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The image attached is a first-order circuit because the two branches of the circuit are uncoupled, but I'm struggling to show that mathematically. From the diagram, we can immediately write two equations from KVL (each loop containing the independent source) that are first order ODEs in the two different capacitors. But I can't think of a relation between the two capacitors.

If anyone can explicitly show why this must be a first-order circuit I would really appreciate it.enter image description here

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  • \$\begingroup\$ What if Vs is time varying? \$\endgroup\$
    – DKNguyen
    Apr 10 '20 at 19:19
  • \$\begingroup\$ I suppose it could be, but does that matter? The order of the circuit doesn't depend on the independent sources. \$\endgroup\$
    – 1729_SR
    Apr 10 '20 at 19:21
  • \$\begingroup\$ Have you combined bot equations to try solving it? You should end up with a second derivative somewhere. I don't have a pencil with me. I think. It's been a long time. \$\endgroup\$
    – DKNguyen
    Apr 10 '20 at 19:37
  • \$\begingroup\$ According to my lecture notes this is a first-order circuit and so no second derivative should appear. That's essentially the crux of my question - why is that so? Because, like you, my first instinct would have been to say that this is second-order. \$\endgroup\$
    – 1729_SR
    Apr 10 '20 at 19:39
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You've defined the circuit, but not the output. Are you looking at, for instance, the voltage across the 1 F cap? Let's assume so. Since your voltage source has zero impedance, the voltage across either capacitor (and you need to pick one point) will be independent of the existence (or lack of same) of the other RC pair.

So the response at either capacitor will a first-order response. In order to calculate it you can remove the other RC, with no effect on your results.

EDIT - OP has asked me to flesh out this answer, so let me try.

Let's assume (just for fun) that Vs has a value of 1 volt. By convention, voltage sources are ideal sources. That is, a 1-volt source will put put 1 volt regardless of the current required.

Now, connect the 4 ohm/.5 F RC network. What is the output of Vs? 1 volt.

Now connect the 4 ohm/1 F network. What is the output of Vs? 1 volt.

So the voltage produced at either capacitor will be independent of the value (or even the existence) of the other capacitor.

Now, about "zero impedance". Vs is shown as a voltage source, able to supply any arbitrary current. If you connect the two outputs together with a 0 ohm resistor, you'll get infinite current. What if, instead of an ideal source, it "really" consists of a 1 volt ideal source in series with a 1 ohm resistor? This is what an output impedance of 1 ohm means. Then shorting the output will result in 1 amp, which is much more in line with real voltage sources such as batteries.

Now consider what happens when we do the connection experiment I mentioned earlier. Just for the sake of illustration, get rid of the capacitors.

If you connect a single 4 ohm resistor across the output, the voltage source will 1 ohm in series with 4 ohms, for a total of 5 ohms, and an output current of 0.2 amps. Ohm's Law will tell you that the voltage across the 4 ohm resistor will be 0.8 volts.

Now add a second 4 ohm resistor across the output. Effectively, this will produce a 2 ohm load. The voltage source will see 1 ohms plus 2 ohms, and produce 0.333 amps of current, and the voltage across the load will be 0.667 volts - not 0.8.

So, the output impedance of a power supply will affect the voltage delivered to a a load - but if the output impedance is zero, the voltage at the load will be independent of the value of the load.

I hope this helps.

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  • \$\begingroup\$ Can you flesh this out a little more. "Since the voltage source has zero impedance"...why does this have any bearing on the question? I'm hoping you can explicitly walk me through the steps that led you to the conclusion that these two branches are decoupled. \$\endgroup\$
    – 1729_SR
    Apr 10 '20 at 20:42
  • \$\begingroup\$ @1729_SR If the voltage source has zero impedance, than the two capacitors just see Vs. They could equally be connected to two different power supplies that happened to have the same voltage. Since you're expecting them to interact, maybe you drew the circuit differently than you meant? \$\endgroup\$ Apr 10 '20 at 21:52
  • \$\begingroup\$ @1729_SR - I've edited the answer to address your question. I hope it helps. \$\endgroup\$ Apr 11 '20 at 12:44
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There is no relationship between the capacitors in your circuit. The two branches are in parallel with a voltage source. Their behavior is independent. Here are the KCL equations:

