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Complicated looking circuit

  1. Apr 25, 2008 #1

    ~christina~

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    [SOLVED] complicated looking circuit

    1. The problem statement, all variables and given/known data

    Consider circuit the below.
    (a) How much energy is delivered to the 10 Ω resistor from the time at which S2 is closed (Note: all other switches are open) to t = 30 µs?

    (b) How much energy is stored in the 2 µF capacitor if S1 and S3 are also closed and the circuit has reached steady-state? Express your results in electron volts.

    (c) What is the current in each branch of the circuit when S1, S2, and S3 are closed and the circuit has reached steady-state?

    (d) What is the reading on the voltmeter connected across the 6.0 Ω resistor when all of the switches are closed and the circuit has reached steady-state?

    [​IMG]

    2. Relevant equations


    3. The attempt at a solution

    (a) How much energy is delivered to the 10 Ω resistor from the time at which S_2 is closed (Note: all other switches are open) to t = 30 µs?

    hm...what equation would I use for this? (I can't seem to find a section discussing the energy "delivered" to a resistor within a time period, unless I calculate this some other way)
    energy would be in Joules as well, right?

    I was thinking that it would be: 0.5CV^2 but that is the energy that is stored in a capacitor and not resistor.

    (b) How much energy is stored in the 2 µF capacitor if S1 and S3 are also closed and the circuit has reached steady-state? Express your results in electron volts.
    I can't figure out which direction is the current going in the first place.
    And I think I need to write the sum equation for current, and sum equation for voltage since that's what my prof said, but I'm not sure how to do this for a circuit complicated looking like this one.

    Yes, I really think I need help on this problem.

    Please help.:frown:
     
  2. jcsd
  3. Apr 25, 2008 #2
    Remember that resistive elements cannot store energy, they only dissipate it. That said, there is a formula for the energy dissipated by a resistor per unit time, and I'm sure you've used it many times before (hint: power = ?)

    So your tools are KCL and KVL, right? More complex circuits simply mean more equations, but the base ideas are all the same. Try writing out all the equations you can and see where that gets you.

    Also, what's the current through a capacitor when you reach steady state?
     
  4. Apr 25, 2008 #3

    Doc Al

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    Just to add a bit to johnw188's advice:
    If you knew the current going through a given resistor, could you tell me the power it "consumes"? (I hope the answer is yes.) The trick is to find an expression for the current--and thus power--as a function of time, then integrate.

    Hint: Since S1 & S3 are open, all you have to worry about is the upper loop. Treat it as a standard RC circuit.

    To find the energy stored in a capacitor, you'll need to find the voltage across it.

    Do it systematically. Assign currents and write Kirchoff's laws for loop voltage and current. Hint: Once steady-state has been reached, what's the current through any segment containing a capacitor?
     
  5. Apr 25, 2008 #4

    ~christina~

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    well I found that[tex] P= \frac{\Delta q V} {\Delta t}= IV[/tex]
    alright.
    I=0
    yes because if you have the Resistance ([tex] \Omega[/tex]) and current (A)
    then I could use this equation to find the power: [tex] P= I^2 R [/tex] right?

    well, when a steady state is reached, the current in the segment is zero since the capacitor is fully charged. (I= 0).

    For part a)
    well... I have to find an expression for the current.
    I am thinking that the current goes from right to left after the circuit is closed.
    And the function for current when a capacitor is charging is
    [tex]I(t)= \frac{ \epsilon} {R} e^{-t/RC} [/tex]

    so would I use that to find the current? (but that wouldn't give me the collective current from t=0 to t= 30) so would this involve integration....(I suppose it would but I really don't know how to set up equations and I find it different to integrate in physics than it was to do in actual calculus so I get sort of lost with that, unfortunately)

    [tex]R_1= 10 \Omega[/tex]
    [tex]R_2= 6.0 \Omega[/tex]
    [tex]C= 3.0x10^-6F [/tex]
    [tex]V= 2.0V[/tex]

    [tex]P=I^2R[/tex]
    [tex]I(t)= \frac{ \epsilon} {R} e^{-t/RC} [/tex]

    I know the R and the current if that is the right equation but as to how to use that to integrate...

    I'm going to work on B and post that.
     
  6. Apr 25, 2008 #5

    Doc Al

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    Right.


    Right.

    Good. That's the current as a function of time, so what's the power as a function of time?

    To get the energy from the power, you'll have to integrate: [itex]\int P(t) dt[/itex].

    (Luckily an exponential function is easy to integrate. :wink:)
     
  7. Apr 25, 2008 #6

    ~christina~

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    Hm...lets see if I understand what your saying.



