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Current out of the battery, capacitor

  1. Feb 26, 2012 #1
    1. The problem statement, all variables and given/known data
    The circuit in the figure has a capacitor connected to a battery, two switches, and three resistors. Initially, the capacitor is uncharged and both of the switches are open.
    26-p-050.gif

    (a) Switch S1 is closed. What is the current flowing out of the battery immediately after switch S1 is closed?
    .015 A
    (b) After about 10 min, switch S2 is closed. What is the current flowing out of the battery immediately after switch S2 is closed?
    .0164 A
    (c) What is the current flowing out of the battery about 10 min after switch S2 has been closed?

    (d) After another 10 min, switch S1 is opened. How long will it take until the current in the 200-Ω resistor is below 1 mA?


    2. Relevant equations
    I(t) = V/R * e^(-t/(RC))

    3. The attempt at a solution
    I am stuck on (c) right now.
    At first I just used the equation above and plugged in the total Resistance and 600s for time. The answer I got was obviously wrong because the problem is asking for the current coming OUT OF the battery. The current will split two ways, one going to the 100 ohm resistor and the other towards the capacitor and the 200 ohm resistor.

    I then figure to only use the equation to solve for the current going to the capacitor and then use the ratio to find the current coming out of the battery. This does not work because the value for current that I get is so close to zero that the current coming out of the battery might as well be zero.

    Does the resistor, R, in the equation mean the total resistance of the circuit or just the 200 ohm resistor in series with the capacitor?
     
  2. jcsd
  3. Feb 26, 2012 #2

    gneill

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    Staff: Mentor

    For (c) why don't you start by determining the time constant for the circuit? How does the time constant compare to 10 minutes?
     
  4. Feb 26, 2012 #3
    ok Tc = R * C
    I am using the total resistance which is 366.667 ohm.
    Tc = 366.667*4*10^-3
    Tc= 1.4667 seconds
    Well comparing the time constant to 600 seconds (10 min) the Tc is a lot smaller...
     
  5. Feb 26, 2012 #4

    gneill

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    Staff: Mentor

    The resistance that you use to find the time constant should be the resistance 'presented to' the capacitor by the network. To find it, remove the capacitor and suppress the voltage source (replace it with a short) and then find the equivalent resistance of the network looking into the open terminals at the capacitor's location. However...

    The time constant that you find using the above will also be very small compared to 10 minutes. So, what does that tell you about the current though the capacitor branch after 10 minutes?
     
  6. Feb 26, 2012 #5
    What exactly do you mean when you say replace the voltage with a short?

    I think the current will be very low almost at zero because with so much time the capacitor is already reaching its maximum charge.
     
  7. Feb 26, 2012 #6

    gneill

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    Staff: Mentor

    To find the effective resistance that a network with sources presents to some port (pair of terminals), suppress the sources and calculate the effective resistance. Supressing sources means replacing voltage sources with short circuits (zero resistance wire) and current sources with opens (simply erase the current source).
    An engineering rule of thumb is that all the excitement is over after five time constants. After five time constants the capacitor will have effectively reached its final charge and current will be effectively nil as far as practical measurements are concerned.

    Does that help you to determine the battery current?
     
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