Self Inductance Question

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  • #1
TFM
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Homework Statement



Give a definition of (self) inductance. Suppose a battery, which supplies a constant EMF ϵ_0 is connected to a circuit of resistance R and inductance L at t = 0. Find an expression for the current as a function of time.

Homework Equations



V = IR

[tex] V = -L\frac{dI}{dt} [/tex

The Attempt at a Solution



I am assuming that this is to be treated as a Kirchoff Loop, thus the total voltage = 0

Voltage providers:

Inductor
Battery

Users:

Resistor

Thus I have the equation:

[tex] \epsilon - L\frac{dI}{dt} - IR = 0 [/tex]

and thus:

[tex] \epsilon - L\frac{dI}{dt} = IR [/tex]

treating like a differential equation:

[tex] \epsilon - L\frac{dI}{dt} = IR [/tex]

[tex] \epsilon dt - L dI = IR dt [/tex]

rearrange:

[tex] \frac{L}{IR} dI = -dt + \epsilon dt [/tex]

Gives:

[tex] \frac{1}{L}ln(IR) dI = -t + \epsilon t [/tex]

multiply by L

[tex] ln(IR) = -Lt + \epsilon t [/tex

take exponentials:

[tex] IR = e^{-Lt} + e^{\epsilon t} [/tex]

Does this look right so far?

TFM
 

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Answers and Replies

  • #2
alphysicist
Homework Helper
2,238
2
Hi TFM,

The Attempt at a Solution



I am assuming that this is to be treated as a Kirchoff Loop, thus the total voltage = 0

Voltage providers:

Inductor
Battery

Users:

Resistor

Thus I have the equation:

[tex] \epsilon - L\frac{dI}{dt} - IR = 0 [/tex]

and thus:

[tex] \epsilon - L\frac{dI}{dt} = IR [/tex]

treating like a differential equation:

[tex] \epsilon - L\frac{dI}{dt} = IR [/tex]

[tex] \epsilon dt - L dI = IR dt [/tex]

rearrange:

[tex] \frac{L}{IR} dI = -dt + \epsilon dt [/tex]

This equation does not follow from the previous one.
 
  • #3
TFM
1,026
0
No it doesn't

so:

[tex] \epsilon dt - L dI = IR dt [/tex]

see, I have to rearrange to get I onto the left side.

does this look better:

[tex] \epsilon dt - L dI = IR dt [/tex]

[tex] - L dI = IR dt - \epsilon dt [/tex]

[tex] - L dI = (IR - \epsilon) dt [/tex]

[tex] - \frac{L}{IR - \epsilon} dI = dt [/tex]

so and so:

[tex] - \frac{1}{L}ln(IR - \epsilon) = t [/tex]

Does this look better?

TFM
 
  • #4
alphysicist
Homework Helper
2,238
2
No it doesn't

so:

[tex] \epsilon dt - L dI = IR dt [/tex]

see, I have to rearrange to get I onto the left side.

does this look better:

[tex] \epsilon dt - L dI = IR dt [/tex]

[tex] - L dI = IR dt - \epsilon dt [/tex]

[tex] - L dI = (IR - \epsilon) dt [/tex]

[tex] - \frac{L}{IR - \epsilon} dI = dt [/tex]

I think this part looks okay.

so and so:

[tex] - \frac{1}{L}ln(IR - \epsilon) = t [/tex]

This isn't quite right. The integration is not right (if you take the derivative of the left side with respect to I you don't get the left side of the previous equation). Also remember that these are definite integrals, so you have to evaluate the limits.
 
  • #5
TFM
1,026
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Would the limits be for I between 0 and I and for t between 0 and t?

I am not sure how to integrate this, because I thought when you integrated:

[tex] \frac{1}{x} dx [/tex]

you got:

[tex] ln(x) [/tex]

and for:

[tex] \frac{b}{x} [/tex]

where b is a constant, you got:

[tex] \frac{1}{b}lnx [/tex]

???

TFM
 
  • #6
alphysicist
Homework Helper
2,238
2
and for:

[tex] \frac{b}{x} [/tex]

where b is a constant, you got:

[tex] \frac{1}{b}lnx [/tex]

No, the integral

[tex]
\int \frac{b}{x} dx \to b \ln x
[/tex]
(plus a constant) because in that case the b can come out of the integral (it is not affected by the integration process at all). However, you do have to take into account that the R is multiplying the I.

[tex] - L\left( \int\limits_0^I \frac{1}{IR - \epsilon} dI \right) = \int\limits_0^t dt [/tex]
 
  • #7
TFM
1,026
0
Okay, so:

[tex] - L\left( \int\limits_0^I \frac{1}{IR - \epsilon} dI \right) = \int\limits_0^t dt [/tex]

[tex] -L(ln(IR - \epsilon) - (0R - \epsilon)) = t - 0 [/tex]

[tex] -L(ln(IR - \epsilon - - \epsilon)) = t [/tex]

[tex] -L(ln(IR)) = t [/tex]

???

TFM
 
  • #8
alphysicist
Homework Helper
2,238
2
Okay, so:

[tex] - L\left( \int\limits_0^I \frac{1}{IR - \epsilon} dI \right) = \int\limits_0^t dt [/tex]

[tex] -L(ln(IR - \epsilon) - (0R - \epsilon)) = t - 0 [/tex]

The R has to be handled like this:

[tex]
\int \frac{1}{IR-\epsilon}\ dI \to \frac{\ln(IR-\epsilon)}{R}
[/tex]
(plus the constant).

