Resistor-Capacitor circuit

Discussion in 'Electronic Basics' started by Joachim, Nov 15, 2005.

  1. Joachim

    Joachim Guest

    How can I solve the following problem: "Show that the voltage Vc (t)
    across the capacitor as a function of time goes as 4.125(1 -
    exp(-t/0.1375)) volts. When the steady state has been reached, what is
    the power dissipated in each resistor and the energy stored in the
    capacitor?"

    Circuit diagram: http://krebs.uk.com/rc.jpg

    I know how to do this for a normal RC circuit, without the 22kOhm
    resistor in parallel around the capacitor. I think I need to somehow
    use Kirchhoff laws to figure this out, but I could really use a hand
    getting started.

    Thanks in advance,

    Joachim
     
    Joachim, Nov 15, 2005
    #1
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  2. Joachim

    AJW Guest

    Try drawing the circuit a little differently.

    Move the cap to the right.

    Now, ask yourself questions about the source the cap is seeing: In
    effect it's a voltage source (how big?) and a series resistance (how
    big?).

    The diagram you supplied is a neat way of confusing the issue.
     
    AJW, Nov 15, 2005
    #2
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  3. Joachim

    AJW Guest

    Try drawing the circuit a little differently.

    Move the cap to the right.

    Now, ask yourself questions about the source the cap is seeing: In
    effect it's a voltage source (how big?) and a series resistance (how
    big?).

    The diagram you supplied is a neat way of confusing the issue.
     
    AJW, Nov 15, 2005
    #3
  4. Joachim

    Christian S. Guest

    On 15 Nov 2005 11:13:58 -0800, "Joachim" <> wrote:

    >How can I solve the following problem: "Show that the voltage Vc (t)
    >across the capacitor as a function of time goes as 4.125(1 -
    >exp(-t/0.1375)) volts. When the steady state has been reached, what is
    >the power dissipated in each resistor and the energy stored in the
    >capacitor?"
    >
    >Circuit diagram: http://krebs.uk.com/rc.jpg
    >
    >I know how to do this for a normal RC circuit, without the 22kOhm
    >resistor in parallel around the capacitor. I think I need to somehow
    >use Kirchhoff laws to figure this out, but I could really use a hand
    >getting started.
    >
    >Thanks in advance,
    >
    >Joachim



    There is only one node in this circuit that is not fixed.
    Let's call this node A and the node voltage va.
    Apply KCL at this node: -(6V - va)/10k + va/22k + 20uF* dva/dt = 0

    Solve the differential equation and you are done.

    Chris
     
    Christian S., Nov 15, 2005
    #4
  5. Joachim

    Old Man Guest

    "Joachim" <> wrote in message
    news:...
    > How can I solve the following problem: "Show that the voltage Vc (t)
    > across the capacitor as a function of time goes as 4.125(1 -
    > exp(-t/0.1375)) volts. When the steady state has been reached, what is
    > the power dissipated in each resistor and the energy stored in the
    > capacitor?"
    >
    > Circuit diagram: http://krebs.uk.com/rc.jpg
    >
    > I know how to do this for a normal RC circuit, without the 22kOhm
    > resistor in parallel around the capacitor. I think I need to somehow
    > use Kirchhoff laws to figure this out, but I could really use a hand
    > getting started.


    The two resistors form a voltage divider such that the
    "steady state" voltage, V_cmax, accross the capacitor
    is given by

    V_cmax = V_supply * [ R_parallel / ( R_series + R_parallel )]

    whereof V_cmax = 4.125 V

    The steady state energy, E, stored in the capacitor is

    E = ( C / 2 ) * ( V_cmax )^2

    The steady state power dissipation in the resistors are
    those of the voltage divider. In the steady state, the
    capacitor draws no current.

    The RC time constant of the circuit is

    R*C = 0.1375 seconds

    Thus, with C = 20 uF, R = 6875 Ohms which is the
    resistance of R_series and R_parallel when hooked in
    parallel

    R = R_series * R_parallel / [ R_series + R_parallel ]

    > Joachim


    [Old Man]
     
    Old Man, Nov 15, 2005
    #5
  6. Joachim

    CWatters Guest

    "Joachim" <> wrote in message
    news:...
    > When the steady state has been reached, what is
    > the power dissipated in each resistor and the energy stored in the
    > capacitor?"


    Hint: Steady state = DC and capacitors behave like an open circuits for
    DC... so remove the capacitor for now...

    Then calculate the steady state current flowing...

    I = V/R
    = 6/(10,000+22,000)
    = 187u5A

    Power in 22K..
    = I^2 x R
    = 773uW

    Power in 10K
    = I^2 x R
    = 352uW

    Voltage on the 22K Res...

    = I x 22K

    or because it's a classic potential divider...

