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Circuit question - RLC

  1. Apr 14, 2013 #1
    1. The problem statement, all variables and given/known data

    https://imageshack.us/scaled/large/708/capturehdx.png [Broken]

    2. Relevant equations

    https://imageshack.us/scaled/large/845/capturevtq.png [Broken]

    V = IR

    https://imageshack.us/scaled/large/18/captureid.png [Broken]

    [itex]I_{C}(t) = C\frac{dV_{x}(t)}{dt}[/itex]

    https://imageshack.us/scaled/large/818/captureplr.png [Broken]

    [itex]V_{x}(t) = L\frac{di_{L}}{dt}[/itex]

    3. The attempt at a solution

    I'll start by solving for the equation of the node voltage [itex]V_{x}(t)[/itex]

    [itex]\frac{V_{s} - V_{x}(t)}{R} = C_{1}\frac{d[V_{x}(t) - V_{s}]}{dt} + C_{2}\frac{d[V_{x} - 0]}{dt} + \frac{1}{L}∫_{0}^{t}[V_{x}(t) - 0]dt + I_{L}(t=0^{+}) [/itex]

    I use the fact that [itex]I_{L}(t=0^{-}) = 0 A[/itex] and that because the current through a inductor can't change instantaneously so [itex]I_{L}(t=0^{-}) = I_{L}(t=0^{+}) = 0 A[/itex]. I also simplify some mathematically.

    [itex]-\frac{V_{x}(t)}{R} = C_{1}\frac{dV_{x}(t)}{dt} + C_{2}\frac{dV_{x}}{dt} + \frac{1}{L}∫_{0}^{t}V_{x}(t)dt[/itex]

    Simplify some more

    [itex]-\frac{V_{x}(t)}{R} = (C_{1} + C_{2})\frac{dV_{x}(t)}{dt} + \frac{1}{L}∫_{0}^{t}V_{x}(t)dt[/itex]

    Take the derivative with respect to time of both sides

    [itex]-\frac{d}{dt}\frac{V_{x}(t)}{R} = \frac{d[(C_{1} + C_{2})\frac{dV_{x}(t)}{dt} + \frac{1}{L}∫_{0}^{t}V_{x}(t)dt]}{dt}[/itex]

    Simplify

    [itex]-\frac{1}{R}\frac{V_{x}(t)}{dt} = (C_{1} + C_{2})\frac{d^{2}V_{x}(t)}{dt^{2}} + \frac{1}{L}V_{x}(t)[/itex]

    Set equal to zero

    [itex]0 = (C_{1} + C_{2})\frac{d^{2}V_{x}(t)}{dt^{2}} + \frac{1}{R}\frac{V_{x}(t)}{dt} + \frac{1}{L}V_{x}(t)[/itex]

    Replace the constants with constants A, B, and C

    [itex]A = C_{1} + C_{2}[/itex]
    [itex]B = \frac{1}{R}[/itex]
    [itex]C = \frac{1}{L}[/itex]

    [itex]0 = A\frac{d^{2}V_{x}(t)}{dt^{2}} + B\frac{V_{x}(t)}{dt} + CV_{x}(t)[/itex]

    This is a second order differential equation so it is known that

    [itex]V_{x}(t) = Ke^{st}, \frac{dV_{x}(t)}{dt} = Kse^{st}, \frac{d^{2}V_{x}(t)}{dt^{2}} = Ks^{2}e^{st}[/itex]

    So plugging this into the equation

    [itex]0 = AKs^{2}e^{st} + BKse^{st} + CKe^{st}[/itex]

    Divide through by [itex]ke^{st}[/itex]

    [itex] 0 = As^{2} + Bs + C[/itex]

    This is a quadratic equation so it's known that

    [itex]s = \frac{-B ± \sqrt{B^2 - 4AC}}{2A}[/itex]

    It can be seen from the equation for s that there are two values for s, I'll call them [itex]s_{1}[/itex] and [itex]s_{2}[/itex]. So my voltage equation becomes

    [itex]V_{x}(t) = A_{1}e^{s_{1}t} + A_{2}e^{s_{2}t}[/itex]

    In order to solve for [itex]A_{1}[/itex] and [itex]A_{2}[/itex] I need to know the initial condition of the node voltage.

