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Simple kirchhoffs law problem

  1. Jun 20, 2009 #1
    1. The problem statement, all variables and given/known data
    q4.jpg

    This is a simlified version of the problem but I will understand it after this part is explained. The path highlighed in yellow, I need the equation for that path. THE CAPACITOR IS FULLY CHARGED.


    2. Relevant equations
    I1=I2+I3 (know how to use this)
    sum of the voltages = 0


    3. The attempt at a solution
    so for the path in yellow:
    12 - I1(5ohm) + 12 = 0 ???? I'm not 100% sure of what to do at the capacitor.
     
  2. jcsd
  3. Jun 20, 2009 #2

    djeitnstine

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    EDIT: At the resistor, [tex]V=IR[/tex] and [tex]I=\frac{dQ}{dt}[/tex]. This is an RC circuit, you need to solve a differential equation and of course at the Capacitor [tex]V=\frac{Q}{C}[/tex]
     
  4. Jun 20, 2009 #3
    after the capacitor is fully charged, no current will flow through it, so I can totally eliminate that portion of the circuit diagram and it would be a simple circuit with 2 resistors... is that correct? that is just my theory
     
  5. Jun 21, 2009 #4

    djeitnstine

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    Yes you are correct. The potential difference across the capacitor is 0 since its fully charged.
     
  6. Jun 21, 2009 #5

    cepheid

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    Exactly. You know that in the DC steady state, the capacitor is just like an open circuit, which means that you have a simple resistor divider circuit.

    No. That makes no sense at all. Think about it from an electrostatics viewpoint. You have a bunch of positive charge on a plate separated in space by a certain distance from an equal amount of negative charge on another plate. What this means is that there will be an electric field (nearly uniform in this case due to the geometry of the plates) and therefore a (NON-ZERO) potential difference (aka voltage) between the plates. Conclusion: the voltage across the capacitor is NOT zero.

    From a circuit standpoint, in the steady state, one would expect the voltage at that node to be given by exactly what you would predict from the resistor voltage divider circuit that remains. I.e.:

    Vc = V0(R2/(R1 + R2)) ​

    where V0 is the battery voltage and which resistor is which is determined by whatever makes this voltage division formula correct.

    If you don't believe me, then by all means set up differential equation. But you have to do it properly, meaning that your starting point should be the equation that you get from Kirchoff's junction rule applied to the node where the capacitor voltage is measured. The result I get for the capacitor voltage as a function of time is:

    [tex] v_c(t) = \frac{V_0}{R_1 + R_2}R_2\left[1 - \exp\left(-\left(\frac{1}{\tau_1} + \frac{1}{\tau_2}\right) t \right)\right] [/tex]​

    Where τ1 and τ2 = R1C and R2C respectively.

    So, in the steady state, the capacitor voltage is exactly what you'd expect from treating it as an open circuit.

    Edit: This solution assumes that we "plug in" the battery at time t = 0, i.e. the capacitor starts out initially uncharged because the circuit has no source in it before t = 0.
     
    Last edited: Jun 21, 2009
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