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

Finding voltage and current in a somewhat unusual circuit

  1. Jul 11, 2011 #1
    [Disclaimer: I'm pretty sure this doesn't belong in the HW questions forum. I'm a linguistics major not taking any EE courses and it's the summertime. I just happen to be using an undergrad EE text.]

    For the life of me I cannot crack the following:

    82qEk.png

    Since I believe the 70 and 30 ohm resistors are in parallel, likewise the 20 and 5 ohm resistors, I tried finding the total current through the circuit by combining them and then combining the two equivalent in series but I'm not even sure I can do that and I really am just stuck at this point.

    Your help would be greatly appreciated.
     
  2. jcsd
  3. Jul 12, 2011 #2
    Yep, you can do that.
     
  4. Jul 12, 2011 #3
    Ok, I did. So I believe the total current is 2 amps. Is that right? Even if I've got that right, I can't seem to figure out how the current is divided up in the network. I tried to apply KCL with incorrect results. Stuck...
     
  5. Jul 12, 2011 #4

    gerbi

    User Avatar
    Gold Member

    Hmm... how to crack it?
    Io is zero, because the line in which it "flows" is just connection between two nodes, You can simplify it by reducing it to one node.
    How to get Vo ? (Vo is the voltage that appears at the same time at 20 ohm and 5 ohm resistors). Look at the circut, distribution of voltage is that: 50=I*(70||30+20||5)
    The amount of voltage on resistors are proportional to the restistance so Vo=50*(R1/R1+R2) where R1=20||5=4ohm, R2=70||30=21ohm. Vo=8V
     
  6. Jul 12, 2011 #5
    The back of the book agrees with you about the voltage, but not about the current. The answer given is 0.2 A. I think there's been a typo. They probably mean 2 A, because that's the total current I found; of course, the current can't be zero, because then the circuit would be open...

    So this might be resolved?
     
  7. Jul 12, 2011 #6

    gerbi

    User Avatar
    Gold Member

    Are You sure about the current Io can't be zero ? The two nodes which are connected with this simply line (where Io "flows").. what is the resistance of this line ? (since ALL the resistances in the circut are represented by a resistor, no other resistances are allowed). Then what is the current Io? No resistance, no voltage drop. Infinity ? Zero ?
     
  8. Jul 12, 2011 #7
    In any case, something's not quite right with the answer given for this problem. Maybe I should tell the publisher.
     
  9. Jul 12, 2011 #8

    gerbi

    User Avatar
    Gold Member

    Chever, the answer given in the book is not correct.. I find it not very unusual.

    Anyway, this thread shouldn't be on this forum. Moderator will move it soon I guess
     
    Last edited: Jul 12, 2011
  10. Jul 12, 2011 #9
    WAIT before you leave.
    your book is wonderful. The answer is Indeed 0.2A.
     
  11. Jul 12, 2011 #10
    Here is How.
    1. First 70 in parallel with 30 is reduced to 21 Ohm.
    2. 20 in Parallel to 5 is reduced to 4 Ohm.
    3. Now, We find total current is 50/(21+4) = 2A.
    4. We Can now use current divisor formula to find current through all the resisters
    Now, it happens that current through 70 Ohm is 0.6A and through 20 Ohm is 0.4A.
    So, rest 0.2A flows through the Path marked Io.
    Find the current through all resisters. It will be consistent with Kirchoffs current law.
     
  12. Jul 12, 2011 #11
    I agree. I solved circuit and got I0=0.2A which is correct answer-

    If EE isnt your major, you probably cant solve circuit like this. We electrical engineer have our specific circuit analysis methods for circuits like that.
     
  13. Jul 12, 2011 #12
    @gerbi How did you come to believe that Io is zero simply because its a connection of two nodes?
    And Its not unusual to find people blaming the book when they can't find the answer. Revise Yourself ten times before blaming on the author.
     
  14. Jul 12, 2011 #13

    gerbi

    User Avatar
    Gold Member

    Really strange way to solve problems. What is the voltage drop in the line where Io flows ? (since it is no resistance). You see, that's the diffrence.. When you will asume there is one node (not two, where Io "flows" between them), what You can surely do, there is no current.
     
  15. Jul 12, 2011 #14

    gerbi

    User Avatar
    Gold Member

    the reason is simple, all the resistances in this circut are represented by resistors. If branche have no resistance it can be simplified to one node. Would You agree ?

    If You will look ath cicrut Your way, two nodes, which is incorect for me, then Your solution is good

    Another edit of this post:I am electrical engineer. My answer might be strange for You guys because I'm from Europe.. and I suppose most of ppl here are US. Maybe you analyze circuts in some other way.. dunno
    One more thing, I was little hasty saying the answer in the book is wrong. It's correct when You assume some.. really strange, and unacceptalbe (in my opinion), assumptions
     
    Last edited: Jul 12, 2011
  16. Jul 12, 2011 #15
    I'm with you up to step 4

    can you explain how current division was used here in more detail?

    apparently Kirchoff's laws and Ohm's law are adequate for solving this circuit

    and I've gotten everything else right up until now
     
  17. Jul 12, 2011 #16

    gerbi

    User Avatar
    Gold Member

    1. If You have two parallel branches, like here 70 and 30 ohm and current 2A, the flowing current will divide into two branches like this I(70ohm)=2A*(30/70+30) and I(30ohm)=2A*(70/70+30) so it's 0,6 A and 1,4 A (2A in total).

    2. Ohm's and two Kirhoff laws are enought for this.. but You still need to remember about some basic pronciples
     
  18. Jul 12, 2011 #17
    Ofcourse those methods come from those 3 basic laws.

    Check http://en.wikipedia.org/wiki/Current_divider for current division. Total current of circuit is 2A so we can solve current flowing through 70 ohm resistor. Same can be done for 20 ohm resistor and then we just use KCL to solve I0.
     
  19. Jul 12, 2011 #18
     
  20. Jul 12, 2011 #19

    gerbi

    User Avatar
    Gold Member

     
  21. Jul 12, 2011 #20
    Yeah, I agree. There is only one node there. I am not assuming two nodes while solving the currents in the resisters. But in this case the node has been stretched out into to points. Still the sum of incoming current to this stretched node is equal to sum of out-going current from this stretched node. I am simply applying kirchoffs current rule to the Point 1 (out of two points of stretched node) to find the current flowing from one point of the node to the other point of same node.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Finding voltage and current in a somewhat unusual circuit
Loading...