RC resistance

Discussion in 'Homework Help' started by poopscoop, Apr 17, 2014.

  1. poopscoop

    Thread Starter Member

    Dec 12, 2012
    139
    16
    Okay, from a conceptual standpoint, I have a simple RC circuit.

    A 12v source passes through a 1k resistor, to a capacitor and a 1k resistor in parallel. Both are in series with the 1k resistor.

    When the capacitor is charging, my textbook tells me the R value for RC is the thevenin resistance of the entire circuit, .5k ohms. However, since an uncharged capacitor is essentially a short, wouldn't all of the current flow into the capacitor at first, completely ignoring the resistor in parallel? Thus, why isn't the resistance for RC charging time just the resistor that 'matters' and limits the current into the capacitor?

    Similarly, if I evaluate said circuit with short circuit current to find thevenin resistance, I get 1k ohms, because the short circuit current is simply going to ignore the parallel resistor.
     
  2. shteii01

    AAC Fanatic!

    Feb 19, 2010
    3,386
    496
    I think part of the exercise might be the assumption that the resistor between power supply and the parallel RC combination might represent the internal resistance of the power supply.
     
    poopscoop likes this.
  3. WBahn

    Moderator

    Mar 31, 2012
    17,720
    4,788
    The charging time is not determined by the initial current into the capacitor, but by the overall time it takes to charge the cap.

    If the parallel resistor were missing, then it is true that all of the current that flows in the series resistor would flow into the cap, but also note that that cap would charge all the way up to the supply voltage.

    With the parallel resistor in place, the initial current that flows will still all go into the cap, as you noted. After that, some of the current will start going through the parallel cap. This alone means that it can't be ignored and that it will have some effect on the effective RC time constant. But the big effect is that the cap is only going to charge up to half of the supply voltage because of the voltage divider action of the two resistors. The net result is that it will reach its final voltage (or the same fraction of it) twice as fast as it would reach the higher final voltage without the parallel cap.
     
    poopscoop likes this.
  4. WBahn

    Moderator

    Mar 31, 2012
    17,720
    4,788
    But so what? What difference would it make whether this is assumed or not?
     
  5. poopscoop

    Thread Starter Member

    Dec 12, 2012
    139
    16
    Ah, I see now.

    The reason I was getting weird results for thevenin resistance: I was dividing short circuit current by supply voltage and not by thevenin voltage. Works now.
     
Loading...