1. Not finding help here? Sign up for a free 30min 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!

Faulty electricity homework question?

  1. Jun 29, 2014 #1
    Hi,
    Hoping someone will be able to help me out with a electricity homework question that sounds a bit unreasonable to me.

    1. The problem statement, all variables and given/known data

    Question 8 (Read the section after Q4 first!)

    The I-V characteristics of a diode and a resistor to be used in a simple circuit with a varaible voltage power supply are shown in the graphs.

    I have a linear graph for the resistor for which I = 200mA at V = 1 V
    I have a graph for the diode for which I = 0 mA at V = 0, I = 20 mA at V = 1, I about 1000mA at V = 2 ... At V = 1.8, the graph is near-vertical.

    The question:
    The diode and resistor are placed in parallel and a variable voltage applied to them.
    a. If a voltage of 4.0 V is applied to the combination, what current will flow through them both?

    2. Relevant equations

    Ohm's law, V=IR. Applicable always to Ohmic devices, and applicable to non-Ohmic devices only for constant V or I (R is always changing).

    3. The attempt at a solution

    If I extrapolate the graph of the I-V characteristic for the diode to V=4, the gradient approaches infinity. Thus the resistance of the diode approaches zero. (R = 1/gradient). Since V = IR, the voltage drop across the diode also approaches zero.

    Therefore the effective resistance of the diode-resistor series pair is equal to the resistance of the resistor, 5 ohm.

    At 4.0V, I = V/R = 4V/5Ohm = 0.8A = 800mA.

    However, the answer states 400mA.

    I have attached a photo of the page with the graphs.
    If my working is wrong, I would really like to know where I went wrong!
    Thanks!
    Stephen
     

    Attached Files:

    Last edited: Jun 29, 2014
  2. jcsd
  3. Jun 29, 2014 #2

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Correct so far.

    Not correct.

    Not correct.

    The resister and diode are in parallel.
    You described a situation where the diode is conducting - therefore it can be replaced by a short circuit.
     
  4. Jun 29, 2014 #3
    Is the connection parallel or series?
     
  5. Jun 29, 2014 #4

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    From post #1: explicit problem statement:
    ... mind you, could be a typo.
     
  6. Jun 29, 2014 #5
    I think it is a typo. that's why I asked.
     
  7. Jun 29, 2014 #6

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    The question asks:
    That implies it's in series. If in parallel, it should say through "the combination", or somesuch.
     
  8. Jun 29, 2014 #7
    Sorry... it was a typo. The diode and resistor are in series, not parallel.
    @Simon Bridge:
    "(R = 1/gradient). Since V = IR, the voltage drop across the diode also approaches zero."
    You said this was incorrect... could you elaborate on that?
    "Therefore the effective resistance of the diode-resistor series pair is equal to the resistance of the resistor, 5 ohm."
    Is this still wrong?
     
  9. Jun 29, 2014 #8

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    The total drop across the pair is 4V, but you do not know how that is distributed.
    Can you think of a way of superimposing the two graphs to represent this?
     
  10. Jun 29, 2014 #9
    Well, in series, the current is constant across the circuit. So if I can find V for which both graphs have the same I, I guess that would be it... something to chew on... looks like I = 400mA for which both V are approximately equal to 2V!

    Thanks guys!
     
  11. Jun 30, 2014 #10

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Yes, that looks about right. But I ask again, can you think of a way of superimposing the graphs that would allow you to read the answer straight off?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Faulty electricity homework question?
  1. Electricity homework (Replies: 2)

Loading...