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Ohm's Law

  1. Nov 12, 2007 #1
    [SOLVED] Ohm's Law

    1. The problem statement, all variables and given/known data

    I need to find the voltages at V1, V2, V3 in the picture, with the given information using Ohm's Law. I never have learned this stuff very well, so help would be appreciated. I'm not sure what to do...
    2. Relevant equations

    E = I/R

    3. The attempt at a solution

    I'm not sure how to find my current(which I'm guessing I need first). I know that in Series, the current is equal and they add in parallel.

    So, my question is...for V1, do I need to find the total current, or just the current through that point(12/100)?

    Thank you, and sorry that I'm not real smart on this stuff....I think its important that I learn it though View attachment Voltage.bmp
    Last edited: Nov 12, 2007
  2. jcsd
  3. Nov 12, 2007 #2
    You should go to a site like imageshack or something and upload your picture, then put the link here so the picture can be seen immediately
  4. Nov 12, 2007 #3
    Well, I tried to do that, as you asked.....not real sure how to do it, but here is the link I think....

  5. Nov 12, 2007 #4
    well, you should apply kirchoff's voltage law to find the current. after which you will know the potential difference across each resistor and then you can find the respective potentials at the different points.
  6. Nov 12, 2007 #5
    I'm not real sure how to apply that law, and furthermore, not real sure what it exactly is.

    I'm thinking that I have two loops, 1 loop with 100,20,100 Ohms. Another loop with 100,20,20, 100 Ohms. So then do I add those resistances up, and use Ohms law to solve for the current? This is the part that confuses me, I'm not sure how to start.....need specific instructions, so I can see what I'm doing I guess.
  7. Nov 12, 2007 #6
    Can someone guide me here....I don't have the answers to base myself, so I'm not sure what to do, our book is absolutely worthless
  8. Nov 12, 2007 #7
    Does the voltage at V1 = 5 volts?

    I found this by finding the resistance in the first circle to be 100+20+100=220 ohms.

    So, 12/220=.05A. So, 100*.05=5Volts.

    Is that correct, I'm totally lost?

    Also, can I use the 12 volts DC like that, or do I need to do something to it?
  9. Nov 13, 2007 #8
    You have a number of unknowns, so you'll need at least that many equations.

    First, the obvious unknowns:

    [tex]V_1, V_2, V_3[/tex]

    You also have current unknowns:

    [tex]I, I_1, I_2[/tex]

    Now let's label the resistors:


    Now we need to come up with 6 equations

    (1) [tex]V=V_1+V_2[/tex]
    (2) [tex]I=I_1+I_2[/tex]
    (3) [tex]V_2=I_1 R_3[/tex]
    (4) [tex]V_2=I_2(R_4+R_5)[/tex]
    (5) [tex]V_3=I_2 R_5[/tex]
    (6) [tex]V=I(R_1+R_2+\frac{R_3 R_4+R_3 R_5}{R_3+R_4+R_5})[/tex]
    (7) [tex]V_1=I_1\frac{R_3 R_4+R_3 R_5}{R_3+R_4+R_5}[/tex]

    That should be enough for you to solve this now.
  10. Nov 13, 2007 #9
    Ok, I understand how you got those equations I guess(after looking them over)....but then how do I solve for each item?

    I can solve for your equation #6 for I, but then how can I solve the rest? Every other way, I have more equations than unknowns.....
  11. Nov 14, 2007 #10
    Linear algebra.

    It's a bit tedious, but you have to start combining equations to eliminate variables.

    For example, using equation 6 you can get I. Next, using equation 2 solve for [tex]I_1[/tex] or [tex]I_2[/tex].

    If you solved for [tex]I_1[/tex], then substitute what [tex]I_1[/tex] is equal to in equation 3. Now equation 3 will have two unknowns: [tex]V_2[/tex] and [tex]I_2[/tex].

    Equation 4 also has the same unknowns. You can then combine/manipulate equations 3 and 4 to obtain the values of [tex]V_2[/tex] and [tex]I_2[/tex].

    One you get the hang of it, you can figure out the rest.
  12. Nov 14, 2007 #11
    Hey thanks a lot Bill!!! I got the right answers finally(what others had). I think I actually understand it, going to do some other examples soon to ensure, but thank you!!! I really appreciated you help!
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