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Given Current and Resistance in a Circuit, find the Voltage of a Battery

  1. Nov 19, 2014 #1
    • Warnings: Thread title too vague; Off-site image as problem statement; Template section headings missing.
    1. Consider the group of resistors in the figure, where R1 = 12.3 omegacap.gif and R2 = 7.95[PLAIN]http://www.webassign.net/images/omegacap.gif. [Broken] The current flowing through the 7.95http://www.webassign.net/images/omegacap.gif resistor is 1.22 A. What is the voltage of the batter?
    http://i.imgur.com/zB2eulo.png

    Fig2.gif

    2. Series - Req=R1+R2+..., Parallel - 1/Req=1/R1+1/R2+..., Ohm's Law - V=IR


    3. I started off by getting the equivalent resistance of the whole circuit, which I did by using the parallel equation for 13.8 and 17.2, which gave me 7.657 as the resistance of those two. Then I added 7.95 and 4.11 because those two are in series, which gave me 12.06. I believe the 12.06 and the 7.657 are now in parallel, which means that 1/12.06+1/7.657=1/Req. This gives me 4.68, which I believe is in series with the 12.3 and the 15 ohm resistors, so adding them all together gives me 31.98.

    I then take that 31.98 and plug it in to Ohm's law with the 1.22 A of current, which gives me 39.01 V. This is incorrect and I'm not sure where I went wrong.

    The obvious places to look would be making sure that each resistor was actually in series or in parallel with the one that I placed it with. I also am unsure if when they say that 1.22 A of current is flowing through the 7.95 resistor, if that means that the current for the entire circuit is 1.22 or if it has decreased by splitting up into separate currents and I have to calculate that.

    I have tried this several times and cannot figure out what I'm doing wrong. Thank you.
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Nov 19, 2014 #2

    gneill

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    Your problem started when you began reducing the resistor network and burying (losing) the resistor for which you actually know the current. Once a component has been combined-away into a reduced equivalent network, any current or voltage values that pertained to it alone become meaningless and inapplicable. So when you eventually tried to apply the current through R2 to the reduced network as a whole it was a meaningless operation.

    So. A better approach would be to apply KVL and KCL and Ohm's law, and work step-by-step through the circuit determining voltages across components and currents until you have filled-in all the currents and voltages.
     
  4. Nov 19, 2014 #3
    Oh, good to know! What exactly do you mean by KVL and KCL?
     
  5. Nov 19, 2014 #4

    gneill

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    Kirchhoff's Voltage and Current Laws.
     
  6. Nov 19, 2014 #5
    Alright so I'm starting at the 7.95 Ohm resistor and going forward, calculating the Voltage. I got 9.7 V for the 7.95 Ohm, and 5.01 V for the 4.11 Ohm. What do I do when I get to the 13.8 and 17.2 one in the middle. How do I know how the current splits up and how much enters those parallel resistors. Can I just combine them?
     
  7. Nov 19, 2014 #6

    gneill

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    You know the voltages across R2 and 4.11 Ω resistors, so what's the total voltage across that branch? What 's in parallel with that branch? What do you know about the voltage across things that are in parallel?
     
  8. Nov 19, 2014 #7
    So I don't need the current for that at all then. I can just add up the voltages to 14.7 V and say that since they are in parallel, the voltage across that middle branch is 14.7 as well. Now, do I apply the 1.22 A of current to the 15 Ohm and 12.3 Ohm resistors and then add up all the voltages?
     
  9. Nov 19, 2014 #8

    gneill

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    No, the 1.22 A applies only to the first branch. That current will contribute to the total current through the 15 and 12.3 Ohm resistors, but it's not all of it.

    You now have the voltage across the middle branch and can work out the current in that branch.
     
  10. Nov 19, 2014 #9
    Ok so I got 1.92 A for that, by taking 14.7 V divided by 7.66. That is the current for the middle branch if the calculations are correct. Now I combine those and that gives me the current for the first two resistors?
     
  11. Nov 19, 2014 #10

    gneill

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    That's right, you're applying KCL at the top middle node. Continue...
     
  12. Nov 19, 2014 #11
    So 3.14 is the circuit current, and I used that to get the voltages at all the resistors. The only thing I don't know now is how to get the battery's voltage. I know that KVL states that all voltages should add to 0, so would that mean that I add up the voltages (47.1+38.622+14.7+14.7)? That gives me 115.122, which seems incorrect.
     
  13. Nov 19, 2014 #12

    gneill

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    KVL applies to the sum of voltages around a closed loop. So, pick one loop that includes the unknown battery.
     
  14. Nov 19, 2014 #13
    Oh duh! I always forget to just stick to one loop. I took off one of the 14.7s and got 100.4 as the correct answer. Thank you so much for sticking with me through that. I really appreciate it.
     
  15. Nov 19, 2014 #14

    gneill

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    Glad I could help. Good luck with your studies.
     
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