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Coils and inductance

  1. Apr 9, 2013 #1
    Hi there I needed someone to check my answers for these questions and see if they are correct and if not if they can point me in the right direction the question is for revision purposes got an exam in a week.

    Thanks.

    A coil of inductance 4H and resistance of 80Ω is connected in parallel with a 200Ω of resistor of negligible inductance across a 200v dc supply. The switch connecting these to the supply is then opened; the coil and resistor remain connected together. State in each case one for immediately before and one for immediately after opening the switch;

    The current Through the resistor

    Switch closed
    200v/200Ω =1a
    Switch open
    200v/80Ω =2.5a

    The current through the coil

    Switch Closed
    200v/80Ω = 2.5a
    Switch open
    200v/(200ohm +80Ω) = 0.714a


    The e.m.f induced in the coil

    Switch closed
    (2.5a*80Ω)/200v = 0v
    Switch Open
    2.5a *(200Ω + 80Ω) = 700v

    The voltage across the coil

    Switch closed
    200v/1a = 200v
    Switch open
    2.5a * 200Ω = 500v
     
  2. jcsd
  3. Apr 9, 2013 #2

    phyzguy

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    These aren't right. You have the 'switch closed' numbers right, but not the 'switch opened' ones. Immediately after the switch is opened, the current through the inductor will be the same as immediately before the switch is opened, since the current through the inductor cannot change instantaneously. All of this current then goes through the 200 ohm resistor, since it has nowhere else to go. Try using this way of thinking to calculate the other quantities.
     
  4. Apr 9, 2013 #3
    Thats where I'm having the problem, I don't seem to understand how to tackle this problems or what formula's to use this unit was covered last year and even though I'm going through my notes I can't seem to tackle the problem.
     
  5. Apr 9, 2013 #4

    phyzguy

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    Try drawing the circuit with the currents immediately before and after the switch is opened.
     
  6. Apr 9, 2013 #5
    Why would I need to draw the circuit? I think i'm making the question more complexed than it sounds I just need to know how to calculate them.
     
  7. Apr 9, 2013 #6

    phyzguy

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    Well, try doing what I told you in Post #2, and answer the following questions, in order:

    (1) What is the current in the inductor before the switch is opened?
    (2) What is the current in the inductor after the switch is opened?
    (3) What is the current in the resistor after the switch is opened?
    (4) What is the voltage across the resistor after the switch is opened?
    (5) Given this, what is the voltage across the inductor after the switch is opened?
    (6) Given this, what is the induced EMF in the inductor?
     
  8. Apr 9, 2013 #7
    (1) 200v/ 80ohms = 2.5a
    (2) 200v/200ohms = 1a
    (3)
    (4) 2.5 * 80 =200v
    (5)
    (6)
    I just had a feeling it would be wrong the questions I have answered I answered them confidently rest of them have been a stab in the dark.
     
  9. Apr 9, 2013 #8

    phyzguy

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    Correct.
    No! The current through the inductor cannot change instantaneously. Since V = L dI/dt, if the current changed instantaneously, dI/dt would be infinite, which would require an infinite voltage. So the current in the inductor after the switch is opened is the same as before the switch is opened, 2.5A.

    Using this, try answering questions 3-6.
     
  10. Apr 10, 2013 #9
    (2) 200v / 80 =2.5a
    (3) 200v/200 ohms = 1a

    Is this correct?
     
  11. Apr 10, 2013 #10

    phyzguy

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    No, After the switch is opened, the 2.5A flowing through the inductor must flow through the resistor, because it has nowhere else to go. Kirchoff's current laws tell you this. So if 2.5A is flowing through the 200 ohm resistor, what is the voltage across this resistor?
     
  12. Apr 10, 2013 #11
    What is the current in the inductor before the switch is opened?
    200v/80hms= 2.5a

    What is the current in the inductor after the switch is opened?
    200v/ 80ohms= 2.5A

    What is the current in the resistor after the switch is opened?
    200v/-80 = -2.5A

    What is the voltage across the resistor after the switch is opened?
    -2.5A * 200ohms = -500V

    Given this, what is the voltage across the inductor after the switch is opened?
    -2.5* 200-500V

    Given this, what is the induced EMF in the inductor?
    -2.5A * (200ohm + 80ohm) = -700V
     
  13. Apr 10, 2013 #12

    phyzguy

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    Great! I think this is now correct.
     
  14. Apr 10, 2013 #13
    Great so to be sure all this is correct now?

    A coil of inductance 4H and resistance of 80Ω is connected in parallel with a 200Ω of resistor of negligible inductance across a 200v dc supply. The switch connecting these to the supply is then opened; the coil and resistor remain connected together. State in each case one for immediately before and one for immediately after opening the switch;

    a) The current Through the resistor
    Immediately before Switch opening
    200v/ 200ohm = 1a
    Immediately after Switch opening
    200v / 80ohms = 2.5a

    b) The current through the coil
    Immediately before Switch opening
    200v / 80ohm = 2.5a
    Immediately after Switch opening
    200v / 80 ohm = 2.5a

    c) The e.m.f induced in the coil
    Immediately before Switch opening
    2.5a * 80ohms -(200v) = 0v
    Immediately after Switch opening
    -2.5a *(200Ω + 80Ω) = -700v

    d) The voltage across the coil
    Immediately before Switch opening
    2.5a * 80 ohm = 200v
    Immediately after Switch opening
    -2.5a * 200ohms = -500v
     
    Last edited: Apr 10, 2013
  15. Apr 10, 2013 #14

    phyzguy

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    The current through the resistor immediately before the switch is opened is 1.0 A, like you had in post #1. I think the rest is correct.
     
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