Inductance 02 - RL Circuits

SAFARI DIRECTOR: Graham

2012 #34

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OPENING QUESTIONS: WITHOUT opening your notes or opening the book OR talking with your crew for the first 90 seconds, see how many different formulate you can list for:

EMF or εMF or ∆V or Voltage

    1. ? = sweep over there → Vb - Va = ∆V to see answers
    2. ?? = sweep over there →-∫E ∙ ds
    3. ??? = -dsweep over there →ΦB/dt
    4. ???? = -sweep over there →Bℓv
    5. ????? = Esweep over there →d
    6. ?????? = sweep over there →
    7. V = IRU/q

     

OBJECTIVE:  I will be able to annotate the diffeq for calculating current in terms of time during today's class

WORDS/FORMULAE FOR TODAY

TERMS

  • self induction/back emf
  • Inductance
  • solenoid ("A coiled wire with n loops per unit length")

CONSTANTS:

 

UNITS:

      • henry = (volt)(sec)/amps

FORMULAE:

      • εL = L(di/dt)
      • Kirchoff Amended: ε - iR - L(di/dt) = 0
      • B = μonI
        • n = number of loops per unit length
        • I = current
      • εL = -L(di/dt) AP = -L(dI/dt)
        • dI/dt = change in current
        • L = propotionality constant depending on the "geometry of the loop and other physical characteristics"
      • L = NφB/i
        • N = number of loops (or coils if you prefer)
        • i = current that may vary with time (hence the lower case)

WORK O' THE DAY

Let's start with including induction into Kirchoff's Rule:

Recall that:

ε - IR = 0

Now let's add in the term for inductance:

ε - IR - L(di/dt) = 0

Which sets us up for a VERY interesting diff eq (which is also a learning target by the way) so let's roll through that now. I'll give you a coupla minutes for each step, then I'll stop you and show you that next step.

Please don't get TOO stressed over the steps... several of which are not at all intuitive. As always, roll with the steps and take from them what you can!

Sooo.... let's have some *fun?* and work on getting current (i) in terms of time (t)

Step #1:

ε - IR - L(dI/dt) = 0

Step #2: We need to get I in terms of t so let's get rid of R:

ε/R - I - (L/R)(dI/dt) = 0

Step #3: This one's a tad bit odd. We'll do a "u" substitution here to clean things up a bit:

Let u = ε/R - I

so

du = -dI

Step #4: Substitute for BOTH u and du:

u = -(L/R)(du/dt)

Step #5 Combine like terms:

-(R/L)dt = du/u

Step #6 Setup integration on both sides with limits of integration from u = ui to uf and t = 0 to t

∫-(R/L)dt = ∫du/u

Step #7 Integrate

-(Rt/L) = Ln(uf) - Ln(ui)

Step #8 Simplify:

-(Rt/L) = Ln(uf/ui)

Step #9 Whip out good ol' ex on both sides

e-(Rt/L) = uf/ui

Step #10: Rearrange:

uie-(Rt/L) = uf

Step #11: Now let's go BACK to our initial conditions see that:

u = I - ε/R

And also note that at t = 0, i = 0 so that:

u = ε/R

So now we can substitute that in for ui:

/R)e-(Rt/L) = uf

and more generally: u = I - ε/R:

/R)e-(Rt/L) = I - ε/R

Step Last (thank the flying lizards!):

I = (ε/R)(1 - e-(Rt/L))

Y*I*K*E*S*

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An RL circuit consists of a resistor and an inductor in series as shown below

Consider the following:

1) The switch is closed and current starts to flow

2) A back EMF is induced in the coil which resists the flow of current

3) The flow of current then cannot be directly attributed using Ohm's Law (V=IR) however the TIME in which it takes the current to reach it's maximum flow is of iterest (at least to the AP)

The AP says we gotta do this:

(1) Apply Kirchhoff's rules to a simple LR series circuit to obtain a differential equation for the current as a function of time.

(2) Solve the differential equation obtained in (1) for the current as a function of time through the battery, using separation of variables.

(5) Calculate the rate of change of current in the inductor as a function of time.

Here's a more detailed analysis of what we just did:

1) Work with a PARTNER to review THIS

and this

And follow the steps

 

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Coursework: Problem 15 and 19 on page 990

STUDY GUIDE: