A voltage drop means a loss in energy in the form of heat for example in a resistor, in what form the voltage drop across an inductor is?

Question

In an RL series circuit where the current rate of change is positive, there is a voltage drop across the inductor, if voltage drop means an energy loss, then in what form the energy is lost in the inductor?

Answer 1

if voltage drop means an energy loss

Voltage drop does not mean an energy loss. Ideal inductors (and ideal capacitors) store energy. They either store it for all time, or release it later (real inductors or capacitors may store energy, then dissipate it as heat).

If you start with a discharged inductor and apply a voltage to it, the inductor's current rises as more and more energy is stored in its magnetic field. If you had an ideal inductor and you could bring the terminal voltage to zero, this current would remain forever. If you reverse the voltage on the inductor, then the current will remain flowing in the same direction, and because the voltage is reversed you will get energy out of the inductor.

Answer 2

The induced emf $\mathcal E = (-)L\frac {dI}{dt}$ and so the instantaneous power is $\mathcal E \,I$.

The energy delivered to the inductor when the current increases from zero to $I$ is $\displaystyle \int \mathcal E \,I \,dt = \int L\,\frac{dI}{dt} \,I \,dt= \int_0^I L\,I \,dI=\frac 12 LI^2$ and this represents the energy stored in an inductor or rather the magnetic field associated with the inductor.

Unlike a resistor where the electrical energy is dissipated as heat, an inductor can be thought of as a store of energy and so when the current through the inductor is decreasing which in turn means the magnetic field is decreasing, energy is returned back to the circuit and assuming that the inductor itself has no resistance, no heat is dissipated.

Having an emf with no current flowing means that the instantaneous power is zero and at that instant the inductor is not storing any (magnetic) energy.

. . . . . if voltage drop means an energy loss is true for a resistor with the "lost" energy becoming heat but what happens when the current through an ideal rechargeable battery enters at the positive terminal and leaves via the negative terminal?
There is a conversion of electrical energy into chemical energy.
Thus in a sense there is a loss of electrical energy but that electrical energy can be "recovered" when current enters the negative terminal and leaves via the positive terminal.

Answer 3

For a RL circuit, the differential equation in the case of a source of power is: $$V = RI + L\frac{dI}{dt}$$ It is analogous to a mass moving in a viscous media, being driven by some force, and where the drag force is proportional to the velocity (what can be the case for small velocities): $$F = k_dv + m\frac{dv}{dt}$$ It is clear that the fraction of the force used to accelerate the mass can be restored in the case of deceleration. So it is not lost. But the part braked by the media is really lost.

In the same way, the energy is not lost in the inductor. It is stored as magnetic field. But it is lost in the resistor, where it is a dissipated as heat.