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Series circuit blues

  1. Apr 6, 2007 #1
    1. The problem statement, all variables and given/known data

    2. For the following circuit, find RT, IT, I3, V1, V3, V5.


    2. Relevant equations

    V = IR, ohms law, Kirchoff's laws

    3. The attempt at a solution

    RT = R1 + R2 + R3 + R4 + R5 + R5
    RT = 20 + 22 + 30 + 27 + 40 + 20
    RT = 159 ohms

    IT = V/RT
    IT = 120 V / 159 ohms
    IT = 0.75 amps

    I have the total resistance and total current.. I'm not sure exactly how to go about finding the current at I3 though. Any help would be appreciated.

    voltage, minus the voltage drop of the 2 resistors before it, / resistance of the 2 resistors before it? I'm not sure.
  2. jcsd
  3. Apr 6, 2007 #2
    In a series circuit, the current is constant all the way round, so the current at I3 is the same as the IT that you calculated.

    As the current is the same through all the resistors, you can calculate the voltage drop across each resistor using V=IR

    Hope that helps
  4. Apr 6, 2007 #3
    That does help a lot, thanks.

    But it raises one question for me, you say the current is the same through all resistors, but as I understand, the resistance causes the current to drop.

    Hence the I = V / R formula

    The previous question I did was "What is the current through each resistor?" in "A circuit with a 6.0 V battery has a 3 ohm resistor and a 15 ohm resistor in series."

    At a 3 ohm resistor

    I = V / R
    I = 6 v / 3 ohm
    I = 2 amps

    At a 15 ohm resistor

    I = V / R
    I = 5 v / 15 ohm
    I = 0.33 (repeating)

    Is this wrong, and the current through each of these resistors is equal to IT?
    Last edited: Apr 6, 2007
  5. Apr 6, 2007 #4


    User Avatar

    Staff: Mentor

    No, as Edward says, the current is the same everywhere in the series circuit. That same current causes a different voltage drop in the different load resistors, and that same current flows through the power supply (battery).
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