Calculating the impedance formula of an inductor

Question

Assume we have a purely inductive circuit such as below:

To calculate the most general case, \$V_1\$ is given as such:

$$v(t) = V_m \cdot \cos(\omega t+\theta)$$

For an inductor, \$v(t) = L \dfrac{\mathrm{d}i(t)}{\mathrm{d}t}\$. Writing it in integral form:

$$i(0)+\int_0^t{v(t)\mathrm{d}t}= i(t)$$

Now consider \$i(0)=0\$, meaning there was no current in the system before the source was on:

$$\int_0^t{V_m\cdot \cos(\omega t+\theta) \mathrm{d}t}= i(t)$$

Moreover, the result of the integral, and hence the current:

$$\dfrac{V_m}{\omega}\cdot(\sin(\omega t+\theta)-\sin(\theta))= i(t)$$

If transformed to their complex counterparts via \$\mathrm{e}^{j\omega t} = \cos(\omega t)+j\sin(\omega t)\$, the result, which is the impedance, is not \$j\omega L\$ but some term similar to \$j\omega L\$, but also containing the phase of the voltage source, which shouldn't be there, since impedance is not dependent on the phase of the source.

Where am I going wrong?

Answer 1

The result, which is the impedance, is not jωL but some term similar to jωL, but also containing the phase of the voltage source, which shouldn't be there, since impedance is not dependent on the phase of the source Where am I going wrong?

The result is correct; you get a term that is the AC impedance of the inductor and, a term that represents the instantaneous phase angle of the applied voltage when it is first powered. Consider this simulation using a sinewave voltage of 1 volt peak across a 1 henry inductor: -

As you can see, the current contains the usually expected alternating part but, it is also raised above 0 amps by a DC value. This is what happens when you drive an inductor with a sinewave voltage where the phase angle is 0° at the point of activation of the sinewave.

In other words, the DC term above represents the phase angle of the voltage source when that voltage source was first applied. 0° means a big DC offset; 90° means no DC offset. Your derivation is probably correct but, you just don't believe it can be so.

So, if you activated the sinewave at 90°, you just get the AC waveform with no DC offset: -

At 45° it looks like this: -

And all of this is caused by the act of integration right back at the start of your formulas. It's also the reason why many transformers heavily saturate when first powered-on (a real-life problem). Anti-inrush current circuits are used in real-life to activate the voltage at the top of its waveform.

Of course, a real inductor (lossy) soon finds equilibrium such as when a sinewave is applied at 0° but in series with 0.3 ohms: -

But for many cycles, this can and does cause breakers to trip on transformer feeds.

Answer 2

Phase requires a reference phase. The input phase is arbitrary, so it can be set to zero without loss of generality. (If the input phase is different from zero, use the trig identities for sin(A+B) and cos(A+B).) You must restrict your integral to period boundaries,

The result of your integral becomes:$$i(t)=\frac{V_{m}}{\omega}sin(\omega t)$$

EDIT (Response to ozgun can comment below):

Impedance is not a time domain quantity. It is a frequency domain qualtity. In fact the most general form for impedance is expressed in the Laplace domain which removes the sinusoidal constraint. It cannot be obtained directly fwom a time domain integral without transformation in to the frequency domain.

END EDIT

A simpler approach is to use Euler's form: $$v_{L}=V_{m}e^{j(\omega t+\phi)}$$ This allows the input phase to be factored out: $$v_{L}=V_{m}e^{j\phi}e^{j\omega t}$$ Then the integral for the inductor current for \$i_{L}(0)=0\$ is: $$i_{L}=\frac{V_{m}}{j\omega}e^{j\phi}e^{j\omega t}$$ The input phase then cancels out of the impedance ratio: $$\frac{v_{L}}{i_{L}} = j\omega L$$

Of course the Laplace transform method works as well. The Laplace transform converts the defining relation from time domain into the complex frequency domain which allows the most general expression for impedance in that it applies to all signals not just sinusoids.

The Laplace transform of the defining relation with zero initial conditions is:

$$V_{L}(s)=sLI_{L}(s)$$ $$Z(s)=\frac{V_{L}(s)}{I_{L}(s)}=sL$$ $$Z(j\omega)=j\omega L$$

Answer 3

You have to use complex voltage source \$V_s(t) = V_m e^{j(ωt+θ)}\$. Then you use kirchoff's voltage law in the circuit.

Current in the the circuit is \$i(t) = I_me^{j(ωt+Φ)}\$.This is a assumption. Let \$Z_L\$ be impedance of the inductor.The values of \$I_m\$ and Φ can be calculated by using KVL, phasor notation and initial conditions.

Applying KVL in the loop,

$$-V_s(t) + L\frac{di}{dt} =0$$

$$V_s(t) = L\frac{di}{dt}$$

$$V_me^{(j(ωt+θ)} = L\frac{di}{dt} I_me^{j(ωt+Φ)}$$

$$V_m e^{j(ωt+θ)} = jωLI_me^{j(ωt+Φ)}$$

Dividing LHS and RHS by \$e^{jwt}\$

$$V_me^{jθ} = jωLI_me^{jΦ} $$

Voltage across the inductor should be same as voltage across the voltage source.

Therefore \$V_L\$ = jωLI which is similar to the expression \$V_L\$ = \$Z_L\$I. So,impedance \$Z_L\$ of the inductor is jωL.