$$\frac{V_S - V_{C1}}{4\Omega} = 0.5\mathrm{F}\cdot\frac{dV_{C1}}{dt}$$ $$\frac{V_S - V_{C2}}{4\Omega} = 1\mathrm{F}\cdot\frac{dV_{C2}}{dt}$$

Note that these are uncoupled equations -- we can solve them separately. Now, look at this circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

Here are the KCL equations:

$$\frac{V_{S} - V_{C1}}{R_1} = C_1\frac{dV_{C1}}{dt} + \frac{V_{C1} - V_{C2}}{R_2}$$ $$\frac{V_{C1} - V_{C2}}{R_2} = C_2\frac{dV_{C2}}{dt}$$

These equations share the \$({V_{C1} - V_{C2}})/{R_2}\$ term, which means we can't solve them separately. To solve this system, you would start by solving for \$V_{C1}\$ in the second equation:

$$V_{C1} = V_{C2} + R_2C_2\frac{dV_{C2}}{dt}$$

and plugging that into \$V_{C1}\$ in the first equation. But the first equation contains \${dV_{C1}}/{dt}\$! When we plug in our formula for \$V_{C1}\$, we also have to use its derivative, which gives us the second derivative of \$V_{C2}\$:

$$\frac{dV_{C1}}{dt} = \frac{dV_{C2}}{dt} + R_2C_2\frac{d^2V_{C2}}{dt^2}$$

That's why it's a second-order circuit, while your circuit (whose equations are uncoupled) is not.

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The order of the circuit? That concept must be agreed before the case can be solved.

One definition: It's first order circuit if you can get all currents and voltages with any initial conditions by solving only 1st order scalar differential equations. The "scalar" limitation is because one can build formally a state variable vector equation of a complex LC circuit with matrices and the 1st order derivative of the state variable vector.

In your circuit capacitor voltages V1 and V2 obey equations dV1/dt=(Vs-V1)/(R1C1) and dV2/dt=(Vs-V2)/(R2C2). Both of these can be solved separately if Vs and the initial value of the capacitor voltage are known. The currents can be calculated from the voltages and resistances.

Actually the differential equations of V1 and V2 together are a state variable vector equation, but solving it as one state variable equations is possible without generating a higher order equation.

If it happens that Vs isn't stiff, but drops more or less due the current, the independecies of the branches vanish and the circuit is of 2nd order.

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  • \$\begingroup\$ What do you mean by "[if] Vs isn't stiff"? Do you mean Vs being a real source with some internal resistance? If it is just about internal resistance of the supply it will not change the order of the ODE to be solved. \$\endgroup\$
    – jDAQ
    Apr 10 '20 at 20:35
  • \$\begingroup\$ Can you expand on this? I see how I have two separate first-order scalar ODEs, but I don't see how they can be solved without 2 initial conditions. How can one initial condition suffice to solve this circuit? \$\endgroup\$
    – 1729_SR
    Apr 10 '20 at 20:37
  • \$\begingroup\$ The number of initial conditions being two does not mean a system will be a second order system, since the highest order derivative will still be one. \$\endgroup\$
    – jDAQ
    Apr 10 '20 at 20:42
  • \$\begingroup\$ @1729_SR I didn't claim anything about the number of initial conditions. 2 is needed like in a circuit where is 2 capacitors in series. I claimed only that there's no 2nd order diff.eq when one state variable is solved at a time. \$\endgroup\$
    – user287001
    Apr 10 '20 at 20:44
  • \$\begingroup\$ So to be crystal clear, this is a first-order system that requires two initial conditions? \$\endgroup\$
    – 1729_SR
    Apr 10 '20 at 20:49
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To my thinking, this is a second order circuit. It's just a special case where the coefficient of the second derivative in the combined ODE happens to be zero, because the state variables don't influence each other. You can see this if (as suggested above) you introduce coupling via a resistor in series with your source, and then look at what happens as that resistance approaches zero.

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It is second order system. You can look at the impedance fuction, Z(s)=V(s)/I(s) which is s^2. Also, order of system is equal to "independent energy storage elements" in that system. This is because each independent energy storage element is associated with one state variable. In the above ckt there are two capacitors which can not be replaced by single equivalent capacitor, hence order is 2.

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Question: If we want to characterize a CIRCUIT, is it correct to ask for the ORDER of a circuit? Can a circuit have an order?

To me, it is more appropriate to analyze a specific transfer function derived fronm the circuit.

For example - asking for the current through each branch or asking for the voltage across one of the capacitors we have, of course, a 1st-order equation (lowpass).

On the other hand, because the total conductance (or the total impedanze Z1||Z2) is of second order (see the answer from "a concerned citizen") the expression of the total current through the circuit will be a 2nd-order expression.

EDIT: Clear and descriptive example:

In some real cases, we have a signal voltage source, which drives at the same time a lowpass and a highpass. Let´s say each of second order.

Would you say that we have one single circuit of 4th order? No - of course, not. Again - a CIRCUIT cannot have a specific order - it is a function derived from this circuit which is described by the order of this function (input resistance, transfer function,..)

Of course, the situation is completely different, when the signal source has an internal source resistance. In this case, both filters are not isolated from each other because the current into one circuit determines the voltage drop across the source resistance and, hence, influences the input voltage for the other circuit.

Fazit: It is not the circuit, but a specific variable or function which has to be analyzed while asking for the order.

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  • \$\begingroup\$ We can find the currents for each branch solving 1st-order equation. So how can the total current be 2nd-order? Total current is sum of them. \$\endgroup\$
    – S.H.W
    Apr 12 '20 at 1:44
  • \$\begingroup\$ When you add both expressions for the currents and apply some math rules to get one expression only, the common denominator is of second order. \$\endgroup\$
    – LvW
    Apr 12 '20 at 10:03
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You seem very sure of yourself that this is a first order circuit. Let's see, maybe presuming is not such a good idea:

$$\begin{align} Z_1 &= R_1 + \frac{1}{sC_1} = \frac{sC_1R_1 + 1}{sC_1}\\ Z_2 &= R_2 + \frac{1}{sC_2} = \frac{sC_2R_2 + 1}{sC_2}\\ Z_1||Z_2 &= \frac{1}{\frac{1}{Z_1}+\frac{1}{Z_2}} = \frac{C_1C_2R_1R_2s^2 + (C_1R_1 + C_2R_2)s + 1}{C_1C_2(R_1+R_2)s^2 + (C_1 + C_2)s} \end{align}$$

The only time this becomes a 1st order is when both resistors and both capacitors are equal. In general, the order of the circuit is given by the number of reactive elements: two capacitors, 2nd order.

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This is a first order equation. It may be simpler to explain if a Fourier or Laplace transform is applied. Once this is done then combining the two in parallel makes it easily apparent that this is a first order circuit. The attached picture demonstrates the math.enter image description here

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The order of the circuit depends on the no. of "effective" storing elements. By the term effective we mean that those elements (inductor or capacitor) which can not be further separated.

As in the given circuit there are 2 capacitors..but we can always solve the two parallel RC branches which will given an equivalent single RC branch..

So basically ckt will consist of 1) source Vs in series with an equivalent resistance Req and equivalent Capacitor Ceq

Hence overall the "effective" storing element is 1. and hence the order is 1.

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  • \$\begingroup\$ I'd call that the order of the transfer function from voltage input to voltage output. The circuit as a whole has an order of 2. \$\endgroup\$
    – John Doty
    Apr 12 '20 at 15:33

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