Circuit Analysis – First-Order Circuit Resembling Second-Order

circuit analysis

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

Best Answer

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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