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Nonlinear resistor problem

  1. Sep 11, 2015 #1

    s3a

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    1. The problem statement, all variables and given/known data
    The problem and its solution are attached in the TheProblemAndSolution.jpeg file.

    2. Relevant equations
    V = RI
    G = 1/R

    3. The attempt at a solution
    I tried to understand what the solution said, but I'm still very lost.

    Here's what I'm stuck on.:
    When the solution says "Since each value of voltage gives only a single resistance value, the resistor is voltage controlled.", I don't see how it is the case that each value of voltages gives only a single resistance value.

    Similarly, the solution also says "[The resistor] is not current controlled since certain values of current give more than one value of resistance.", and I also don't see how it is the case that .

    I also don't see how the solution goes from Fig. 3 to Fig. 4; in an attempt to reverse-engineer the solution, I see that 5V / 2Ω = 2.5 A, which gives the vertical intercept, and I also see that the horizontal intercept is the value of the voltage source, which is 5 V, but I'm not sure whether or not those are coincidences.

    Also, am I missing something, or is the solution expecting me to get the (precise) values of 2.2 A, 1.9 A and 0.3 A, by just looking at Fig. 5 (which is Fig. 1 and Fig. 4 overlapped)?

    Any help in figuring out how to do this problem would be GREATLY appreciated!
     

    Attached Files:

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  3. Sep 11, 2015 #2

    Simon Bridge

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    See figure 1. Work out the resistance vs voltage graph. There is only one current for each voltage value.
     
  4. Sep 13, 2015 #3

    s3a

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    Oh, I think I see! Each voltage only has one current associated to it and therefore one resistance (since one voltage and one current can only yield one resistance as shown by the equation V = RI), and because a particular current value can give more than one voltage value, since one current value and two voltage values gives two resistance values, as shown by the equation V = RI, it follows that certain values of current give more than one value of resistance and each value of voltage gives only one value for current. Right?

    So, assuming what I said above is correct, that solves my first confusion, so thanks for that. The next thing I'm confused with is how the solution proceeds from Fig. 3 to Fig. 4. Could you please help get me unstuck with that?
     
  5. Sep 13, 2015 #4

    Simon Bridge

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    Well the physical resistor won't have two values at the same time. Strictly speaking: knowing there are two possible resistances that could give a particular current, so knowing the current does not tell you the resistance uniquely, but knowing the voltage does.

    They do not get figure 4 from figure 3, and they do not say they do.
    How do you find the v-i characteristic for a linear resistor?
     
  6. Sep 14, 2015 #5

    s3a

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    Assuming I'm correct, the v-i characteristic for a linear resistor is a linear function, i(v) = 1/R * v (where R is a constant value and i is a function of v). I obtained this result by noting that v and i must be proportional, which makes the R in v = iR a constant.

    I'm still confused about figures 3 and 4, though. :(
     
  7. Sep 14, 2015 #6

    rude man

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    Fig. 4 IS derived from Fig. 3. Fig. 4 is the graphical equivalent of fig. 3.
    Take fig. 3, put I = 0 amps thru the network, what terminal voltage do you get? What does fig. 4 say the terminal voltage is?
    Put I = 2.5A thru the network, what terminal voltage do you get? What does fig. 4 say the terminal voltage is now?
    See how fig. 4 depicts what happens when you put various values of I thru the network of fig. 3, 0 < I < 2.5A?
     
  8. Sep 14, 2015 #7

    s3a

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    Looking at figure 3, if I put I = 0 A through the network, I get V = (2)(0) = 0 (which is not 5V, as stated in figure 4), and if I put I = 2.5 A through the network, I get V = (2)(2.5) = 5V (which is not 0V, as stated in figure 4).

    Looking at figure 4 along with the interval given by 0 A < I < 2.5 A, I see that, as the current increases, the voltage decreases (so, I'm assuming that the (variable) resistance is decreasing more than the current is increasing, in order to make the equation V = RI hold).
     
  9. Sep 14, 2015 #8

    rude man

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    You'red looking at just the resistor whereas fig. 3 depicts a 5V source and a 2 ohm resistor. Try again putting 0A and 5A thru the network of fig. 3, not just the resistor, then what is the voltage at the top of the network? (Make sure you run the current OUT OF the network since that's what it does in the circuit.)

    Looking ahead, you should also notice that the validity of fig. 5 rests on the fact that the voltage and current thru the fig.3 network are the same as for the non-linear resistor as depicted in fig. 1.
     
  10. Sep 16, 2015 #9

    s3a

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    I think I understand figure 3 and figure 4 now. Basically, since the portion of the circuit shown in figure 3 is in series (as is the whole circuit), the voltages add, and since voltage drops occur across resistors when the current goes through them, there is a voltage drop of V = (2Ω)(0A) = 0 V and V = (2 Ω)(2.5 A) = 5 V across the 2Ω resistor, such that we get 5 V - 0 V = 5 V and 5V - 5V = 0 V. Also, since we know that the resistor is linear, we can form the line of figure 4 using the two points (v,i) = (0, 2.5) and (v,i) = (5, 0).

    By that, do you mean that figures 3 and 4 can simply be overlapped since figures 3 and 4 each show a range of values from 0 to 5 volts and 0 to 3 amperes (i.e., the same range of values)?

    Lastly, how are the specific values of 2.2 A, 1.9 A and 0.3 A obtained? If they just expect me to look at the graph and come up with those exact values, that makes me a little uncomfortable.
     
  11. Sep 16, 2015 #10

    rude man

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    By that I mean the i-V characteristic of fig.3 network is i = (5-V)/2 and for the nonlinear resistor it's i = i(V) as shown in fig. 1 so since both have the same i and V for any i and V in the loop then when you graph i = (5-V)/R and i = i(V) the points of intersection are the only points satisfying commonality of i and V.
    Fig. 5 tells you what the three current points are, but you're right, you otherwise couldn't have told that just by looking at the graph.
     
  12. Sep 17, 2015 #11

    s3a

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    I see why figure 3's current is the same as the non-linear resistor's current (because all currents in a series circuit, such as figure 2's circuit, are equal), but I don't see why the voltage has to be the same.

    Also, does figure 5 look bad to you? I say that because, for example, instead of i = 1.9 A, it visually seems like i = 1.5 A.
     
  13. Sep 17, 2015 #12

    rude man

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    Look at the circuit (fig. 2). Isn't it obvious that it comprises the network of fig. 3 plus the non-linear resistor? Look at the voltage at the top of fig. 3 and see if you don't see that that's the same voltage as the voltage at the top of the non-linear resistor when you combine fig.3 and the non-linear resistor to make up fig. 2.
    I agree completely. The three currents on the curve look like 0.5, 1.5 and 2.1A if you go by the dashed lines. On top of that, the dashed lines don't look too well aligned to begin with ...
     
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