Difference between RC and RRC circuits

[greg_rack]

TL;DR

I get the practical difference between the two circuits, but can't really understand why
(Pictures below)

Circuit A(RC) Circuit B(RRC)

Hi guys, the thing I don't get is why the maximum charge of the A capacitor is $q_{0}=C\cdot EMF$ whilst in the other case you must consider the drop in tension caused by the first resistor, so the maximum charge of the cap is $q_{0}=C\cdot (EMF-V_{R1})$.
Aren't the ends of the capacitor subject to the same potential difference in both cases? And since the charge is proportional to voltage and capacity, shouldn't it be the same in both cases?

 

[vanhees71]

I'm not sure, whether you talk about an AC or a DC circuit. From the drawings I guess it's DC.

Then you just need to use Kirchhoff's rules for time-independent fields. For circuit A there's no current and the capacitor voltage equals the applied voltage, i.e., $U_{\text{C}}=V_1=Q/C$ (with $Q$ the charge on the capacitor).

For circuit B Ohm's Law tells you that there's a current through the two resistors, $I=V_1/(R_1+R_2)$. Then taking the line integral of $\vec{E}$ along the loop containing the resistor tells you that $U_C=Q/C=I R_2=V_1 R_2(R_1+R_2)$.

 

[greg_rack]

I'm not sure, whether you talk about an AC or a DC circuit. From the drawings I guess it's DC.

Yup, it's DC... I'm sorry for the imprecision :)

Then you just need to use Kirchhoff's rules for time-independent fields. For circuit A there's no current and the capacitor voltage equals the applied voltage, i.e., $U_{\text{C}}=V_1=Q/C$ (with $Q$ the charge on the capacitor).

For circuit B Ohm's Law tells you that there's a current through the two resistors, $I=V_1/(R_1+R_2)$. Then taking the line integral of $\vec{E}$ along the loop containing the resistor tells you that $U_C=Q/C=I R_2=V_1 R_2(R_1+R_2)$.

Oh okay, got it!
So it's just a matter of considering the circuit in its "final stage" when the capacitor is fully charged:
in circuit A, there will be no current flowing, whilst in the other case, there will still be a current flowing through both resistors, causing a drop in the tension applied to the cap.
Am I right?

 

[vanhees71]

Yes, exactly. If it's considered an AC problem, you have to write down the corresponding differential equations and solve them for the given initial condition.

 

[sophiecentaur]

Yes, exactly. If it's considered an AC problem, you have to write down the corresponding differential equations and solve them for the given initial condition.

This is true if you want to get a deeper understanding of what's happening but using the concept of Impedance takes solving such problems into fairly simple algebra. Complex arithmetic is involved but it's nearly always a lot simpler than using Integration.

Ideal Capacitors (and Inductors) have Reactance, which depends on the frequency involved. The Impedance of a 'real' component consists of Resistance and Reactance
You use the formula
Z = R + jX. ( j =√(-1) )
Z's can be treated in the same way as R's. The results of Series and Parallel connections of Z's use the same formulae but you use Complex Arithmetic. It's not a trivial step to move from Real quantities to Complex quantities but it becomes second nature and there are even some Calculators that can do it for you.
@greg_rack do a search on the terms of Complex Impedance and you will find something of an appropriate level to suit your taste.

 

[vanhees71]

Well, yes, but impedances make sense for the long-term behavior of AC circuits after all transients have been damped out. They don't answer the question about the transient behavior describing, e.g., what happens shortly after switching on a DC circuit and how, in the long-time limit, it goes into the static result.

 

[sophiecentaur]

Well, yes, but impedances make sense for the long-term behavior of AC circuits after all transients have been damped out. They don't answer the question about the transient behavior describing, e.g., what happens shortly after switching on a DC circuit and how, in the long-time limit, it goes into the static result.

Which goes to prove that are no easy answers in EM.
In ‘real life’ I would think there are more AC based problems than step change problems. So both aspects would be relevant.

 

[anorlunda]

Well, yes, but impedances make sense for the long-term behavior of AC circuits after all transients have been damped out. They don't answer the question about the transient behavior describing, e.g., what happens shortly after switching on a DC circuit and how, in the long-time limit, it goes into the static result.

That's very true. I like to explain it as AC applies to sinusoidal signals. A true sinusoid is defined for $-\infty \leq t \leq \infty$ whereas most examples (as in this thread) consider only $u(t)sin(\omega t)$ where $u(t)$ is the unit step function. That is aperiodic.

 

[DaveE]

Well, yes, but impedances make sense for the long-term behavior of AC circuits after all transients have been damped out. They don't answer the question about the transient behavior describing, e.g., what happens shortly after switching on a DC circuit and how, in the long-time limit, it goes into the static result.

Unless you prefer Laplace transforms...

 

[sophiecentaur]

consider only u(t)sin(ωt) where u(t) is the unit step function.

This is true in nearly every real life situation but, as long as (in your example) ω is significantly greater than unity, we can neglect u(t). We rarely look at the clock when we turn on the equipment and make a note of how long the experiment has run for. But this u(t) factor can be really confusing for people making a first stab at the Fourier transform. We all use DFTs, which make big assumptions and which produce a non-continuous result - which we accept and gloss over any problems.

But I fear we are going a bit off topic.

 

[vanhees71]

Unless you prefer Laplace transforms...

Sure, that's the standard method to solve the corresponding initial-value problems (at least for engineers; physicists don't use Laplace transforms as often though it's indeed more convenient than Fourier transforms for initial-value problems).

 

[anorlunda]

It's hard to find words that won't confuse someone. I especially don't like the use of "static solution", and I don 't like "transients die out".

We use AC analysis to study AC systems, both in the steady state and during transients on the power grid. It's an approximation of course because the amplitude and phase are not constant for $-\infty < t < \infty$. So I'm trying to avoid the self contradiction of "We study transients in the power system using AC analysis that is valid only when the transients have died out." There are two kinds of transients in that same sentence. Similarly, there are multiple meanings for "static solution"

Power engineers use the phrase "switching surges" to model the kinds of transients that happen just after the switch is closed, or when lightning strikes. We use Maxwell's Equations, not DC analysis to do that.

So yes, there are a number of ambiguities in nomenclature depending on context.

 

[vanhees71]

It's not that complicated. AC circuit theory is based on the quasistationary approximation(s) of Maxwell's Equations. They are valid for "compact" circuits, i.e., where the spatial dimensions of the circuit are small compared to the wave-length $\lambda=c/f=2\pi c/\omega$, where $f$ is the frequency (or a typical scale of frequencies) of the AC.

The result are ordinary linear differential equations with constant coefficients. Typically they are of a damped-harmonic oscillator type. For the usual house-hold-current application you have a harmonic source, i.e., something like
$$\ddot{f} + 2 \gamma \dot{f} +\omega_0^2 f=A \exp(-\mathrm{i} \omega t).$$
The typical solution is of the form
$$f(t)=C_1 f_{1 \text{hom}}(t) + C_2 f_{2 \text{hom}}(t) + D \exp(-\mathrm{i} \omega t),$$
where $C_1$ and $C_2$ are indetermined constants and $f_{1/2 \text{hom}}$ are two linearly independent solutions of the homogeneous equation (i.e., the ODE with A=0) and the final term, harmonic with the frequency of the driving source, a particular solution of the inhomogeneous equation with a determined value of $D$. The integration constants $C_1$ and $C_2$ are determined by the initial conditions, $f(0)=f_0$, $\dot{f}(0)=v_0$.

If $\gamma>0$ (which is the usual case, because in the real world there's always some dissipation/ohmic losses) then the $f_{1/2\text{hom}}(t) \propto \exp(-\gamma t)$, i.e., falling off with a "relaxation time", $\tau =1/\gamma$. After this "life-time of the transients" $f$ is in the "steady state", i.e., oscillating harmonically with the frequency imposed by the external source.

For this latter case you can determine $D$ in terms of "impedances", i.e., complex quantities similar to resistances $R$ in DC and obeying the same rules for parallel and series connections as resistances in DC circuit theory. For the electrician's practical purpose usually that's all he needs to know.

 

[alan123hk]

For similar circuits, Thevenin's theorem can be used to simplify the calculation of the transient response of the voltage across the capacitor.