RC Circuit: Urgent Help

Discussion in 'Homework Help' started by lisa3412, Sep 6, 2014.

  1. lisa3412

    Thread Starter Member

    Sep 6, 2014
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    RC Circuit.png Hey there. .

    I have this problem on RC Circuit. .which I have uploaded.. and I have solved it using some concepts which aren't very clear to me. .I need you guys to help me conclude if what I am doing is right.

    1) Solve for Initial Condition:
    Here I got the Vc (Voltage across capacitor) to be the voltage across 3kohms. That comes to be 4v

    2) Solve for Final Condition:
    Here Vc=0, since we assume the circuit to be in stay state.

    3) Find Time Constant:
    here Rth=3
    C=100uF
    hence RC= 300ms

    4) Hence Vc(t)= 0+(4-0)e^(-t/300)

    Guys, please let me know if what I did till now is correct.

    Looking forward to your replies.

    Regards
     
  2. crutschow

    Expert

    Mar 14, 2008
    13,026
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    The answers to 1) and 2) are correct but your value for Rth is incorrect (I assume you meant Rth = 3k). Since R1 and R2 both go to ground when the switch goes to ground, Rth is the parallel value of R1 and R2.
     
  3. lisa3412

    Thread Starter Member

    Sep 6, 2014
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    Oh. .i see where i went wrong. .but looking at it again, won't Rth be the series value of 3&6?
     
  4. djsfantasi

    AAC Fanatic!

    Apr 11, 2010
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    Redrawing the original diagram may help. The voltage source is disconnected by the switch, so I omitted it. The switch then connects on end of R1 to ground, so I drew it there directly connected. I also slid the resistor on the schematic line to highlight the parallel connection.
    RC Circuit-2.png
     
  5. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    No, how can it be ? From the capacitor perspective, those two resistor are connected in parallel.
     
  6. lisa3412

    Thread Starter Member

    Sep 6, 2014
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    Ok. .Got it. .Thank you :). .

    I still need help in two more cases though:

    1) In the Initial Condition, Vc(t)= 4v. .so, how to calculate Ic(t)?

    2) what is the energy stored in the circuit at 1ms? How to calculate this?
     
  7. Jony130

    AAC Fanatic!

    Feb 17, 2009
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  8. lisa3412

    Thread Starter Member

    Sep 6, 2014
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    ok. .so Jony, just to confirm:

    Initial Condition: Ic(t) = 4/2=2A?
     
  9. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Are you sure that Rth = 2 ohm??
    R1 = 6kohm = 6000 ohm and R2 = 3kohm = 3000.
     
  10. lisa3412

    Thread Starter Member

    Sep 6, 2014
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    Oh yes. I meant 2kOhms.
    My professor has asked all of us to go ahead with kilo ohms only.
    Jony I had this outlook, please correct me if I am wrong:

    During the Initial Condition, Capacitor behaves like an open right? so, doesn't that mean I(t) should be equal to 12/(6+3)??
     
  11. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Yes, I(t) for t=0 is equal 4V/3Kohm but Ic(t) is a capacitor current. So for Ic(0) = 0A. But for time >0 Ic(t) = Vc(t)/Rth
     
  12. lisa3412

    Thread Starter Member

    Sep 6, 2014
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    I made a mistake. .the question asks you to find out I(t) at initial & final condition.
    Final condition, I(t) will be zero
    Initial Condition, I(t) will be 12/9 right?
     
  13. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    12 Volts/9000 Ohm
    You keep loosing the k in 6 kOhm and 3 kOhm. Not good. If you keep this up, you will never get right answer in your homework, quizes and tests.

    You concept is correct. In the initial condition stage the 12 volt voltage source is connected to the circuit. The capacitor is fully charged. What does that mean? A fully charged capacitor act as an open circuit. So for initial condition you can redraw the circuit and it will have: voltage source and two resistors in series. The equivalent resistance is then R1+R2. And the current in the circuit is voltage source divided by equivalent resistance. Which is what you did. That is why your concept is correct. It would have been good if you got the actuall numbers right too.
     
  14. lisa3412

    Thread Starter Member

    Sep 6, 2014
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    Yes! Thank you shteii01. I will keep it in mind. Thank you all of you.
     
  15. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    I agree with the units problem. If you are going to state that I=12/9 and it is really I=12/9000 then you should at least indicate that the units are not Amps but milliamps, as in:
    I=12/9 ma.

    The reason myself and others stress this point is because people often make mistakes where they think it really is 12/9 amps, and they then would be totally wrong and i believe they would want to know about this for future tests and for future life on the job.
     
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