Self Inductance Question

1. Nov 13, 2008

TFM

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

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.

2. Relevant equations

V = IR

$$V = -L\frac{dI}{dt} [/tex 3. The attempt at a solution I am assuming that this is to be treated as a Kirchoff Loop, thus the total voltage = 0 Voltage providers: Inductor Battery Users: Resistor Thus I have the equation: [tex] \epsilon - L\frac{dI}{dt} - IR = 0$$

and thus:

$$\epsilon - L\frac{dI}{dt} = IR$$

treating like a differential equation:

$$\epsilon - L\frac{dI}{dt} = IR$$

$$\epsilon dt - L dI = IR dt$$

rearrange:

$$\frac{L}{IR} dI = -dt + \epsilon dt$$

Gives:

$$\frac{1}{L}ln(IR) dI = -t + \epsilon t$$

multiply by L

$$ln(IR) = -Lt + \epsilon t [/tex take exponentials: [tex] IR = e^{-Lt} + e^{\epsilon t}$$

Does this look right so far?

TFM

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2. Nov 14, 2008

alphysicist

Hi TFM,

This equation does not follow from the previous one.

3. Nov 14, 2008

TFM

No it doesn't

so:

$$\epsilon dt - L dI = IR dt$$

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

does this look better:

$$\epsilon dt - L dI = IR dt$$

$$- L dI = IR dt - \epsilon dt$$

$$- L dI = (IR - \epsilon) dt$$

$$- \frac{L}{IR - \epsilon} dI = dt$$

so and so:

$$- \frac{1}{L}ln(IR - \epsilon) = t$$

Does this look better?

TFM

4. Nov 14, 2008

alphysicist

I think this part looks okay.

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. Nov 14, 2008

TFM

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:

$$\frac{1}{x} dx$$

you got:

$$ln(x)$$

and for:

$$\frac{b}{x}$$

where b is a constant, you got:

$$\frac{1}{b}lnx$$

???

TFM

6. Nov 14, 2008

alphysicist

No, the integral

$$\int \frac{b}{x} dx \to b \ln x$$
(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.

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

7. Nov 14, 2008

TFM

Okay, so:

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

$$-L(ln(IR - \epsilon) - (0R - \epsilon)) = t - 0$$

$$-L(ln(IR - \epsilon - - \epsilon)) = t$$

$$-L(ln(IR)) = t$$

???

TFM

8. Nov 14, 2008

alphysicist

The R has to be handled like this:

$$\int \frac{1}{IR-\epsilon}\ dI \to \frac{\ln(IR-\epsilon)}{R}$$
(plus the constant).

Also when you go to apply limits,

$$\ln (x)\Big|_{x_0}^{x_f} \neq \ln (x_f - x_0)$$

Instead, it is:

$$\ln (x)\Big|_{x_0}^{x_f} = \ln (x_f) - \ln(x_0)$$

9. Nov 14, 2008

TFM

Okay, so:

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

This will integrate to:

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

$$-L \left(\frac{IR - \epsilon}{R} - \frac{0R - \epsilon}{R}\right) = t$$

$$-L \left(\frac{IR - \epsilon}{R} - \frac{- \epsilon}{R}\right) = t$$

does this look better?

???

TFM

10. Nov 14, 2008

alphysicist

You seemed to have dropped your $\ln$'s when you put in the limits.

11. Nov 14, 2008

TFM

Sorry, copied through typo

It should be:

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

$$-L \left(\frac{ln(IR - \epsilon)}{R} - \frac{ln(0R - \epsilon)}{R}\right) = t$$

$$-L \left(\frac{ln(IR - \epsilon)}{R} - \frac{ln(- \epsilon)}{R}\right) = t$$

???

TFM

12. Nov 15, 2008

TFM

Does this look correct

???

TFM

13. Nov 15, 2008

alphysicist

What does that give for I?

14. Nov 15, 2008

TFM

Well, if we rearrange it:

$$-L \left(\frac{ln(IR - \epsilon)}{R} - \frac{ln(- \epsilon)}{R}\right) = t$$

$$\left(\frac{ln(IR - \epsilon)}{R} - \frac{ln(- \epsilon)}{R}\right) = -\frac{t}{L}$$

Multiply by R:

$$ln(IR - \epsilon) - ln(-\epsilon) = -\frac{Rt}{L}$$

take exponentials:

$$IR - \epsilon + \epsilon = e^{-\frac{Rt}{L}}$$

$$IR = e^{-\frac{Rt}{L}}$$

$$I = \frac{^{-\frac{Rt}{L}}}{R}$$

Does this look correct?

TFM

15. Nov 15, 2008

alphysicist

This line is not correct, because:

$$\exp \left\{ \ln x + \ln y\right\} \neq x + y$$

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. Nov 15, 2008

TFM

so:

$$ln(IR - \epsilon) - ln(-\epsilon) = -\frac{Rt}{L}$$

isn't $$ln(a) - ln(b) = ln(\frac{a}{b})$$

???

If so:

$$ln(IR - \epsilon) - ln(-\epsilon) \equiv ln\left(\frac{IR - \epsilon}{-\epsilon} \right)$$

giving:

$$ln\left(\frac{IR - \epsilon}{-\epsilon} \right) = \frac{-Rt}{L}$$

taking exponentials:

$$\frac{IR - \epsilon}{-\epsilon} = e^{\frac{-Rt}{L}}$$

Does this look better?

TFM

17. Nov 15, 2008

alphysicist

Yes.

18. Nov 16, 2008

TFM

Excellent. So:

$$\frac{IR - \epsilon}{-\epsilon} = e^{\frac{-Rt}{L}}$$

$$IR - \epsilon = \epsilon e^{\frac{-Rt}{L}}$$

$$IR = \epsilon - \epsilon e^{\frac{-Rt}{L}}$$

factorise out:

$$IR = \epsilon \left(1 - e^{\frac{-Rt}{L}} \right)$$

divide by R:

$$I = \frac{\epsilon}{R} \left(1 - e^{\frac{-Rt}{L}} \right)$$

Does this look correct?

TFM

19. Nov 16, 2008

alphysicist

That looks right to me.

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