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In the topic of pure A.C. Circuits, we have $I = V/\omega L$ and $I = V/\omega C$.

I know that $\omega C$ & $\omega L$ does the work of Resistor, but what does it mean and what do we mean by saying that $I$ is inversely proportional to $\omega$ and $C/L$? Edit: I want to know the behaviour of I with $\omega$ and L.

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  • $\begingroup$ Check you 2nd formula and also, are V and I magnitudes or phasors? $\endgroup$
    – Hal Hollis
    Commented Jun 12, 2017 at 18:20
  • $\begingroup$ One thing that helped me understand this was the realization this in an AC circuit, the current and the voltage could be out of phase. $\endgroup$
    – zeta-band
    Commented Jun 12, 2017 at 18:24

4 Answers 4

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Basics in AC circuits

Since we are using AC circuits, we have inputs which can be described as:

$$I_{in} = A\sin(\omega t + \phi)$$

And:

$$V_{in} = B\sin(\omega t + \phi)$$

Note that all sinusoid input signals can be written like this! (If the signal is not a pure sinusoid, we can use Fourier decomposition, but this is beyond the scope of this post)

The differential equations for inductors and capacitors are:

$$V_L = L\frac{dI_L}{dt}$$

$$I_C = C\frac{dV_C}{dt}$$

If we would take a single inductor or capacitor and hook it up to a periodic input signal, then $I_L = I_{in}$ and $V_C = V_{in}$. We can now calculate $I_L$ and $V_L$:

$$V_L = AL\omega\cos(\omega t + \phi)$$

$$I_C = BC\omega\cos(\omega t + \phi)$$

Here the terms $L\omega$ and $C\omega$ can be identified as the reactances $X_L$ and $X_C$ (note $\omega = 2 \pi f$).

As you can see, current leads voltage in the capacitor, and voltage leads current in the inductor. This can be remembered via "CIVIL".

We can now go to phasor notation, as the only thing important in those circuits is the phase. This is keep track of the phases. As we will see, phases are very important in AC circuits, especially when dealing with power.

Phasor notation

We can convert a periodic signal, like $V_L$ to phasor notation, which is:

$$V_L = AL\omega\cos(\omega t + \phi) = \Re(L\omega e^{i\phi}e^{i\omega t})$$

You can check this for yourself. In phasor notation we drop some terms; the signal in phasor notation is:

$$V_L = AL\omega e^{i\phi}$$

Complex power is defined as:

$S = UI^* = P + iQ = \Re(UI^*) + i\Im(UI^*)$

Note the complex conjugate! Here P is the active power and Q the reactive power (to be elaborated below).

In a resistor, where the current and voltage are in-phase, we can calculate the power by integrating over one period of the signal. This power is called active power, and yields the familiar (?) $\frac{1}{\sqrt{2}}$ term.

In inductors and capacitors, it is a different story, however. Since we are integrating a sine and cosine over one period, where the sine and cosine are orthogonal functions, this yields zero power! This can be interpreted in the case of the inductor as first storing power into the magnetic field and then retrieving this power again. This doesn't look like a problem, right?

In power networks, this is a problem! This power actually has to flow through the load lines, and then flows back again! We want to use $active$ power, not this "weird" power, which is called $reactive$ power.

The answer

Now to come back to the original question:

If we would express our phasors in terms of the voltages (so a current phasor in a capacitor would have a 90 degrees positive phase with respect to the voltage phasor, and a current phasor in an inductor would have a 90 degrees negative phase with respect to the voltage phasor), then we have that: (assuming phi is in the range (-180, 180] degrees

If $\phi = 0$ degrees or $\phi = 180$ degrees we have $active$ power.

If $\phi < 0$ degrees we have inductive reactive power. If $\phi > 0$ degrees we have capacitive reactive power.

In other words, the reactances of capacitors are $iX_c$ and those of inductors have $-iX_l$, where the i's deal with the positive/negative 90 degrees phase shift of the phasor.

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    $\begingroup$ $UI^*$ is the complex power $S = P + jQ$, i.e., $P = \mathcal{Re}\{UI^*\}$ $\endgroup$
    – Hal Hollis
    Commented Jun 12, 2017 at 19:59
  • $\begingroup$ @HalHollis Good point. I will update my answer. $\endgroup$
    – JBrouwer
    Commented Jun 12, 2017 at 20:03
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    $\begingroup$ Given the introductory level of the post "Note that all input signals can be written like this!" may be too terse. Every signal can be treated in terms of a linear combination of pure sinusoids, and the total result built up from the results on those simple problems but a reader who needs the answer you have written won't know that right off. $\endgroup$ Commented Jun 12, 2017 at 21:13
  • $\begingroup$ @dmckee While I agree with this, what I mean in my answer is in the general sense of a RLC circuit, where all voltages and currents can be written as a single sinusoid. $\endgroup$
    – JBrouwer
    Commented Jun 12, 2017 at 21:18
  • $\begingroup$ Hmmm. Perhaps just change to "All sinusoidal input signals ..." then. Which I think is a good match for the overall level of the post (a nice job, BTW). $\endgroup$ Commented Jun 12, 2017 at 21:19
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Could you elaborate further? I'll try and answer though:

These laws are equivalent to Ohm's law. It was observed that voltage and current ratio was constant in certain points, depending only on the material in which the current flows. We just extended that law to A.C. system and found these ratios had these values.

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I want to know the behaviour of I with ω and L.

If the voltage across an inductor is sinusoidal, e.g., $$v_L = V\cos (\omega t + \phi)$$

then, since $v_L = L\frac{di_L}{dt}$, the current through the inductor is also sinusoidal

$$i_L = \frac{V}{\omega L} \;\sin(\omega t + \phi)$$

The amplitudes of these voltage and current sinusoids are proportional and the proportionality factor is called the reactance of the inductor $X_L$:

$$X_L(\omega) = \omega L$$

Importantly, since the voltage and current of an inductor are in quadrature, an inductor stores energy during half of the sinusoidal cycle and then releases that stored energy during the other half; the source of emf does no net work on the inductor

Contrast with a resistor which does not store energy but rather converts the electrical energy to heat and so the source of emf does net work on the resistor.

This is what distinguishes a reactive component from a resistive; in AC steady state, no net work is done on a reactive component while net work is done on a resistive component.

Moving to the phasor (complex amplitude) domain, we have the notion of impedance which is a complex number that is the ratio of voltage and current phasors.

The impedance of an inductor is simply related to its reactance:

$$Z_L = jX_L = j \omega L,\qquad j \equiv\sqrt{-1}$$

You might wonder what the imaginary unit $j$ is for. The voltage and current sinusoids given above are in quadrature, i.e., have a relative phase of $90^\circ$, and multiplication by $j$ in the phasor domain is a $90^\circ$ phase shift in the time domain.

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Keep it simple. Reactance means a reaction rather than a resistance.

A Capacitance is an electric field that reacts to a changing voltage. The electric field is a result of and opposes a varying voltage.

An Inductance is a magnetic field that reacts to a changing current. The magnetic field is a result of and opposes a varying current.

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