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RC Circuit with a Switch

  1. Apr 29, 2015 #1
    1. The problem statement, all variables and given/known data
    P26_46.jpg Two resistors and two uncharged capacitors are arranged as shown above. Then a potential difference of V0 = 12 V is applied across the combination as shown.

    a. What is the potential at point a with S open? (Let V = 0 at the negative terminal of the battery.)
    b. What is the potential at point b with the switch open?
    c. When the switch is closed, what is the final potential of point b?
    d. How much charge flows through the switch S after it is closed?

    2. Relevant equations


    Voltage of Capacitor = Q/C
    V=IR
    Kirchhoff's Laws (Junction Rule and Loop Rule)

    3. The attempt at a solution

    I combined the two resistors so I would have one equivalent resistor and I combined the capacitors to have one equivalent capacitor.

    Now, the circuit is just the voltage source, a single resistor (13.2 Ohms) and a single capacitor (0.16 microFarads)

    I found the current through the resistor to be .909 Amps.

    Applying Kirchhoff's Laws, I got three equations:
    i1-i2-i3=0
    12-13.2(i2)=0
    13.2(i2)-Q/.16microF(i3)=0 --> not sure how to solve for Q since there's two unknowns here (i2 and Q)

    From here, I'm not sure what I need to do.
     
  2. jcsd
  3. Apr 29, 2015 #2

    gneill

    User Avatar

    Staff: Mentor

    If you combine the resistors and the capacitors, you "lose access" to points a and b where you need to find the potentials.

    Note that the two resistors form a potential divider, as do the two capacitors. See if you can find information on how to deal with potential dividers of both types (Resistive, Capacitive). It's very handy to know the simple, easy to remember formulas for them as they recur over and over again in problems (and in real life circuit analysis!).
     
  4. Apr 29, 2015 #3
    I completely forgot about those rules. I just thought point a and b still stayed after combining the resistors and the capacitors. Thanks for the tip!

    Now, for finding the charge on the capacitor after switch is closed, I'm a bit confused because here is the hint for it:

    Hint: Figure out the combined total charge on the bottom plate of the top capacitor and the top plate of the bottom capacitor - both before and after the switch is closed.

    Not sure what that is saying and not sure how to proceed. Would I be able to apply the equation Q=C*V here? and if so, is the C the sum of the capacitors? and the V the total Vo (12V)?
     
  5. Apr 29, 2015 #4
    Nevermind, I forgot that I already had the potential needed to solve for Q.
     
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