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Time variant RC and RL circuits

  1. Mar 29, 2012 #1
    Ok so I have some topic overview questions. I think I am on the right track but would like to make sure I am thinking about this the correct way.

    When I say time variant I simply mean there is a simple switch in the circuit. This is link to a were I found my supplemental material and it has a switch like I am talking about. http://www.allaboutcircuits.com/vol_1/chpt_16/2.html

    Question 1:
    Under what conditions does a Capacitor act like (A) an open circuit? (B) a short circuit? (C) a voltage source?

    Question 2:
    Under what conditions does an Inductor act like (A) a short circuit? (B) an open circuit? (C) a current source?

    MY ANSWERS:
    1a) at t=0-
    because initially fully discharged thus empty and open circuit

    1b) at t=0+
    because the V across would be 0 thus short circuit

    1c) at ∞
    because capacitors remember/store voltage thus voltage source

    2a) at t=0-
    because initially like nothing is there thus short circuit

    2b) at t=0+
    because the I across would be 0 thus an open circuit

    2c) at ∞
    because inductors remember all current sources thus would act like a current source

    SOME ADDITIONAL QUESTIONS:
    When we look at initial-final response equations:
    y(t)=yF + (yi - yF)e-t/τ
    y=v,i as asked fot
    yi=y(0+)
    yF=y(∞)
    τ= RC or L/R depending on the circuit

    3) How do you find y(0+) for a capacitor?
    4) How do you find y(0+) for current through an inductor?
    5) How do you find y(0+) for voltage on a resistor?
    6) How do you find y(∞)?


    I do not know the answer to 3-6. I am obviously kind of confused when it comes to the addition of the on/off switch into the circuit network.

    HELP on the 0+, 0- and ∞ times would be great!
     
    Last edited by a moderator: Mar 29, 2012
  2. jcsd
  3. Mar 29, 2012 #2

    psparky

    User Avatar
    Gold Member

    Simple my friend.......
    Resistance or reactance of a capacitor is 1/JWC
    Reactance of an inductor is JWL
    Current thru a capacitor is C*dv/dt=i(t)
    Voltage across an inductor is L*di/dt=v(t)

    W= radians per second.....or 2*pi*frequency.

    Draw a circuit with a battery source and a capacitor and resistor in series.. Find the current at steady state.......W=0 in a battery.....so you have infinite resistance across capacitor at steady state. Voltage across cap will equal battery since there is no voltage drop across resistor.

    Draw same circuit with inductor and resistor in series. W=0 in battery........zero resistance....short circuit. Zero voltage across inductor....voltage across resistor will equal your battery.

    Draw cap circuit now with AC source (leave out resistor for ease of learning) Plug in your frequency into 1/JWC.....use V=IR to find current. Record your answer.

    Draw inductor circuit with AC source. (leave out resistor for ease of learning) Plug in your frequency into JWL.....use V=IR to find current. Record your answer.

    Now draw the same cap and inductor circuits again with an AC source....try 170sin(377t) Now use C*dv/dt=it and L*di/dt=v(t). IN other words take the derivative of that voltage source....You should get the exact same answers as the ones you recorded.

    To gain a clear understanding of what you are doing, you need to absorb what I just wrote. If you get my drift....all your above questions will be answered. Understanding steady state is key.....then the transient part is just a matter of plugging and chugging into the transient formulas available to you.
     
    Last edited: Mar 29, 2012
  4. Mar 29, 2012 #3
    00330.png

    for that circuit, imagine that the capacitor is fully discharged.
    Q=VC says that the voltage across a discharged capacitor is zero, and since the switch is open no current flows, so no voltage across R. when the switch is closed, at t=0 theres now 15V across R and C, but no charge has built up in C, so it must all be across R.
    its at this point that C acts like a short circuit, and I flows through R to satisfy ohms law.

    that current charges C, so over time a V builds across the capacitor. note that this voltage counteracts the supply voltage, and V across R is reduced, consequently I is reduced, and the rate that the capacitor charges is reduced. this is the i(t) = C*dv/dt differential equation.

    after a long time I is (almost) 0. this happens when the capacitor has enough charge that
    the voltage matches the supply voltage. at this point, there is no voltage across the resistor, so no current flows. at this point the Capacitor acts like an open circuit.
     
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