However, if the voltage source is suddenly turned off (pictured in the diagram with the switch before the inductor) then you have the case that the "charged"inductor will be transiently decreasing its current towards 0 in the opposite direction.

You would see this happening if you connect an oscilloscope to the resistor, which in this case would act as a load.

From a physical perspective, here you would analyze this through Biot Savart's law where a magnetic field is induced by a current inside of the loop of the coil (inductor). When the current is suddenly cut (by opening the circuit) there will be a residual magnetic field but now it is reversed in direction (due to right-hand convention) thus effectively "reversing" the direction of the current at least for the transient moment (the coil will only store some energy in the form of a magnetic field which is proportional to the coil loops, wire thickness, etc).

In this simple circuit, as the current increase, the inductor will induce an emf and a current point up from B to A.

That is not correct. The inductor will not cause the current to point up from A to B. It will induce an emf to oppose the increase in current pointing down from A to B. That means the polarity of the voltage across the inductor upon closing the switch will be as shown in FIG 1 below.

The voltage across the inductor should be -L*dI/dt

Per Faraday's law, the voltage across the inductor is $$LdI/dt$$. If the current is increasing in time then $$dI/dt\gt 0$$ and the voltage is positive, as shown in FIG 1. It it were decreasing in time then $$dI/dt\lt 0$$ and the polarity would be opposite that shown in FIG 1.

then the battery voltage plus this emf equal to RI, voltage across the resistor.

The instant after closing the switch $$RI=0$$ because $$I=0$$ the instant after closing the switch. And that's because you can't change the current through an ideal inductor in zero time. Therefore the instant after closing the switch it follows from equation 7.65 that

$$L\frac{dI(t)}{dt}=\epsilon_{0}$$

and the polarity across the inductor is as shown in FIG 1. As time progresses the current builds, the voltage across the resistor increases, and voltage across the inductor decreases. But the polarity across the inductor remains the same. After a long time, the voltage across the ideal (zero resistance) inductor is zero and all the battery voltage is across the resistor.

Hope this helps.