RLC circuit - transient event

Discussion in 'Homework Help' started by manoh, Jan 1, 2013.

  1. manoh

    Thread Starter New Member

    Jan 1, 2013
    Hi, guys.

    I've actually done what I had to do, but I have a somewhat easy, fundamental question for you (mind you, I'm not good at Electronics, so please bear with me).

    I'm using PSpice to run a simulation on the following circuit:


    (A curious thing is that the original circuit of my homework had 2 extra resistors - one above the inductor, and one above the capacitor - but their resistance was 0. I decided not to include them here, since PSpice doesn't allow a circuit to have a resistor with no resistance... Plus, a resistor with 0 resistance sounds silly to me.)

    Here we have a circuit with a source of direct current (100V). We have a switch as well, which is initially opened and then closes. I've placed two markers - one to mark the current of the inductor, one to mark the voltage of the capacitor.

    The capacitor is initially charged to 100V (that's how I interpret what my homework sheet says: Uc(0-)=100V). That's why I've set the IC variable of the capacitor to 100. I've calculated the initial current through the inductor by dividing the input voltage (100V) by the resistance R1 (0.1k) - which results in 10A initial inductor current. That's why I've set the IC variable of the inductor to 10.

    What I wanted to do in PSpice is visualize the transient event itself - I want to see what happens to the iductor current and the capacitor voltage when the switch closes.

    Here's what I got...

    Inductor current: click, large image.

    Hmm, starts at 10 Amps and gradually sets itself to something below zero, I guess that's acceptable. What bothers me is the...

    Capacitor voltage: click, large image.

    Starts at -100V (I guess I can fix that by setting the IC variable of the capacitor to -100V), jumps up to 700V, then silences itself forever with a voltage of 0.

    My question is... is this normal? Can a 5uF capacitor stand a voltage of 700V? Or am I doing something wrong?

    I am by no means certain in the correctness of the things I did, nor my own scarce knowledge of electrical circuits. I'd be very happy if someone shed some light on me.

    P.S. I have yet to go deeper into this assignment, since I have to calculate the current and voltage analytically... But I want to be sure I'm on the right path here.
  2. WBahn


    Mar 31, 2012
    The first thing to do is ask if the basic parts of the answer make sense.

    You've calculated the steady state current through the inductor and got a value of 10A. You then ran a simulation that forced the initial current in the inductor to match that value and let the simulation run until it reached steady state again. In steady state, a capacitor looks like an open circuit, meaning that after a long time the circuit will look like the switch is open again and, therefore, the inductor should settle to the same current that it started at.

    Does it?

    It looks like it settles to -1A.

    Figure this out and you will spot a math error and and spot a misconception regarding the polarity of initial conditions.

    But, most importantly, you should learn something about the value of always asking if the answer makes sense in as many ways you can.
    manoh likes this.
  3. manoh

    Thread Starter New Member

    Jan 1, 2013
    Yes, I see what you mean!

    The steady state current through the inductor is, of course, 1A.

    As for the polarity of the initial conditions - I completely ignored the fact that the capacitor and inductor have polarities, meaning that the initial current through the inductor, for example, flows from its + pin to its - pin. In the schematic of the circuit (as shown in the first post), I had no idea which pin is which.

    The simulation proves that I got the polarities wrong and I should have either set the initial conditions to -1A and -100V, or flipped the elements themselves in the circuit.

    I did the second, and here's what I got:

    Inductor current: click.

    Starts with a current of 1A, jumps a bit during the transient event, then settles at 1A again, just like it should. Nice!

    Capacitor voltage: click.

    Now THIS makes sense! Initially charged to 100V, the capacitor discharges after the switch is closed. What bothers me is that it's not a steady discharge from 100V to 0V - at one point, there's a drop below 0V... (If memory serves, the discharging capacitor graphs in my textbooks never go below 0V. My circuit is more complex than the ones in the textbooks, having a DC source and all that, a drop below 0V might be normal?)

    Thanks for all your help, WBahn!

    P.S. Ehhh, I've got a looot of reading to do to fully understand these circuits... But textbooks never suffice. I'm glad I found this place, where people gently push you towards the right direction, making you think about your mistakes.

    P.P.S. I'll soon start calculating this circuit analytically... some questions may surface - in that case, I might as well use this same thread for asking them.
  4. WBahn


    Mar 31, 2012
    It's not that inductors and capacitors are polarized; most of them aren''t (polarized capacitors have to do with the electrochemistry involved and not the capacitance itself). But current and voltage are both intrinsically polarized quantities. It's not enough to say that the current in a resistor is 10A, you must also indicate which direction that current is flowing. The same with voltage.

    If you change the polarity of one of the initial conditions in your problem and not the other, you will see the voltage on the capacitor initially increase in magnitude.

    To see what is happening, look at the conditions in the circuit just after the switch closes. Before the switch closes, the inductor has 1A of current through it (in a particular direction) and the capacitor has 100V across it (again, in a particular direction). When the switch closes, conservation of energy requires that neither the inductor current nor the capacitor voltage can change instantaneously, but there is no such restriction on inductor voltage and capacitor current. Thus, as soon as the switch changes, the voltage across the inductor changes to 100V. This causes an instantaneous change in the rate at which the current in the inductor changes, but does not cause an instantaneous change in the current itself; the current may start to increase or may start to decrease, depending on the relative polarities. Similarly, depending on the polarity of the capacitor voltage, the resistor in the circuit either has 0V across it or 200V across it, but in either case you can calculate the current through it and, since you know the inductor current, can use KCL to calculate the capacitor current. Depending on the relative polarities of things, this current will either be discharging the capacitor or charging it even more, with the voltage across it starting to change accordingly.

    As the circuit settles, it is entirely possible that the inductor will force enough of a change in the capacitor charge to make it's voltage change polarity. It's even possible it might oscillate for a while, but eventually the presence of the voltage source and the resistor will dampen the response out leaving only the steady state solution.
  5. bountyhunter

    Well-Known Member

    Sep 7, 2009
    The reason those were there is because all real world caps and inductors have some resistance.
  6. manoh

    Thread Starter New Member

    Jan 1, 2013
    Yes, I see. So even though the resistors have 0 resistance, they're put there to symbolize that real inductors and capacitors have resistance.

    By the way, guys, an update on my problem solving. It turned out I had to calculate and plot the current through the capacitor and the inductor voltage. I spent a few hours today solving matrices and differential equations, and I think I got the correct solutions.

    I calculated the equations of the capacitor current and inductor voltage, I plotted them into PSpice and what do you know, the graphs of the equations I calculated analytically were almost the same as the simulated ones (with an acceptable margin of error). So I guess that's problem solved for me!

    Many thanks to WBahn, I don't think I would've done it without you.
  7. WBahn


    Mar 31, 2012
    You're more than welcome. Hope to see you around the boards some more.