    [tex]\int P(t) dt= \int I^2 R= R \int (\frac{ \epsilon} {R} e^{-t/RC})^2= \frac{\epsilon^2} {R} \int e^{-2t/RC}[/tex]

    b)How much energy is stored in the 2 µF capacitor if S1 and S3 are also closed and the circuit has reached steady-state? Express your results in electron volts.

    assuming what I think is correct.

    since it says the system reaches a steady state I think that the current in that segment would be zero so it would be like it was cut, and the only segment that would be left is what I drew below.

    [​IMG]

    Is it right or not?

    Thanks
     
  8. Apr 25, 2008 #7

    Doc Al

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    Good! (Don't drop off the "dt".) Now do the integration.

    I think you've got the right idea. The loop you outlined in yellow is the only one with a non-zero current. (But you can still find and analyze a loop that contains the capacitor.) Find the current and you'll be able to figure out the voltage drops across the resistors (the ones with current through them).
     
  9. Apr 26, 2008 #8

    ~christina~

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    [tex]\int P(t) dt= \int I^2 R= R \int (\frac{ \epsilon} {R} e^{-t/RC})^2 dt= \frac{\epsilon^2} {R} \int e^{-2t/RC}dt= [/tex]

    that's where I got stuck. I tried to look in my book for a hint (cal book) but no luck there.

    I was thinking of the chain rule but not sure if that is correct. (I think I could change the
    [tex]-2t/RC[/tex] to [tex]-2t(RC)^{-1}[/tex] then use product rule on that)

    how can I analyze the one with the capacitor if there is no current through it at the time that they say I have to analyze it at? (steady state= I= 0)

    what I'm thinking (even though the circuit looks funny if I eliminate the 2 capacitors since the voltage sources are + =>+ and -=> - and I don't think that is possible.

    what I'm thinking is : (going from left to right on the circuit with current

    [tex] I_1 = I_2 [/tex] since in each resistor there is the same ammount of current

    [tex] \Sum V= \epsilon_1 - IR_1 - \epsion _2 - IR_2= 0[/tex]

    [tex]I= \frac{ \epsilon _1 - \epsilon_2} {R2+ R_2} [/tex]

    but I don't know the [tex]\epsilon_1 [/tex] (not given) so how do I find the voltage?
    ________________________________________________
    I've just noticed that it says "how much E is stored in 2micro F capacitor at the time that circuit reaches steady state and I've been thinking of finding something else.

    AND I'm going in a loop right now.

    Help please? (not sure where to start)
     
  10. Apr 26, 2008 #9

    ~christina~

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    Help anyone?
     
  11. Apr 26, 2008 #10
    I haven't looked all of it. But you are just complicating it too much.
    1) Reduce your thing to Thevian (two times - intial and final)
    2) use x(t) = x(f) - (x(i)-x(f))*e^ .. equation for 'i' through your thing
    3) find V as a function of time
    3) use 1/2*C*v^2
    4) find change in energy

    no messy integrals.
     
  12. Apr 26, 2008 #11

    Doc Al

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    This is much easier than you think. For one thing, RC is a constant.

    Hint: What's the derivative of [itex]e^x[/itex]?


    You can always take any loop and add up the voltage drops--even a loop that has a capacitor. (Hint: If a resistor has no current through it, there's no voltage drop across it.)

    To find the current going through the only loop that has current, just add up the voltage drops and set equal to zero. Let's say you start just below the left battery and move clockwise. (I can't read the voltage on that battery, so I don't know which battery is bigger--I'll just assume the current goes clockwise.) Here's what I get: +V - (10)I -2 -(20)I = 0. Solve for I.
    Once you've got the current you know what the voltage drops are across the 10 ohm and 20 ohm resistors. To find the voltage across the capacitor (which will tell you the energy), pick a loop that contains the capacitor and add up the voltages, as usual. For example, pick the loop that contains the 2.0V battery, the 20 ohm resistor, the 5 ohm resistor (no current here!), and the capacitor.
     
  13. Apr 27, 2008 #12

    ~christina~

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    Hint: What's the derivative of [itex]e^x[/itex]?

    I know it's e^x but if something is e^2x it would be 1/2e^2x right?

    BUT if that is right in the above derivative then wouldn't

    [tex]\int P(t) dt= \int I^2 R= R \int (\frac{ \epsilon} {R} e^{-t/RC})^2 dt= \frac{\epsilon^2} {R} \int e^{-2t/RC}dt= [/tex]
    be equal to [tex] \frac{RC} {2} e^{-2t/RC} [/tex]??

    I don't think that's right though.
     