    At time [itex]0^{-}[/itex] the circuit looks like the one below

    https://imageshack.us/scaled/large/833/capturetiru.png [Broken]

    [itex]\sum Q = 0 = 4 uF(V_{x}(t = 0^{-}) - 10) + 1 uF(V_{x}(t = 0^{-}) - 0)[/itex]

    Divide through by uF

    [itex]0 = 4V_{x}(t = 0^{-}) - 4(10) + V_{x}(t = 0^{-})[/itex]

    simplify

    [itex]40 = 5V_{x}(t = 0^{-})[/itex]

    so

    [itex]V_{x}(t = 0^{-}) = 8 V[/itex]

    I use the fact that the voltage across a capacitor can't change instantaneously so [itex]V_{x}(t = 0^{-}) = V_{x}(t = 0^{+})[/itex].

    So when you plug in 0 into the voltage equation you get the following equation

    [itex]V_{x}(t = 0^{+}) = A_{1} + A_{2}[/itex]

    Being that I have two unknowns I need another equation. I use the fact that the voltage through a capacitor can't change instantaneously

    So it can be seen that [itex]\frac{dV_{x}(t)}{dt}[/itex] represents the change in voltage across [itex]C_{2}[/itex] which is zero. So I have the following second equation

    [itex]0 = A_{1}s_{1} + A_{2}s_{2}[/itex]

    So I choose to solve for [itex]A_{1}[/itex] and [itex]A_{2}[/itex] using matrix and putting them in reduce row echelon form

    1, 1, [itex]V_{x}(t = 0^{+})[/itex]
    [itex]s_{1}, s_{2}, 0[/itex]

    I divide row 2 by [itex]s_{1}[/itex]

    1, 1, [itex]V_{x}(t = 0^{+})[/itex]
    [itex]1, \frac{s_{2}}{s_{2}}, 0[/itex]

    I multiply row 1 by negative one and add it to row two to get a new row 2

    1, 1, [itex]V_{x}(t = 0^{+})[/itex]
    [itex]0, \frac{s_{2}}{s_{2}}-1, -V_{x}(t = 0^{+})[/itex]

    I divide row two through by [itex]\frac{1}{\frac{s_{2}}{s_{2}}-1}[/itex]

    1, 1, [itex]V_{x}(t = 0^{+})[/itex]
    [itex]0, 1, -\frac{V_{x}(t = 0^{+})}{\frac{s_{2}}{s_{2}}-1}[/itex]

    I multiply row two by negative one and add to row one to get a new row one

    1, 0, [itex]V_{x}(t = 0^{+}) + \frac{V_{x}(t = 0^{+})}{\frac{s_{2}}{s_{2}}-1}[/itex]
    [itex]0, 1, -\frac{V_{x}(t = 0^{+})}{\frac{s_{2}}{s_{2}}-1}[/itex]

    So I get the following equations, I factored the first one

    [itex]A_{1} = V_{x}(t = 0^{+})(1 + \frac{1}{\frac{s_{2}}{s_{2}}-1})[/itex]

    [itex]A_{2} = -\frac{V_{x}(t = 0^{+})}{\frac{s_{2}}{s_{2}}-1}[/itex]

    The quantity [itex]\frac{s_{1}}{s_{2}}[/itex] seems to be an interesting quantity.

    [itex]\frac{s_{1}}{s_{2}} = \frac{-B - \sqrt{B^{2} - 4AC}}{-B + \sqrt{B^2 - 4AC}}[/itex]

    I multiply by a form of one [itex]\frac{-B - \sqrt{B^{2} - 4 AC}}{-B - \sqrt{B^{2} - 4 AC}}[/itex] and get the following.

