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Voltage and Current in RC circuits

  1. Feb 28, 2012 #1
    1. The problem statement, all variables and given/known data
    A circuit includes a spark gap that connects to the ground. The width of the gap is adjusted so that the capacitor discharges to the ground each time the voltage reaches 99% of the maximum.

    (The circuit is basically a simple series circuit with an 80v power supply, connected to a 2500[itex]\Omega[/itex] resistor. Then comes a junction, one way leads to a 400[itex]\mu[/itex]F capacitor, and the loop starts again in the power supply. The other way leads to the grounded spark gap).



    Find the current flowing through the resistor at time t=[itex]\tau[/itex]


    2. Relevant equations

    V=IR
    V=V0(1-e-t/[itex]\tau[/itex])


    3. The attempt at a solution
    First I found the constant [itex]\tau[/itex] for the circuit by doing (2500)(400X10-6)=1

    Then found V at time [itex]\tau[/itex] by doing

    V=80(1-e-[itex]\tau[/itex]/[itex]\tau[/itex])=51




    Then I thought that simply dividing 51 by 2500 would do it, but apparently not, any ideas?
    Thanks...
     
  2. jcsd
  3. Feb 28, 2012 #2

    gneill

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    Staff: Mentor

    The voltage across the resistor is not the voltage across the capacitor.

    HINT: You can modify your voltage formula slightly to give the voltage across the resistor.
     
  4. Feb 28, 2012 #3
    should i subtract/add the voltages?
     
  5. Feb 28, 2012 #4

    gneill

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    Staff: Mentor

    That's one way to do it; you know one end of the resistor is held at 80V and the other is determine by the capacitor voltage for which you've written an expression.

    On the other hand, you should recall that all the voltages and currents in the circuit will be changing according to very similar exponential curves. The current through the resistor is going to start at some maximum value when the capacitor is initially empty, and then it will decay towards zero as the capacitor fills. So you should be able to write by inspection a function for the current.
     
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