Also when you go to apply limits,

[tex]
\ln (x)\Big|_{x_0}^{x_f} \neq \ln (x_f - x_0)
[/tex]

Instead, it is:


[tex]
\ln (x)\Big|_{x_0}^{x_f} = \ln (x_f) - \ln(x_0)
[/tex]
 
  • #9
TFM
1,026
0
Okay, so:

[tex] - L\left( \int\limits_0^I \frac{1}{IR - \epsilon} dI \right) = \int\limits_0^t dt [/tex]

This will integrate to:

[tex] -L\left[\frac{ln(IR - \epsilon)}{R}\right]^I_0 = t [/tex]

[tex] -L \left(\frac{IR - \epsilon}{R} - \frac{0R - \epsilon}{R}\right) = t [/tex]

[tex] -L \left(\frac{IR - \epsilon}{R} - \frac{- \epsilon}{R}\right) = t [/tex]

does this look better?

???

TFM
 
  • #10
alphysicist
Homework Helper
2,238
2
You seemed to have dropped your [itex]\ln[/itex]'s when you put in the limits.
 
  • #11
TFM
1,026
0
Sorry, copied through typo :redface:

It should be:

[tex] -L\left[\frac{ln(IR - \epsilon)}{R}\right]^I_0 = t [/tex]

[tex] -L \left(\frac{ln(IR - \epsilon)}{R} - \frac{ln(0R - \epsilon)}{R}\right) = t [/tex]

[tex] -L \left(\frac{ln(IR - \epsilon)}{R} - \frac{ln(- \epsilon)}{R}\right) = t [/tex]

???

TFM
 
  • #12
TFM
1,026
0
Does this look correct

???

TFM
 
  • #13
alphysicist
Homework Helper
2,238
2
What does that give for I?
 
  • #14
TFM
1,026
0
Well, if we rearrange it:

[tex] -L \left(\frac{ln(IR - \epsilon)}{R} - \frac{ln(- \epsilon)}{R}\right) = t [/tex]

[tex] \left(\frac{ln(IR - \epsilon)}{R} - \frac{ln(- \epsilon)}{R}\right) = -\frac{t}{L} [/tex]

Multiply by R:

[tex] ln(IR - \epsilon) - ln(-\epsilon) = -\frac{Rt}{L} [/tex]

take exponentials:

[tex] IR - \epsilon + \epsilon = e^{-\frac{Rt}{L}} [/tex]

[tex] IR = e^{-\frac{Rt}{L}} [/tex]

[tex] I = \frac{^{-\frac{Rt}{L}}}{R} [/tex]

Does this look correct?

TFM
 
  • #15
alphysicist
Homework Helper
2,238
2
Well, if we rearrange it:

[tex] -L \left(\frac{ln(IR - \epsilon)}{R} - \frac{ln(- \epsilon)}{R}\right) = t [/tex]

[tex] \left(\frac{ln(IR - \epsilon)}{R} - \frac{ln(- \epsilon)}{R}\right) = -\frac{t}{L} [/tex]

Multiply by R:

[tex] ln(IR - \epsilon) - ln(-\epsilon) = -\frac{Rt}{L} [/tex]

take exponentials:

[tex] IR - \epsilon + \epsilon = e^{-\frac{Rt}{L}} [/tex]

This line is not correct, because:

[tex]
\exp \left\{ \ln x + \ln y\right\} \neq x + y
[/tex]

Before you take the exponential of both sides, you just want a single natural log on the left . That is, you want the left side to be just:

ln(something)
 
  • #16
TFM
1,026
0
so:

[tex] ln(IR - \epsilon) - ln(-\epsilon) = -\frac{Rt}{L} [/tex]

isn't [tex] ln(a) - ln(b) = ln(\frac{a}{b}) [/tex]

???

If so:

[tex] ln(IR - \epsilon) - ln(-\epsilon) \equiv ln\left(\frac{IR - \epsilon}{-\epsilon} \right) [/tex]

giving:

[tex] ln\left(\frac{IR - \epsilon}{-\epsilon} \right) = \frac{-Rt}{L} [/tex]

taking exponentials:

[tex] \frac{IR - \epsilon}{-\epsilon} = e^{\frac{-Rt}{L}} [/tex]

Does this look better?

TFM
 
  • #17
alphysicist
Homework Helper
2,238
2
Yes.
 
  • #18
TFM
1,026
0
Excellent. So:

[tex] \frac{IR - \epsilon}{-\epsilon} = e^{\frac{-Rt}{L}} [/tex]

[tex] IR - \epsilon = \epsilon e^{\frac{-Rt}{L}} [/tex]

[tex] IR = \epsilon - \epsilon e^{\frac{-Rt}{L}} [/tex]

factorise out:

[tex] IR = \epsilon \left(1 - e^{\frac{-Rt}{L}} \right) [/tex]

divide by R:

[tex] I = \frac{\epsilon}{R} \left(1 - e^{\frac{-Rt}{L}} \right) [/tex]

Does this look correct?

TFM
 
  • #19
alphysicist
Homework Helper
2,238
2
[tex] I = \frac{\epsilon}{R} \left(1 - e^{\frac{-Rt}{L}} \right) [/tex]

Does this look correct?

TFM

That looks right to me.
 

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