    [itex]\frac{s_{1}}{s_{2}} = \frac{(B + \sqrt{B^{2} - 4AC})^{2}}{B^{2} - B^{2} + 4AC} = \frac{B^{2} + 2B\sqrt{B^{2} - 4AC} + B^{2} - 4AC}{4AC} = \frac{2B^{2} + 2B\sqrt{B^{2} - 4AC} - 4AC}{4AC} = \frac{B^{2}}{2AC} + \frac{B}{2AC}\sqrt{B^{2} - 4AC} - 1[/itex]

    So then

    [itex]\frac{s_{2}}{s_{1}} - 1 = \frac{B}{2AC}\sqrt{B^{2} - 4AC} + \frac{B^{2}}{2AC} - 2[/itex]

    Now in my formulas I have the following relation [itex]\frac{1}{\frac{s_{2}}{s_{1}}-1}[/itex] So I solve for this relation.

    [itex]\frac{1}{\frac{s_{2}}{s_{1}}-1} = \frac{1}{\frac{B}{2AC}\sqrt{B^{2} - 4AC} +( \frac{B^{2}}{2AC} - 2)}[/itex]

    I multiply by a form of one [itex]\frac{-\frac{B}{2AC}\sqrt{B^{2}-4AC} + \frac{B^2}{2AC} - 2}{-\frac{B}{2AC}\sqrt{B^{2}-4AC} + \frac{B^2}{2AC} - 2}[/itex]

    [itex]\frac{1}{\frac{s_{2}}{s_{1}}-1} = \frac{-\frac{B}{2AC}\sqrt{B^{2}-4AC} + \frac{B^2}{2AC} - 2}{-\frac{B^{2}}{4A^{2}C^{2}}(B^{2} - 4AC) + (\frac{B^{2}}{2AC} - 2)^{2}} = \frac{-\frac{B}{2AC}\sqrt{B^{2}-4AC} + \frac{B^2}{2AC} - 2}{-\frac{B^{4}}{4A^{2}C^{2}} + \frac{B^{2}}{AC} + \frac{B^{4}}{4A^{2}C^{2}} - \frac{4B^{2}}{2AC} + 4} = \frac{-\frac{B}{2AC}\sqrt{B^{2}-4AC} + \frac{B^{2}}{2AC} - 2}{-\frac{B^2}{AC} + 4}[/itex]

    At this point I multiply through by a form of one [itex]\frac{AC}{AC}[/itex]

    [itex]\frac{-\frac{B}{2}\sqrt{B^{2} - 4AC} + \frac{B^{2}}{2}-2AC}{4AC - B^{2}}[/itex]

    I multiply through by another form of one [itex]\frac{2}{2}[/itex]

    [itex]\frac{-B\sqrt{B^{2} - 4AC} + B^{2} - 4AC}{2(4AC - B^{2})}[/itex]

    Simplify

    [itex]\frac{B\sqrt{B^{2}-4AC} + B^{2}}{2(4AC - B^{2})} - \frac{2AC}{4AC - B^{2}}[/itex]

    My equations for [itex]A_{1}[/itex] and [itex]A_{2}[/itex] then become

    [itex]A_{1} = V_{x}(t = 0^{+})(1 + \frac{B\sqrt{B^{2}-4AC} + B^{2}}{2(4AC - B^{2})} - \frac{2AC}{4AC - B^{2}})[/itex]

    [itex]A_{2} = -V_{x}(t = 0^{+})(\frac{B\sqrt{B^{2}-4AC} + B^{2}}{2(4AC - B^{2})} - \frac{2AC}{4AC - B^{2}})[/itex]

    My voltage equation then becomes

    [itex]V_{x}(t) = V_{x}(t = 0^{+})(1 + \frac{B\sqrt{B^{2}-4AC} + B^{2}}{2(4AC - B^{2})} - \frac{2AC}{4AC - B^{2}})e^{(\frac{-B + \sqrt{B^2 - 4AC}}{2A})t} - V_{x}(t = 0^{+})(\frac{B\sqrt{B^{2}-4AC} + B^{2}}{2(4AC - B^{2})} - \frac{2AC}{4AC - B^{2}})e^{(\frac{-B - \sqrt{B^2 - 4AC}}{2A})t}[/itex]

    Plugging back in the values for A, B, and C

    [itex]V_{x}(t) = V_{x}(t = 0^{+})(1 + \frac{\frac{1}{R}\sqrt{(\frac{1}{R})^{2}-4(C_{1}+C_{2})\frac{1}{L}} + (\frac{1}{R})^{2}}{2(4(C_{1}+C_{2})\frac{1}{L} - (\frac{1}{R})^{2})} - \frac{2(C_{1}+C_{2})\frac{1}{L}}{4(C_{1}+C_{2})\frac{1}{L} - (\frac{1}{R})^{2}})e^{(\frac{-\frac{1}{R} + \sqrt{(\frac{1}{R})^2 - 4(C_{1}+C_{2})\frac{1}{L}}}{2(C_{1}+C_{2})})t} - V_{x}(t = 0^{+})(\frac{\frac{1}{R}\sqrt{(\frac{1}{R})^{2}-4(C_{1}+C_{2})\frac{1}{L}} + (\frac{1}{R})^{2}}{2(4(C_{1}+C_{2})\frac{1}{L} - (\frac{1}{R})^{2})} - \frac{2(C_{1}+C_{2})\frac{1}{L}}{4(C_{1}+C_{2})\frac{1}{L} - (\frac{1}{R})^{2}})e^{(\frac{-(\frac{1}{R}) - \sqrt{(\frac{1}{R})^2 - 4(C_{1}+C_{2})\frac{1}{L}}}{2(C_{1}+C_{2})})t}[/itex]

    That's kind of ugly. But anyways if my equation

    [itex]V_{x}(t) = V_{x}(t = 0^{+})(1 + \frac{B\sqrt{B^{2}-4AC} + B^{2}}{2(4AC - B^{2})} - \frac{2AC}{4AC - B^{2}})e^{(\frac{-B + \sqrt{B^2 - 4AC}}{2A})t} - V_{x}(t = 0^{+})(\frac{B\sqrt{B^{2}-4AC} + B^{2}}{2(4AC - B^{2})} - \frac{2AC}{4AC - B^{2}})e^{(\frac{-B - \sqrt{B^2 - 4AC}}{2A})t}[/itex]

    is correct then I have developed a generic formula for circuits in which you can find a second order homogeneous equation. Does it look right? Thanks for your help!

    When I plug the values pack in I get the following equation

    [itex]V_{x}(t) = \frac{4}{49}e^{2000t}((7j - 401)e^{14000jt} - (7j - 499)e^{-14000jt})[/itex]

    Which dosen't seem right. Not sure what I did wrong though.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Apr 15, 2013 #2
    ok so I guess I have found my mistake. At time [itex]t = 0^{+}[/itex] the current that goes through the resistor where does it go? Does it go through the first capacitor or the second capacitor. I'm aware that it dosen't go through the inductor because initially the current through the inductor is zero and that current through an inductor can't change instantaneously. So I'm not really sure were this current goes and if someone could help in figuring this out than that would be great.

    Originally in the post above I have that the current is initially zero which is wrong.

    Thanks!
     
  4. Apr 15, 2013 #3

    NascentOxygen

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    All that maths .... too much for me to check. mad0270.gif

    If we can neglect the inductor, then at t=0- the top capacitor presents a voltage difference of 2V and the lower capacitor has a voltage of 8V. Then (without the inductor), after the switch is closed, the top capacitor eventually reaches a voltage of 0V, and the bottom capacitor carries all the supply voltage, 10V. So, with that scenario, you could say after t=0 the top capacitor discharges via R and the lower capacitor simultaneously charges via R. The voltage across one can't change without the voltage across the other also changing, in order to maintain their sum as 10V.

