Current and Potential difference in a circuit

  1. For the circuit shown find the current through and the potential difference across each resistor.

    i was using the loop rules to make equations i could substitute into each other to solve this.. but i got stuck when i was making the last one for the currents...

    thus far i have
    I3= I4+I5
    I1= I2+I3
    and thus I1= I2+I4+I5
    which then is 0=-I1+I2+I4+I5
    but i need to find one with all of the currents but i cant figure out what to substitute in to get an equations with I3 in it as well. Help please

    loop 1 0= -I1R1 -I3R3 - I5R5 +24
    loop 2 0= -I1R1 -I2R2 +24
    loop 3 0= -I2R2 +I3R3 +I4R4 +24
    loop 4 0= -I4R4 +I5R5
    and then i need 0= the equation for the currents..
    that way i can plug it into a simultaneous solver.. but im stuck
     

    Attached Files:

    Last edited: Oct 10, 2007
  2. jcsd
  3. I can't see the attachment. maybe you should upload to imageshack or something.
     
  4. Ok, all this loop stuff is not necessary and according to me it adds confusion.

    Use your series-parallel formulae to reduce the circuit and systematically find the voltage drops and currents.

    circuits are usually reduced from the right to left when a single source is on the left.

    See what you can do.
     
  5. well i tried that at first. i determined the equivalent resistance of the circuit to figure out the current the first resistor would receive. i determined this to be 2 amps. the Req= 12 ohms.. but then i just seemed to get confused.. the current going into the first resistor equals that of the total current going into the rest of the resistors.. hmm
     
  6. Ok so 2 amps through R1 so VR1 = 8V correct?

    24V-8V=16V which is the voltage across the rest of the circuit.

    hmmm can you see what to do from there?
     
  7. I think i have an idea :) 16V for the rest of the circuit, but doesn't the rest of the circuit also receive a total of 2amps? oh i see when it splits they both receive the same Voltage because they are in parallel. so the 24ohm resistor receives 16volts as does the Req of the other line.
     
    Last edited: Oct 11, 2007
  8. Req should be 8 ohms not 12 ohms. go back and do it again and see where you went wrong.

    and yes current from the 16 volts gives the rest of the circuit a total of 2 amps.
     
  9. im afraid i dont see

    1/(1/8+1/24)+6= 12

    1/(1/12+1/24)= 8

    4+8=12?

    where did i go wrong?
     
  10. sorry. I assumed you didn't include R1 in Req. Nevermind what I said you are on the right track.
     
  11. so from there i would gather that .... 16 volts goes to the 24ohm resistor 16V=Ix24 I=2/3A
    16 volts goes to the other wire with a Req of 12 ohms. 16V= I 12ohms I= 1 1/3 amps..

    1 1/3 amps is the current that would go to each the 6 ohm resistor and the 6Req of the others. V= 1 1/3 amps x 6ohms V=8 volts ... 8 volts goes to each the 24 ohm resistor and the 8 ohm resistor since they are in parallel (had series typed). 8v= I24ohms I= 1/3 amps, and lastly 8V= I 8ohms I= 1 amp.. confusing possibly. but what do you think?
     
    Last edited: Oct 11, 2007
  12. I havent worked the values out myself but I do not see anything wrong with your method. everything looks good. amazing!
    A good way to confirm your values is to use simulation software like circuit maker. Search for it. I think it is freeware since the company that made it went bankrupt.

    Simulation software will become very important as you progress through your course.
     
  13. yeah i think its right tried circuit maker earlier and got confused :) thanks for the help
     
  14. This is what i got:
    R1= 4 ohms, I1= 2A, V1= 8V drop
    R2= 24 ohms, I2= 2/3A, V2= 16V drop,
    R3= 6 ohms, I3= 4/3A, V3= 16V drop
    R4= 8 ohms, I4= 1/3A , V4= 16V drop
    R5= 24 ohms, I5= 1A , V5= 16V drop

    They should be right but im still not sure but i hope these answer will help.
     
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