Why can't we calculate capacitive reactance like u/i?

[Lotto]

TL;DR

When we have a capacitor in an AC circuit, we can write $i(t)=I_\mathrm m \sin {\left(\omega t+\frac{\pi}{2}\right)}$ and $u(t)=U_\mathrm m \sin{ \omega t}$. It should be truth that $u=X_C i$, but it isn't.

When I write $X_C=\frac ui=\frac{U_\mathrm m \sin \omega t}{I_\mathrm m \sin \left(\omega t+\frac{\pi}{2}\right)}$, it should be $\frac{U_\mathrm m}{I_\mathrm m}$, but it isn't. Why?

 

[vanhees71]

This works in "Fourier space", i.e., you use $u(t)=u_0 \exp(\mathrm{i} \omega t)$ and $i(t)=i_0 \exp(\mathrm{i} \omega t)$. Then the linear equations of circuit theory translate into algebraic equations, relating $i_0$ and $u_0$. The resistors, capacitances and inductances translate into complex-valued "impedances".

As an example take the real coil (i.e., the series of a resistance and an ideal inductance). The equation for this circuit is
$$R i + L \dot{i}=u$$
Plugging in the exponential ansatz for $u$ and $i$ leads to
$$(R + \mathrm{i} \omega L) i_0 =Z i_0= u_0.$$
As you see, you can use just the rules for resistances for the $Z$'s. Here it's a series of $Z_R=R$ and $Z_L=\mathrm{i} \omega L$. You can convince yourself easily that this works also for parallel circuits.

Take also the series of $R$ and a capacitance. The equation reads
$$\frac{Q}{C}+R i=u.$$
Since $i=\dot{Q}=\mathrm{i} \omega Q$ you have $Q=i/(\mathrm{i} \omega)$, and the equation reads
$$\left (R + \frac{1}{\mathrm{i} \omega C} \right ) i_0=u_0,$$
and thus for a capacitance you have to set
$$Z_C=\frac{1}{\mathrm{i} \omega C}.$$
From the $Z$'s you can calculate the relation between the amplitudes of $u$ and $i$ as well as the phase shift between $u$ and $i$.

All this of course refers to the stationary state, i.e., after you switched on the circuit for a sufficiently long time, after which the transients have damped out.

 

[Lotto]

But why doesn't it work? Why can't we use Ohm's law here? Where is the problem?

 

[vanhees71]

In a way it's a kind of Ohm's Law for the complex amplitudes with the impedances instead of resistors in DC circuit theory.

 

[Dale]

But why doesn't it work? Why can't we use Ohm's law here? Where is the problem?

I don’t understand the question. @vanhees71 showed how it does work. So why are you asking why it doesn’t work after he showed you that it does?

 

[vela]

TL;DR Summary: When we have a capacitor in an AC circuit, we can write $i(t)=I_\mathrm m \sin {\left(\omega t+\frac{\pi}{2}\right)}$ and $u(t)=U_\mathrm m \sin{ \omega t}$. It should be truth that $u=X_C i$, but it isn't.

The bolded part of your statement is wrong. The reactance $X_C=\frac{1}{\omega C}$ relates only the amplitudes to each other, i.e., $U_m = X_C I_m$.

The impedance $Z=\frac{1}{i\omega C}$ includes a phase factor which allows you to relate the current and voltage in the frequency domain using Ohm's law. That's what @vanhees described in his post.

 

[Baluncore]

The ratio of a sinewave, to the same sinewave shifted by Pi/2, is effectively sin/cos, which is a tan function of time.
It has zeros and asymptotes of ±∞, during each cycle.

X_c ; is a parameter independent of time, it varies with frequency.
u/i ; is a function of time.

The two sides may have the same dimensions, but being from different domains, cannot be equated without an explicit transform.

 

[vanhees71]

The trick with the impedances doesn't work with cos and sin but with the (complex) exponential function. This is a mathematical trick. You can take the real part as the physical quantities.

 

[gleem]

u/i is time-varying, as pointed out above, but you expected it to be a constant. You are trying to apply Ohm's law defined for DC to AC which is of course time varying. How can you explain the disappearance of the frequency dependence of X_c if u/i is supposed to equal U_M/I_M?

u/i =(U_M/I_M) tan(ωt).

Every time ωt =nπ/2 where n is an integer u/i → ∞ since the current at that instant =0 and every time ωt =nπ u/i = 0 since at that instant the current → ∞.

Ohm's law applied to AC works if you consider the effective voltage and current instead of the instantaneous. The effective voltage or current is the DC voltage or current that has the same energy dissipation effects as the actual AC voltage or current.

V_eff = V_M/√2 same with current.

So that V_eff =|Z| I_eff where |Z| is the magnitude of the complex impedance.

This should be explained in any elementary discussion of AC.

 

[vanhees71]

The ratio of a sinewave, to the same sinewave shifted by Pi/2, is effectively sin/cos, which is a tan function of time.
It has zeros and asymptotes of ±∞, during each cycle.

X_c ; is a parameter independent of time, it varies with frequency.
u/i ; is a function of time.

The two sides may have the same dimensions, but being from different domains, cannot be equated without an explicit transform.

There are no singularities in these solutions and no tan. In the stationary state $i$ and $u$ are both cos functions with different phases. The frequency is that of the external EMF. Take my first example with the "real coil", i.e., a resistance and and ideal inductance in series. Then for the complex amplitudes you get
$$u_0=Z i_0=(R+\mathrm{i} \omega L) i_0.$$
Setting $u(t)=u_0 \exp(\mathrm{i} \omega t)$ you get
$$i(t)=i_0 \exp(\mathrm{i} \omega t)=\frac{u_0}{R+\mathrm{i} \omega L} \exp(\mathrm{i} \omega t).$$
Now you can write
$$\frac{1}{Z}=\frac{1}{R+\mathrm{i} \omega L}=\frac{R-\mathrm{i} \omega L}{R^2+\omega^2 L^2}.$$
This you can write in "polar form"
$$\frac{1}{Z}=\frac{1}{|Z|} \exp(\mathrm{i} \varphi),$$
where
$$|Z|=\sqrt{R^2+\omega^2 L^2}, \quad \varphi=-\frac{R}{\sqrt{R^2+\omega^2 L^2}}.$$
This means that
$$i(t)=\frac{u_0}{\sqrt{R^2+\omega^2 L^2}} \exp[\mathrm{i} (\omega t + \varphi)].$$
Since $\varphi<0$ this implies that the current's phase is behind that of the EMF.

You can always think of the real parts of these complex emf's and currents being the physical quantities. Then the meaning is clear, i.e., in the above example
$$u_{\text{phys}}(t)=u_0 \cos(\omega t), \quad i_{\text{phys}}(t)=\frac{u_0}{\sqrt{R^2+\omega^2 L^2}} \cos(\omega t+\varphi),$$
where I have assumed $u_0 \in \mathbb{R}$ for simplicity.