    A changing voltage across a capacitor is an indication that current is flowing in it.
     
  5. Apr 15, 2013 #4

    rude man

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    The most I'm willing to do is verify your answer. No way am I going to wade thru all that math. I can do it more quickly by going to Laplace transforms which I don't suppose you've had.
     
  6. Apr 15, 2013 #5
    So then if a capacitor can't change voltage instantaneous then the current through a capacitor is always zero at time t=0- and at time t=0+ because if the circuit has been sitting for a while the capacitors act as opens and there's current flowing through them at t=0-

    So like I'm still confused then. If the voltage across a capacitor can't change then which way does the current from the resistor go?

    Vs/R

    at time t=0+ there's a current flowing through the resistor, but if it can't go through the inductor or either cap then were does it go?

    And I completely forgot how to do laplace transforms =( in only a year sense taking differential equations.
     
  7. Apr 16, 2013 #6

    NascentOxygen

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    So it seems. :uhh:

    The voltage across a capacitor can change only after some time has passed to permit more charge to accumulate (or be removed) by current flowing into/outof its terminals. This is a process that takes time.

    In general, in a CR circuit at t=0+ there can be current into/outof capacitors. For a brief time after t=0, current flows without significant change in capacitor voltage. Only after more charge has built up will the capacitor show an increase in voltage.
     
  8. Apr 16, 2013 #7
    So then were does the current go? Through capacitor one or two or both and how to find out which? Like I get that at time t=0- there is no current flowing either capacitor and that voltage across a capacitor can't change instantaneously, a change in voltage across a capacitor allows for a change in current. So based on this information the current can't go through either capacitor at time t=0+, so then were does it go? The current through an inductor can't change instantaneously and initially there is no current flowing through the inductor so it can't flow through the inductor at t=0+ either.

    I think I'm still confused. I understood your post above but I still am unsure as to which capacitor it goes through.
     
  9. Apr 16, 2013 #8

    rude man

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    At t(0-) C1 and C2 are charged up to 10V between them (so what would V_x(0+) be?).

    At t(0+) there is instantaneous current flowing into C2, limited by the 50 ohm resistor. At the same time C1 begins to discharge, also into C2. Meanwhile current begins to build up in the inductor L.

    At t = infinity C1 is charged to 10V, C2 has no charge, and current thru the inductor is 10V/50 ohms.

    EDIT:
    Armed with this knowledge you can hopefully write the pertinent DE's with the relevant initial conditions on C1, C2 and L.
     
  10. Apr 16, 2013 #9

    rude man

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    I finally solved this thing & it was a chore. V_x(t) is a damped sinusoid. I have the results if you want to compare.
     
  11. Apr 16, 2013 #10
    well then V_x(0+) would be 10 volts as well

    At time t=(0+) the current flows through C2, why not C1? So current begins to flow the C1 into C2 as well? C2 discharges to ground?
     
  12. Apr 17, 2013 #11

    NascentOxygen

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    Yes, it's the voltage across a capacitor, and that can't change in an instant.

    As I said in my first reply here: So, with that scenario, you could say after t=0 the top capacitor discharges via R and the lower capacitor simultaneously charges via R. The voltage across one can't change without the voltage across the other also changing, in order to maintain their sum as 10V.

    Because the inductor starts to affect things shortly after 0+ you can't say a capacitor discharges to ground because with the inclusion of inductance you should anticipate things will be oscillatory.

    With the reasoning I outlined, current flows in both capacitors from the word go.
     
  13. Apr 17, 2013 #12

    rude man

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    No. V_x(0+) is the same as V_x(0-) which is not 10V.
    Current flows thru C1 and C2 starting at t(0+). As I said, the movement of V_x(t) is sinusoidal with exponential decay. Eventually, C2 discharges to ground and V_x(∞) = 0.
     
    Last edited: Apr 17, 2013
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