    Last edited: Apr 27, 2008
  14. Apr 27, 2008 #13

    ~christina~

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    okay.
    But that battery voltage is not given. But solving for I anyway, I get,

    [tex]+V - 10 \Omega (I) - 2V - 20 \Omega (I)= 0[/tex]

    [tex]-V -30 \Omega (I)= 0 [/tex]

    [tex]I= -V/30 \Omega [/tex]

    not really since I don't know the current since the other V was not given.

    but subsituting anyway I get

    [tex] V= -10 \Omega (-V/30 \Omega)= V/3 [/tex]
    [tex]V= - 20 \Omega (-V/30 \Omega)= 2V/3 [/tex]

    I think I have to solve for the V..

    Thanks Doc.
     
    Last edited: Apr 27, 2008
  15. Apr 28, 2008 #14

    Doc Al

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    The derivative of e^(ax) is ae^(ax); the integral is (1/a)e^(ax)

    You left out the minus sign and the [itex]\epsilon^2/R[/itex] factor.

    Before we go any further, please double check that that battery voltage isn't given. (It looks like the diagram may have been truncated.) It doesn't make sense that they would not give that voltage.
     
  16. Apr 29, 2008 #15

    ~christina~

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    [tex]-\frac{\epsilon^2} {R} (\frac{RC} {2}) e^{-2t/RC} = \frac{\epsilon^2 C} {2} e^{-2t/RC}[/tex]

    My professor said that we had to solve for 3 unknowns including that voltage, when he was asked about this.
     
  17. Apr 30, 2008 #16

    Doc Al

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    Good, but don't drop the minus sign. You'll need to evaluate this over the interval from t = 0 to t = 30 µs, once you have epsilon.
    That makes no sense to me as you are given no data (such as a current or voltage somewhere). I'd say he's wrong. (Are you sure he wasn't referring to the voltage across the 6.0 Ω resistor, as asked in part d?)

    Is this taken from your text? Or is it a problem your professor created himself?

    (Sorry I've been slow in responding; I've been away for a few days.)
     
  18. Apr 30, 2008 #17

    ~christina~

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    [tex]-\frac{\epsilon^2} {R} (\frac{RC} {2}) e^{-2t/RC} = -\frac{\epsilon^2 C} {2} e^{-2t/RC}[/tex]

    so I would have to find the voltage drop across the resistor, correct?

    no, other people asked the same question about the voltage and whether he forgeot to give it to us, but no. My professor actually drew the circuit on the board and said that it could be done.

    I really don't know. It was part of a set of problems given to us to do. He may have created it himself, or he may have gotten it from somewhere.

    yes, I noticed. And the assignment was do a few days ago, and I'm not sure what I wrote, but I think I stopped at the unknown. :frown:
     
  19. Apr 30, 2008 #18

    Doc Al

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    I think I messed you up a bit on this one by not distinguishing total resistance from the 10 ohm resistance. Let's start over.

    You need to find the current as a function of time:
    [tex]I = \frac{ \epsilon} {R} e^{-t/RC}[/tex]

    Note that R is the total resistance in that RC circuit; R = 10 + 6 = 16 ohms. Epsilon is the battery voltage; epsilon = 2 V.

    To find the power through the 10 ohm resistor, use [itex]P = I^2 r[/itex] where r is now just 10 ohms. Integrate that over the time interval.

    [tex]\int P(t) dt= \int I^2 r= r \int (\frac{ \epsilon} {R} e^{-t/RC})^2 dt= \frac{r \epsilon^2} {R^2} \int e^{-2t/RC}dt= -\frac{r \epsilon^2} {R^2} (\frac{RC} {2}) e^{-2t/RC} = -\frac{r \epsilon^2 C} {2 R} e^{-2t/RC}
    [/tex]

    Sorry about that! (You'd better double check it. :redface:)

    Note that you have all the data needed to get the actual answer for this part.
    I'd love to see that solution.
     
  20. May 3, 2008 #19

    ~christina~

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    Sorry for the late reply Doc, I just saw this now.
    So the current as a function of time, R refers to the total resistance..okay, and since I'm trying to find the energy delivered to only 1 resitor it would mean that the r in power is only refering to the resistance in that one resistor.
    It's fine, I think it's right.
    yes, since I have
    r= 10 Ohms
    R= 16 Ohms
    C= 3x10^-6 F
    and it would be solved from t= 0 to t= 30
    is this correct?
    I would give you the solution but I don't think my professor put it up yet.

    Thanks Doc Al
     
  21. May 3, 2008 #20

    Doc Al

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    Looks like you got it.
     
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