Electromagnetism – Understanding AC Through a Pure Inductor

electromagnetisminduction

I've studied the AC circuit for an ideal inductor in many physics books. After deriving the final equation for current the integration constant $C$ is assumed to be $0$ by giving inadequate reasons. In this question I seek for an adequate reasons.

Suppose an ideal AC voltage source is connected across a pure inductor as shown:

i

The voltage source is$$V=V_0\sin(\omega t).$$ From Kirchhoff’s loop rule, a pure inductor obeys
$$V_0 \sin(\omega t)=L\frac{di}{dt},$$ so
$$\frac{di}{dt}=\frac{V_0}{L} \sin(\omega t)$$
whose solution is
$$i=\frac{-V_0}{\omega L}\cos(\omega t)+C$$

Consider the (hypothetical) case where the voltage source has zero resistance and zero impedance.

In most of elementary physics books $C$ is taken to be $0$ for the case of an ideal inductor.

$$\text{Can we assume that } C \neq 0?$$

  1. (To me this is one of the inadequate reasons). This integration constant has dimensions of current and is independent of time. Since source has an emf which oscillates symmetrically about zero, the current it sustain also oscillates symmetrically about zero, so there is no time independent component of current that exists. Thus constant $C=0$.

  2. (Boundary condition) there might exist a finite DC current through a loop of wire having $0$ resistance without any Electric field. Hence a DC current is assumed to flow through the closed circuit having ideal Voltage source and ideal inductor in series if the voltage source is acting like a short circuit like a AC generator which is not rotating to produce any voltage. When the generator starts, it causes the current through the circuit to oscillate around $C$ in accordance with the above written equations.

Best Answer

The fact is, in the context of ideal circuit theory, the inductor voltages are equal in the circuits below:

enter image description here

In the lower circuit, the inductor current has a constant component, i.e., the lower circuit is equivalent to your $C \ne 0$ case.

But, there's nothing remarkable or surprising about this. Is there something else to your question that I'm missing?


[I] am asking that why in in elementary physics text books the author directly writes C=0 for ideal inductor without mentioning the initial conditions?

The initial conditions are mentioned when the context is transient analysis. For example, from Wikipedia:

enter image description here

However, when the context is AC (sinusoidal) circuit analysis, the underlying assumptions are (at least):

(1) All sources are sinusoidal and of the same frequency

(2) The circuit is in sinusoidal steady state, i.e., all transients have decayed.

When these conditions hold, can we use voltage and current phasors and the notion of impedance to analyze circuits.

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