Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Circuit analysis - Potential differences of resistors

  1. Mar 12, 2010 #1
    1. The problem statement, all variables and given/known data
    circuit_4_pic.jpg
    [tex]R_1 = 70 \Omega[/tex], [tex]R_2 = 144 \Omega[/tex], [tex]R_3 = 237 \Omega[/tex], [tex]R_4 = 117 \Omega[/tex], [tex]R_5 = 246 \Omega[/tex]; [tex]V_{ab}=43 V[/tex]
    I am looking for the voltage of each resistor as well as the current passing through each one.


    2. Relevant equations
    Kirchoff's Voltage/Current Laws: The sum of any potential differences across a closed path is zero; current (charge) is conserved, so any current exiting a node is equal to the sum of the currents entering it.
    [tex]V = IR[/tex]
    [tex]I = \frac VR[/tex]


    3. The attempt at a solution
    I know [tex]R_{eq}=339\Omega[/tex]. Thus, [tex]I = \frac VR = \frac{43}{339} = .127 A[/tex].

    [tex]V_5 = I*R_5=.127*246=31.2 V[/tex]

    [tex]V_1+V_2+V_3-V_4=0[/tex]. This is Kirchoff's voltage equation.

    [tex]I_1=I_2=I_3[/tex] b/c current is conserved, so [tex]\frac{V_1}{70}=\frac{V_2}{144}=\frac{V_3}{237}[/tex]. I solve for [tex]V_1[/tex] and [tex]V_2[/tex] in terms of [tex]V_3[/tex]: [tex]V_1=\frac{70V_3}{237}[/tex] and [tex]V_2=\frac{144V_3}{237}[/tex].

    I know the current through [tex]V_3[/tex] and [tex]V_4[/tex] should sum to .127, so: [tex]\frac{V_3}{237}+\frac{V_4}{117}=.127[/tex].
    [tex]V_4=117\left(.127-\frac{V_3}{237}\right)[/tex].

    [tex]V_4=V_1+V_2+V_3[/tex] from above, so plugging in everything I get:
    [tex]117\left(.127-\frac{V_3}{237}\right)=\frac{70V_3}{237}+\frac{144V_3}{237}+V_3[/tex].

    I solve for [tex]V_3[/tex] and then plug that in to everything else.
    My final answers are: [tex]V_1 = 1.83[/tex], [tex]V_2=3.77[/tex], [tex]V_3=6.2[/tex], [tex]V_4=12.8[/tex], [tex]V_5=31.2[/tex]. However, these answers are incorrect, so I was hoping to find out what I'm doing wrong.
     
  2. jcsd
  3. Mar 12, 2010 #2

    rl.bhat

    User Avatar
    Homework Helper

    Hi Baou, welcome to PF.
    Voltage across R4 = V4 = 43 - V5 = 43 - 31.2 = 11.8 V
     
  4. Mar 12, 2010 #3
    Could you explain how you arrived at that? I don't quite follow...

    Edit: Oh, you mean it should be 11.8 instead of 12.8? Was everything else that I did correct?

    Edit2: Okay, I was just reading the answer on the calculator wrong. Thanks for the help!
     
    Last edited: Mar 12, 2010
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook