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Rate of change of current with time, galvanometer deflection

  1. Oct 20, 2016 #1
    1. The problem statement, all variables and given/known data
    (a) What is meant by the statement that a solenoid has an inductance of 2.0 H?

    A 2.0 H solenoid is connected in series with a resistor, so that the total resistance is 0.50 Ω, to a 2.0 V DC supply. Sketh the graph of current against time when the crrent is switched on.

    What is:
    (i) the final current,
    (ii) the initial rate of change of current with time, and
    (iii) the rate of change of current with time when the current is 2.0 A?

    Explain why an EMF greatly in excess of 2.0 V will be produced when the current is switched off.

    (b) A long air-cored solenoid has 1000 turns of wire per metre and a cross-sectional area of 8.0 cm2. A secondary coil, of 2000 turns, is wound around its centre and connected to a ballistic galvanometer, the total resistance of coil and galvanometer being 60 Ω. The sensitivity of the galvanometer is 2.0 divisions per microcoulomb. If a current of 4.0 A in the primary solenoid were switched off, what would be the deflection of the galvanometer?

    Answers: (a) (i) 4.0 A, (ii) 1.0 A s-1, (iii) 0.5 A s-1, (b) 268 divisions.

    2. The attempt at a solution
    (a) (i) I = V / R = 2 / 0.5 = 4 A.

    (a) (ii) E = L [d I / d t] → d I / d t = 2 / 2 = 1 A s-1.

    (a) (iii) This one can't get right. E = L [d I / d t] → 2 = 2 [2 / d t] → 2 d t = 4 → d t = 2 s. d I / d t = 2 / 2 = 1 A s-1.

    I also tried: P = I E = 2 * 2 = 4 W. P = L I [d I / d t] → 4 = 2 * 2 [d I / d t] → d I / d t = 1 A s-1.

    Maybe it should be like: 2 A is half the 4 A, so 1 A s-1 / 2 = 0.5 A s-1?

    (b) As I understand we need to find θ: θ = a N A B / R, where a is the division per microcoulomb, N = turns, A = cross-sectional area, B = magnetic field and R = resistance. If this formula is correct, I'm not sure how to find B. Maybe something like: B = μ0 n I = 4 π * 10-7 * 1000 * 4 = 5 * 10-3 T. Then plug into the formula: θ = 2 * 2000 * (8 * 10-4) * (5 * 10-3) / 60 = 2.68 * 10-4. It has the 2.68 in it as in the answer 268 divisions. But still I'm not sure whether I'm in the right direction.
     
    Last edited: Oct 20, 2016
  2. jcsd
  3. Oct 20, 2016 #2

    kuruman

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    One question at a time.
    Can you find I(t), the current as a function of time?
    Can you find the time t2 such that I(t2) = 2.0 A?
    Can you find dI/dt as a function of time?
     
  4. Oct 20, 2016 #3

    cnh1995

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    You need to study the step response of a series R-L circuit. Look it up. Then try to answer kuruman's questions in #2.
     
  5. Oct 21, 2016 #4
    Isn't it part (a) (ii)?
    d I / d t = 1 A s-1.

    E = L (d I / d t)
    2 = 2 (2 / d t)
    d t = 2 s.

    d I / d t = 2 / 2 = 1 A s-1.
     
  6. Oct 21, 2016 #5

    cnh1995

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    No. Part a)(ii) is the general formula for voltage across an inductor. You need to find current i(t). How does the current grow in a series R-L circuit when excited with a dc voltage?
     
  7. Oct 21, 2016 #6

    cnh1995

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    How is E=2V? Voltage across the inductor is not 2V when the current is 2A. At which instant is E=2V? Look up 'step response of a series R-L circuit' and post if anything is unclear.
     
  8. Oct 21, 2016 #7
    The "step response of a series R-L circuit" looks extremely difficult. Is there an easier / different way to approach this?
     
  9. Oct 21, 2016 #8

    cnh1995

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    I'm afraid there is no other way. But instead of going for the entire derivation, you can use the formulae directly here.
    Current in an RL circuit grows as i(t)=(V/R)*(1-e-tR/L).
    You can use this equation to find answers to the questions given in your problem.
     
  10. Oct 21, 2016 #9
    I have that formula in my book.
    I used the second formula, since we don't have I = 0 but I = 2 A and go t = 2.7 s and then d I / d t = 2 / 2.7 = 0.72 A s-1, not 0.5 though...

    Update
    Hm, I used the first formula and got the same result. Maybe I'm wrong somewhere. Here are both steps:

    We have: L = 2 H, R = 0.5 Ω, V = 2 V, I = 2 A.

    (i) I = E / R (1 - e- R t / L)
    2 = 2 / 0.5 (1 - e- 0.5 t / 2)
    0.5 = 1 - e- 0.5 t / 2
    -0.5 = - e- 0.5 t / 2
    0.5 = e- 0.5 t / 2
    ln 0.5 = - 0.5 t / 2
    - 0.69 = - 0.5 t / 2
    1.39 = 0.5 t
    t = 2.77 s.

    d I / d t = 2 / 2.77 = 0.72 A s-1.

    (ii) I = (E / R) e- R t / L
    2 = (2 / 0.5) e- 0.5 t / 2
    0.5 = e- 0.5 t / 2
    ln 0.5 = - 0.5 t / 2
    - 0.69 = - 0.5 t / 2
    - 1.39 = - 0.5 t
    t = 2.77 s

    d I / d t = 2 / 2.77 = 0.72 A s-1.
     
    Last edited: Oct 21, 2016
  11. Oct 21, 2016 #10

    cnh1995

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    That's not how you compute dI/dt.
    You have the expression for i(t). Differentiate it w.r.t time and then plug in the values of R, L and E in it.
     
  12. Oct 21, 2016 #11
    t = 2.77 s

    d I / d t = n * 2.77n - 1 A s-1.

    Like this?
     
  13. Oct 21, 2016 #12

    cnh1995

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    Since you have not studied calculus yet in school, I think I will have to do it for you here.
    I(t)=(E/R)*(1-e-tR/L).
    ∴ dI/dt= (E/L)*(e-Rt/L).
    Plug in the values of E, R, L and t. You have already calculated t.
     
  14. Oct 21, 2016 #13
    Ha, that indeed gets 0.5 A s-1. But how did you change the first formula into the second one?
     
  15. Oct 21, 2016 #14

    cnh1995

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    Derivative of Aekt w.r.t. time= Ak*ekt and derivative of a constant=0. You have the equation in the form,
    i(t)=A(1-ekt)=A-A*ekt.
    Derivative of the first term will be zero and derivative of the second term will be as above. This all will be taught to you in the calculus class, from the very basic concepts.
     
  16. Oct 21, 2016 #15
    Thank you.

    What's wrong with the last part?
     
  17. Oct 21, 2016 #16

    cnh1995

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    Also, if you see the graph of the current as a function of time, you'll see it is an exponentially increasing curve starting from i=0 at t=0. At t=0, the current is zero, meaning the inductor acts as an open circuit. Then the current starts rising slowly and at t=infinity (or after a long enough time) it settles to a value I=E/R. Inductor opposes a sudden change in current through it.
     
  18. Oct 21, 2016 #17
    Yes, this part I understand. The iron-cored coil slows down the current (the current is opposed by the back EMF in the coil) and when the circuit is closed the current is zero at first.

    But how do we approach part (b)?
    Is some part of my initial calculation correct?
     
  19. Oct 21, 2016 #18

    cnh1995

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    I think you have mistaken B for the field inside the galvanometer. It is actually the magnetic field in the inductor and has nothing to do with the galvanometer. As the inductor opposes sudden change in current through it, it means it opposes a sudden change in flux linkage too.
    Hence, when you turn off the primary supply, the secondary should maintain the initial B-field at the instant of switching off the primary current.
    I have not verified the math yet but this should be the approach IMO.
     
  20. Oct 21, 2016 #19

    cnh1995

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    Does it have the same area of 8cm2? I guess it does because IMO there's no way we can solve the question without knowing the area or length of the second coil.
     
  21. Oct 21, 2016 #20
    In terms of A that's how it is said in the first post. I'd guess it's the same and is equal to 8 cm2.

    I used the formula for the magnetic flux of coil. If it's not applicable, how do we find the magnetic field then?

    θ = a N A B / R -- is this formula correct? I mean we do need to find theta, right? The deflection of the galvanometer?

    I used N I = N I to find the current in the coil: 4 * 1000 = I * 2000 so I = 2 A. Then B = μ0 n I = 4 π * 10-7 * 2000 * 2 = 5 * 10-3 T. With this in mind I'll get θ = 2.68 * 10-4 (same as in the first post).
     
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