1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Electronics question

  1. Nov 29, 2009 #1
    1. The problem statement, all variables and given/known data

    Hi,

    One end of a metal wire (say, NiCr) of a given length and cross-sectional area is attached to an end of another NiCr wire of of a given length and cross-sectional area. If the free end of the longer wire (R1) is at a given electric potential, and the free end of the shorter wire (R2) is at a lower electric potential, what is the potential at the junction of the two wires?


    2. Relevant equations

    V1 = E*R1/(R1+R2)

    3. The attempt at a solution

    I take this to be a kind of voltage divider. The wires themselves serving as the relevant resistors (R1) and (R2). The impressed voltage (E) is the difference between the potential at the ends of the two wires. The problem I am having is that in a voltage divider the voltage is measured across the resistor (either R1 or R2), but in this problem is the junction part of R1 or R2 or neither?

    I believe the answer is V2 = E*R2/(R1+R2), i.e. the voltage across the wire with the lower potential (plus the difference between the lower potential and ground), but I am not sure why.

    Thanks.
     
  2. jcsd
  3. Nov 30, 2009 #2
    Neither--and your approach is correct. Unless there is something funny about the junction between the wires that acts like a component, it can be treated as an ideal zero-resistance wire between the two NiCr "unwound wire resistors". It just happens to be zero-length as well as zero-resistance. :)
     
  4. Nov 30, 2009 #3
    I'm still somewhat confused. If we put into the equation a value for the resistor reflecting zero-length and zero-resistance, the voltage would also be 0.

    0 = E*(0)/R(total)

    I guess my fundamental question is why the voltage for the junction is the same as the voltage across the wire with the lower potential, rather than the wire with the higher potential.

    Thank you for the response.
     
  5. Dec 5, 2009 #4
    The voltage across the junction alone should be considered zero. All the voltage drop occurs in the measured lengths of wire. Or enough of it that the behavior at the junction can be neglected. The point of the exercise is to calculate the resistance of the nichrome wire, and see how that does behave as a component in a circuit.

    In an electronic circuit with resistors measured in hundreds of ohms and up with only 1% accuracy, typically, the contribution of 30-gauge copper wire at 0.1 ohm/foot can safely be ignored. In a toaster, however, most of the resistance and most of the voltage drop does occur across those glowing red wires.
     
    Last edited: Dec 5, 2009
  6. Dec 14, 2009 #5
    Is the distinction across the junction, rather than at the junction? Your response doesn't clarify for me why the voltage for the junction is the same as the voltage across the wire with the lower potential, rather than the wire with the higher potential.

    Thanks
     
  7. Dec 15, 2009 #6

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Voltages are always taken between or across two points. Just talking about the voltage at a point is meaningless, unless it is understood that you mean the voltage between that point and some other reference point. Sometimes the reference point is specifically identified, and sometimes (as is the case here) it is not specified.

    So...
    We may talk about the voltage across a wire, meaning the voltage difference between the two ends of the wire. In saying that the voltage across the junction is zero, we are just saying the voltage at each end of the junction (i.e., the ends of the wires that are in contact with each other) is the same.

    It is important to understand what V1 and E are in this equation.

    E is the voltage across the entire 2-resistor combination. In other words, it is the difference between the voltages at the free ends.

    Example:
    If one end is at 10 volts (meaning it is 10 V higher than some unspecified reference point), and the other end is at 2V (i.e. 2V higher than the same, unspecified reference point), then

    E = 10V - 2V = 8V​

    Similarly, V1 is the voltage across resistor R1.
     
  8. Dec 17, 2009 #7
    Thank you. This is helpful.

    Viewing the problem as though it were a circuit with two resistors in series (and using your example):

    "E is the voltage across the entire 2-resistor combination. In other words, it is the difference between the voltages at the free ends.

    Example:
    If one end is at 10 volts (meaning it is 10V higher than some unspecified reference point), and the other end is at 2V (i.e. 2V higher than the same, unspecified reference point), then

    E = 10V - 2V = 8V"


    We would find a voltage across R1 which was lower than 10V (but higher than 2V) and a still lower voltage across R2 (but still higher than 2V) and finally 2V at the other end. If we measure the voltage for the circuit at a point between R1 and R2, what would we measure and why? I know the answer is the same voltage as across R2, but I am having a hard time understanding analytically why this is so.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Electronics question
Loading...