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Homework Help: Combination Circuit Question: Determining Current

  1. Mar 9, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the current flowing through each of the resistors in the circuit shown, where [tex]\xi[/tex] = 14V and R=6.3[tex]\Omega[/tex].


    2. Relevant equations

    where I=current (Amps), V=potential different (Volts), R=resistance ([tex]\Omega[/tex]).

    Resistors in Series:


    Resistors in Parallel:


    3. The attempt at a solution

    Step 1: Determine the overall resistance of the circuit.

    6.3[tex]\Omega[/tex] and 5.8[tex]\Omega[/tex] are in series, so Req=12.1[tex]\Omega[/tex] so far.

    Next, 12.1[tex]\Omega[/tex] and 3.2[tex]\Omega[/tex] are in parallel, so the resulting Req is: 1/Req=1/12.1[tex]\Omega[/tex]+1/3.2[tex]\Omega[/tex]=2.531[tex]\Omega[/tex].

    Finally, 2.531[tex]\Omega[/tex], 1.0[tex]\Omega[/tex] and 4.5[tex]\Omega[/tex] are in series. So Req=2.531[tex]\Omega[/tex]+1.0[tex]\Omega[/tex]+4.5[tex]\Omega[/tex]=8.031[tex]\Omega[/tex]

    The total resistance of the circuit is 8.031[tex]\Omega[/tex].

    Step 2: Determine the overall current, I.

    Using Ohm's equation, I=V/R, I found that I=1.743A

    Step 3: Determine the individual currents at each of the resistors.

    This is the part that I can't figure out. I know that resistors 4.5[tex]\Omega[/tex] and 1.0[tex]\Omega[/tex] have current, I=1.743A because the current travels from positive to negative, but I don't understand how to solve for the currents for the rest of the resistors.

    I also understand that the resistors 6.3[tex]\Omega[/tex] and 5.8[tex]\Omega[/tex] will also have the same current.

    (I have been given the correct answers for all of the resistors, but I can't reverse-solve the problems either:)
    6.3[tex]\Omega[/tex] and 5.8[tex]\Omega[/tex], I=0.365A
    3.2[tex]\Omega[/tex], I=1.38A

    Thanks for any insight!
  2. jcsd
  3. Mar 10, 2009 #2
    Your REQ calculation and source current calculation looks correct.

    Notice that the voltage across (R = 3.2 ohms) is the same as the voltage across the combination (R = 6.3 + R5.8 ohms). You combined those in series, then parallel in your earlier work and got (R = 2.53) ohms as a result. The voltage across that combination resistor will tell you the voltages across the re-expanded elements (R = 3.2 ohms) and
    (R = 6.3 + 5.8 ohms). Let me know if I need to clarify, I know that probably sounds confusing.
  4. Mar 10, 2009 #3
    I understand (I think!) what your saying, but I don't know the math in order to solve the problem. Is it a set of ratios to get the Current values for each resistor?
  5. Mar 11, 2009 #4
    You just need to use Ohm's Law V = IR. You can determine the voltage across the resistor combination I was talking about in the previous post because you will know the current through it and the equivalent resistance. You solved for those values earlier, ISource = 1.74 A and REQ = 2.53 ohms, from your calculation below.

    Use those values to determine the voltage across that equivalent resistive element, then expand the circuit back to its original form. Then you will have the voltages across the resistor (R = 3.2 ohms) and the resistor combination (R = 6.3 + 5.8 = 12.1 ohms). You can use the calculated voltage value and the resistor values to determine the two unknown branch currents.
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