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In the case of a pure inductor ac circuit the source emf is equal to the induced emf hence the current is blocked.

In a pure inductance circuit

V = V°sin(wt)

In the very next step we assume that,

-L(di/dt) = V°sin(wt)

Here is where the induced emf equals the source emf.

By integrating and re arranging

I = I°sin(π/2 - wt)

I completely understand the derivation and how it's done. But it couldn't have been done without the initial assumption, and under the initial assumption there can be no current

Now where are my assumptions wrong ?

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    $\begingroup$ "Under the initial assumption there can be no current" You say source emf being equal to inductor emf means there can be no current, but it's not clear why you think that, so it's hard to write an answer that will address your confusion. How is this different from connecting a battery to a resistor? There, the battery voltage is equal to the resistor voltage. What exactly is it that you object to in the derivation you yourself provided? $\endgroup$
    – Puk
    Commented May 30, 2023 at 5:54
  • $\begingroup$ If you could point out how the derivation equates induced emf and source voltage I would be very much pleased. $\endgroup$
    – YaMe
    Commented May 30, 2023 at 6:35
  • $\begingroup$ $\oint \vec E \cdot d\vec \ell = - L\ di/dt$, by Faraday's law. $\vec E=0$ in an ideal inductor so the left hand side is only due to the voltage source, and equal to $-V_\text{source}$, so $L\ di/dt=V_\text{source}$. The signs here depend on how you define the source polarity and current direction, but it doesn't sound like that's the problem here. $\endgroup$
    – Puk
    Commented May 30, 2023 at 6:51

4 Answers 4

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Saying inductors block AC is like saying resistors block current. It doesn't mean that inductors don't permit any AC current at all. In particular, I'm not sure why you think an AC voltage applied to an inductor will lead to no current: the current will be the source voltage divided by the inductor impedance ($j\omega L$).

An RLC circuit is useful because it can be tuned such that at a particular resonance frequency (at which the inductor and and capacitor impedances are equal in magnitude), the circuit presents either a very low (for series RLC) or a very high (for parallel RLC) impedance. This can be used e.g. in bandpass filters to select a narrow range of frequencies and reject others, like in a radio receiver circuit.

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  • $\begingroup$ Well I the derivation of the equation you provided above. Starts of by assuming that the source Voltage is equal to the induced emf. This is were my initial doubts came from $\endgroup$
    – YaMe
    Commented May 30, 2023 at 5:34
  • $\begingroup$ @YaMe I don't follow. If you connect a resistor to a voltage source, the source voltage is equal to the resistor voltage. That doesn't mean there is no current through the resistor. The same applies to an inductor. Please edit your question to clarify in more detail where your confusion lies. $\endgroup$
    – Puk
    Commented May 30, 2023 at 5:40
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The induced emf, $\mathcal E_{\rm L}$, in an instructor of inductance, $L$ depends on the rate of change of current, $\frac{dI}{dt}$, the relationship being $\mathcal E_{\rm L} =- L\frac{dI}{dt}$.
So the current being zero is immaterial, it is the changing current which relates to the emf.

In the case of a pure inductor ac circuit, the source emf is equal to the induced emf hence the current is blocked.

is a misleading statement.

The answer to resolving this apparent contradiction lies in the fact that you are dealing with two different types of electric field (conservative and non-conservative) and that the non-conservative electric field owes its existence to a changing magnetic flux produced by a changing current.

The definition of self-inductance is $L=\dfrac {\Phi}{I}$ where $\Phi$ is the magnetic flux and $I$ is the current.

Differentiating the defining equation with respect to time and then rearranging the equation gives $$\dfrac{d\Phi}{dt} = L\dfrac{dI}{dt} \Rightarrow \mathcal E_{\rm L} = - L\dfrac{dI}{dt} $$ after applying Faraday's law where $\mathcal E_{\rm L}$ is the induced emf produced by a changing current.
The electric field associated with the changing magnetic flux is non-conservative.

Consider a circuit consisting of an ideal cell of emf $V{\rm s}$, a switch and an ideal inductor all in series with one another.

At time $t=0$ the switch is closed.
The initial current must be zero which you can understand with an appreciation of the fact that mobile charge carriers have inertia and thus cannot undergo an infinite acceleration.

The conservative field produced by the cell is trying to increase the current from zero, but the non-conservative field produced by the inductor is trying to stop the current changing.
Which field wins?
At $t=0$ there is no current so it would appear that it is a draw between the two fields, but the non-conservative field can only stop a current flowing at $t=0$ on condition that the current changes.
So the current has to increase despite the opposition of the non-conservative field and so it continues with the current increasing due to the conservative field despite the opposition of the non-conservative field.
All that the non-conservative field can do is slow down the rate at which the current changes; it can never stop the current changing as then it (the non-conservative field) would no longer exist.

For this example in which a battery is applied across an inductor, the current $I = \dfrac{V_{\rm s}}{L}\,t$.

Now if an alternating current is applied to the inductor you find that the ratio of the peak voltage, $V_{\rm peak}$, to the peak current, $I_{\rm peak}$, is $\omega L$ where $\omega$ is the frequency.

Thus you will see that, for a given peak voltage, the magnitude of the inductance plays a part in determining the magnitude of the peak current.

In the case of a series $LCR$ circuit in which the current through each component is the same, it is the phases of the voltages across each of the components which are important as the sum of the voltages must equal the applied voltage.

The fact that the voltages across the capacitor and the inductor are $\pi$ out of phase with one another means that they have "opposite effects" on the circuit which can materially affect the current flowing in the circuit.

I completely understand the derivation and how it's done. But it couldn't have been done without the initial assumption, and under the initial assumption there can be no current.

I hope that I have shown you that there is nothing wrong with the assumption that at a given time the current is zero, but another very important consideration is the value of the rate of change of current at that time.

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You say:

In the case of a pure inductor ac circuit the source emf is equal to the induced emf hence the current is blocked.

But this isn't true. The equation relating the voltage and current for an inductor is:

$$ V = L \frac{dI}{dt} $$

At time zero, i.e. when you first turn close the switch in your circuit, the current is indeed zero, but $dI/dt$ is not zero so $I$ increases with time. The current is not blocked.

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I dont think the current induced in the inductor is equal to the source emf the inductor only applies a hinderance, a reactance an inductor does not block i believe

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