    = 6 x 22/(10+22)

    = 4.125V

    Now put the capacitor back...

    Voltage on capacitor is also 4.125V

    Energy stored in a capacitor

    = 0.5 C V^2

    = 0.5 x 20E-6 x 4.125^2

    = 170uJ

    Assuming I haven't made a typo.

    In the real world you might want to check the leakage current through the
    capacitor can be ignored.
     
    CWatters, Nov 15, 2005
    #6
  7. Joachim

    Black Knight Guest

    "Joachim" <> wrote in message
    news:...
    > How can I solve the following problem: "Show that the voltage Vc (t)
    > across the capacitor as a function of time goes as 4.125(1 -
    > exp(-t/0.1375)) volts. When the steady state has been reached, what is
    > the power dissipated in each resistor and the energy stored in the
    > capacitor?"
    >
    > Circuit diagram: http://krebs.uk.com/rc.jpg
    >
    > I know how to do this for a normal RC circuit, without the 22kOhm
    > resistor in parallel around the capacitor. I think I need to somehow
    > use Kirchhoff laws to figure this out, but I could really use a hand
    > getting started.
    >
    > Thanks in advance,
    >
    > Joachim


    Whoa.... is this a homework problem? The only person who
    should be doing your homework is you, but I don't mind helping.

    Break the problem down.

    The 10k and 22k resistors form a voltage divider.
    What is the final voltage across the capacitor?
    (Pretend the capacitor isn't there.)

    When the current is first applied (switch not shown)
    what was the voltage across the capacitor?

    At switch on (t = 0), the capacitor is effectively a short circuit.
    At t = infinity, it is effectively an open circuit.

    6 * 10/22 = ?
    Looks like your 4.125 isn't quite right. Maybe that's the problem.

    What is exp(-t) when t is large?
    You can use a calculator for that.
    Hint: On Microsoft Windows calculator, use Inv (the check box) ln
    (the natural logarithm function)

    What does the 0.1375 RC time constant represent?

    Hope that helps.
    Androcles
     
    Black Knight, Nov 15, 2005
    #7
  8. Joachim

    Joachim Guest

    Thanks for the help guys, I'm slowly getting to grips with this
    textbook.
     
    Joachim, Nov 15, 2005
    #8
  9. Joachim

    Bob Monsen Guest

    On Tue, 15 Nov 2005 11:13:58 -0800, Joachim wrote:

    > How can I solve the following problem: "Show that the voltage Vc (t)
    > across the capacitor as a function of time goes as 4.125(1 -
    > exp(-t/0.1375)) volts. When the steady state has been reached, what is
    > the power dissipated in each resistor and the energy stored in the
    > capacitor?"
    >
    > Circuit diagram: http://krebs.uk.com/rc.jpg
    >
    > I know how to do this for a normal RC circuit, without the 22kOhm
    > resistor in parallel around the capacitor. I think I need to somehow use
    > Kirchhoff laws to figure this out, but I could really use a hand getting
    > started.


    One way to look at it is that the cap is being fed from a voltage source
    that has a series resistance of 10k || 22k, and a voltage that is equal to
    the voltage divider voltage.

    Notice that 6*22/(22+32) = 4.125, and that (10k||22k)*20e-6 = 0.1375.

    If your textbook has mentioned Thevenin's theorem, that is simply the more
    formal way to do this. It states that if you have a two-terminal
    passive element in a circuit, you can find another circuit that consists
    of the device, a single resistor, and a single current source,
    all in series. The value of the voltage source is the thevenin equivalent
    voltage, and the value of the resistor is the thevenin equivalent
    resistance. The method to do this is trivial:

    1) Remove the component you are analyzing from the schematic (in this case
    C), and determine the voltage between the place where the leads previously
    connected. In this case, it is V = 6 * (22/(10+22)). That is the thevenin
    equivalent voltage.

    2) Now, in that modified circuit, replace the voltage source with a wire,
    and compute the resistance between those same two points. In this case, it
    is 10k || 22k = 6.875k. This is the thevenin equivalent resistance.

    Your circuit is then equivalent to

    4.125V ------[6.875k ohms]------[20uF]-----GND

    from the point of view of the cap, which is far easier to analyze.

    Regarding the resistors, when the cap is in its 'steady state', it
    effectively disappears from the circuit. So, use the equation for power
    given voltage and resistance: (6-4.125)^2/10000 and 4.126^2/22000.

