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Circuit 1 with Resistors and Battery

  1. Feb 13, 2012 #1
    1. The problem statement, all variables and given/known data
    A circuit is constructed with five resistors and a battery as shown. The battery voltage is V = 12 V. The values for the resistors are: R1 = 75 Ω, R2 = 127 Ω, R3 = 129 Ω, and R4 = 130 Ω. The value for RX is unknown, but it is known that I4, the current that flows through resistor R4, is zero.

    What is I1, the magnitude of the current that flows through the resistor R1?

    2. Relevant equations

    V=IR

    3. The attempt at a solution
    I've tried writing down everything I know. The strategy that I'm trying to take is to find Req, and Ieq, and then find the voltage across R1 to use the relevant equation to get the current.

    Rx, R2, and R4 are in parallel, so I2R2=I4R4=IxRx, and 1/R2 + 1/R4 + 1/Rx = 1/R(24x)

    After I added those resistors, it seemed that all the resistors (1, 3, 24x) would be in series, but I'm not sure. If this is true, R1 + R(24x) + R3 = Req

    Veq=Ieq(Req)

    Veq=V1+V24x+V3

    And V1/R1=I1

    I feel like I probably have too many equations, but I don't know how to put them all together.
     

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  3. Feb 13, 2012 #2

    PeterO

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    Rx, R2 and R4 are not in parallel!!

    Note: if no current is passing through R4, then the PD across it is zero, so the PD across R1 must be the same as the PD across R2 so that the potential at each end of R4 is the same.
     
  4. Feb 13, 2012 #3

    gneill

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    Staff: Mentor

    What does it imply when the current through a resistor is zero?

    EDIT: Ha! PeterO beat me to it!
     
  5. Feb 13, 2012 #4
    Okay, I get that. So, V2=V1 and R1I1=R2I2, but I don't know what I2 is. Do I have to calculate Req to get Ieq?
     
  6. Feb 13, 2012 #5

    gneill

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    You won't need to find any currents. Instead you should recognize that the way the battery voltage is divided across the resistors R1 and R3 must be the same as the way it divides across R2 and Rx. What determines the way voltage divides across a voltage divider consisting of two series resistors?
     
  7. Feb 13, 2012 #6
    Voltage adds in series, and current is the same. Does this mean that V1 + V3 = V2 + Vx?

    I could be totally off, but from there I get R1V1 + R3I3 = R2I2 +RxIx. That doesn't really seem to help me.

    If R1 and R3 are in series, then I3=I1 and I2=Ix. It still seems like I have a lot of equations but too many unknowns. I feel like I'm missing something about the symmetry of the circuit.
     
  8. Feb 13, 2012 #7

    gneill

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    Yes, symmetry is important. It's important because of the way that voltage dividers work, and you're still not quite embracing that concept. The ratio of the resistances in a voltage divider determines how the supply voltage divides between the two resistors. This must be true because both resistors share the same current and direct application of Ohm's law determines the voltages. The actual current value is not so important as that resistance ratio.

    You should be able to state what resistor pairs in the circuit must have the same ratio...
     
  9. Sep 26, 2012 #8
    Have read through this thread and still don't understand how to obtain the ratio of resistors OR how to figure out what type of combination R4 is in...Any clarification would be appreciated.
     
  10. Sep 26, 2012 #9

    PeterO

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    This circuit is an example of a Wheatstone Bridge. It is used to measure an unknown resistanc value.

    If R4 was not there, you would have two simple arm in parallel with each other.

    On each arm, there are two resistors connected in Series, forming what is often called a Voltage Divider - meaning that part of the the supply Voltage drops over one of the resistors, while the rest drops over the other.

    if the left branch consisted of a 10Ω and 30Ω resistor, we would have 1/4 the supply voltage dropping over one, and 3/4 dropping over the other.

    If the second branch consisted of a 45Ω and 135Ω resistor, the voltage drops along that arm would also be 1/4 and 3/4 of the supply voltage.

    In that case, the potential at the mid-point of each arm would be the same, and when R4 is connected between those mid-points, there would be no current flow in R4.

    In practice, we often set the first arm to be two equal resistors [dividing the supply voltage in half on that branch], then on the second arm we have an adjustable, but known, resistor plus the unknown resistor.

    After adjustment so no current flows in R4, we might note our known resistor is 137Ω . That means our unknown resistor must also be 137Ω since the second branch is also dividing the supply voltage in half.
     
  11. Sep 26, 2012 #10
    Thank you that definitely helps. But I still am having trouble just beginning the problem by finding I1.

    From what you stated I thought that I would find V1 by (75/202)*12. I've calculated R12eqiv to be 47.15 ohms which gives a current going into both R1 and R2 to be .2545 Amps.

    I honestly am not sure where to go from here. The supply voltage for both branches is 12 v correct?
     
  12. Sep 26, 2012 #11

    gneill

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    You're overcomplicating things. If the current through R4 is zero then it can't affect either branch. So remove it! Discard it! Erase it from the schematic! Pretend it never existed!

    What then is the current through the R1 branch if the voltage across the branch is 12V?
     
  13. Sep 26, 2012 #12
    I appreciate your patience. V1 is [R1/(R1+R3)]*Vbatt and from that can find I1 from V1/R1.

    Thank You
     
  14. Sep 26, 2012 #13
    But you do know I1. If the current through R4 is zero, then R1 and R3 are in series across the voltage V. Use this to calculate I1. Then use your equation above to calculate I2.
     
  15. Jan 28, 2013 #14
    Please Help - Circuit 1 with Resistors and Battery

    Hi, I'd like to revive this thread.
    I'm solving the exact same problem but with different values.
    The last question is asking me the following:

    6) If the value of the resistor R2 were doubled, how would the value of the resistor R3 have to change in order to keep the current through R4 equal to zero?

    The available options are as follows:
    1) R3 would need to be increased
    2) R3 would need to be decreased
    3) R3 would not need to be changed
    4) There is no change that could be made to R3 to keep the current through R4 equal to zero.

    I realize this is a Wheatstone bridge and I read on wikipedia that the ratio between R2 and R1 is equal to the ratio between Rx and R3.
    So, from that I can say that if R2 is increased, R3 should be decreased by a factor of 2.
    Now, firstly, is this correct? ...and if it is, how else can I come to this conclusion?

    Thanks in advance
     
  16. Jan 28, 2013 #15

    PeterO

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    Re: Please Help - Circuit 1 with Resistors and Battery

    That change in R3 is correct, and why do you need another way to come to that conclusion?
     
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