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Homework Help: Closing a circuit

  1. Apr 19, 2010 #1
    1. The problem statement, all variables and given/known data


    It's the very first one.

    2. Relevant equations

    R_eq = delta V / I = [ 1/R_1 + 1/R_n] ^ -1

    I = V / R

    3. The attempt at a solution

    a) So, since the resistors are in parallel, the potential different across the three different resistors have to be same.

    The given, I have 6V.

    Therefore, delta V from a to b should be 0 V to 6 V.

    b) Current between a and b :

    I_a to b= V / R_ab = 6 V / 12 ohm = 1/2 A

    I_c to d= 6V / 6 ohm = 1 A

    c) Since it is just a wire, current point at g should be 6 V / 0 ohm = 0 A

    d ) before the switch is closed, I do not have a current flow, so .. resistance is zero.

    After the switch is closed,

    R_eq = [1/12 + 1/6 + 1/30]^-1 = 3.529 ohm

    Does everything sounds good to anyone? :( man. I suck at circuit.



    um.. can somebody help me with the number 6 also?

    I solved a:

    R_a to b & c to d = [1/12 + 1/6]^-1 = 4ohm
    then R_eq = 4ohm + 8 ohm = 12 ohm

    But how can I find the each resistor's current passing and delta V ( b and c )??

    Since the Power = I x delta V , I guess I can help myself if I find b and c.

    HELP~~~~ please ~~~
    Last edited by a moderator: Apr 25, 2017
  2. jcsd
  3. Apr 19, 2010 #2
    Questions (1a) and (1b) are asking about the difference in voltage and current when the switch is open or closed. For (1c), in order for the current to get from the source to the resistors (and back) it must pass through point g. In question (1d), while the switch is open, current can flow through the 12 Ohm and 6 Ohm resistors, when the switch is closed, current can flow through all three resistors.
  4. Apr 19, 2010 #3
    In question (6a) you calculated the equivalent resistance of the whole circuit, and are given the voltage. From this you should be able to calculate the current through the whole circuit, then the current through each resistor, and finally the [itex]\Delta V[/itex] and power dissipated across each resistor.
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