    ---
    Regards,
    Bob Monsen

    Nature does not at once disclose all Her mysteries. - Lucius Seneca (Roman
    philosopher)
     
    Bob Monsen, Nov 16, 2005
    #9
  10. Joachim

    CWatters Guest

    "Black Knight" <> wrote in message
    news:qCtef.8033$...
    > Whoa.... is this a homework problem? The only person who
    > should be doing your homework is you, but I don't mind helping.
    >
    > Break the problem down.
    >
    > The 10k and 22k resistors form a voltage divider.
    > What is the final voltage across the capacitor?
    > (Pretend the capacitor isn't there.)
    >
    > When the current is first applied (switch not shown)
    > what was the voltage across the capacitor?
    >
    > At switch on (t = 0), the capacitor is effectively a short circuit.
    > At t = infinity, it is effectively an open circuit.
    >
    > 6 * 10/22 = ?
    > Looks like your 4.125 isn't quite right. Maybe that's the problem.


    Nope you made an error. The equation is ..

    6 * 10/(10+22) = 4.125
     
    CWatters, Nov 16, 2005
    #10
  11. Joachim

    CWatters Guest

    "CWatters" <> wrote in message
    news:w7Cef.49826$-ops.be...
    >
    > Nope you made an error. The equation is ..
    >
    > 6 * 10/(10+22) = 4.125
    >


    Now you got me making mistakes!...

    The voltage drop accross the 22K is...

    6 * 22/(10+22) = 4.125

    The voltage drop accross the 10K is

    6 * 10/(10+22) = 1.875
     
    CWatters, Nov 16, 2005
    #11
  12. Joachim

    Black Knight Guest

    "CWatters" <> wrote in message
    news:w7Cef.49826$-ops.be...
    >
    > "Black Knight" <> wrote in message
    > news:qCtef.8033$...
    > > Whoa.... is this a homework problem? The only person who
    >> should be doing your homework is you, but I don't mind helping.
    >>
    >> Break the problem down.
    >>
    >> The 10k and 22k resistors form a voltage divider.
    >> What is the final voltage across the capacitor?
    >> (Pretend the capacitor isn't there.)
    >>
    >> When the current is first applied (switch not shown)
    >> what was the voltage across the capacitor?
    >>
    >> At switch on (t = 0), the capacitor is effectively a short circuit.
    >> At t = infinity, it is effectively an open circuit.
    >>
    >> 6 * 10/22 = ?
    >> Looks like your 4.125 isn't quite right. Maybe that's the problem.

    >
    > Nope you made an error.

    Yep... My apologies. I realised it right after I posted, so I'll have
    to eat it. I 'fess up, I made a boo-boo.My granddaughter called
    as I was about to re-read what I'd written, so I posted in haste.

    The equation is ..
    >
    > 6 * 10/(10+22) = 4.125


    Of course. I didn't engage brain before opening my big mouth,
    so call me an idiot (this time). :)

    Androcles.
     
    Black Knight, Nov 16, 2005
    #12
  13. Joachim

    Black Knight Guest

    "CWatters" <> wrote in message
    news:laCef.49827$-ops.be...
    >
    > "CWatters" <> wrote in message
    > news:w7Cef.49826$-ops.be...
    > >
    >> Nope you made an error. The equation is ..
    >>
    >> 6 * 10/(10+22) = 4.125
    >>

    >
    > Now you got me making mistakes!...
    >
    > The voltage drop accross the 22K is...
    >
    > 6 * 22/(10+22) = 4.125
    >
    > The voltage drop accross the 10K is
    >
    > 6 * 10/(10+22) = 1.875


    6 - 4.125 = 1.875 :)

    BTW, the speed of light from
    (AB+BA)/( t'A-tA) = c =0 <> 300,000,000 meters/second.

    In SR, c = 0.
    That's the biggest boo-boo in the history of physics, has cost
    billions of m.o.n.e.y, so I don't feel all that bad.
    Androcles.
     
    Black Knight, Nov 16, 2005
    #13
  14. On 15 Nov 2005 11:13:58 -0800, "Joachim" <> wrote:

    >How can I solve the following problem: "Show that the voltage Vc (t)
    >across the capacitor as a function of time goes as 4.125(1 -
    >exp(-t/0.1375)) volts. When the steady state has been reached, what is
    >the power dissipated in each resistor and the energy stored in the
    >capacitor?"
    >
    >Circuit diagram: http://krebs.uk.com/rc.jpg
    >
    >I know how to do this for a normal RC circuit, without the 22kOhm
    >resistor in parallel around the capacitor. I think I need to somehow
    >use Kirchhoff laws to figure this out, but I could really use a hand
    >getting started.
    >
    >Thanks in advance,
    >
    >Joachim

    Look up Thevenin's theorem. It says that your circuit can be replaced
    by one with a series resistor equal to the two of them in parallel and
    a battery voltage that is reduced by the voltage divider effect. Old
    man had the right idea but Thevenin's is simpler.

    John Polasek
     
    John C. Polasek, Nov 16, 2005
    #14
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