Solving Series Expansion of Current in Resistor/Inductor Circuit

[CaptGoodvibes]

Homework Statement

A resistor and inductor are connected in series to a battery. The current in the circuit ( in A ) is given by I = 2.7(1-e^-0.1), where t is the time since the circuit was closed. By using the series for e^x, find the first three terms of the expansion of this function.

Homework Equations

e^x = 1 + x + x²/2! + ... + xⁿ/n!

The Attempt at a Solution

Would I substitute -0.1 for x in this expansion like so: e^x = 1 - 0.1 - 0.1²/2! ? And then place this expansion into the original function 2.7(1-(1 - 0.1 - 0.1²/2!)) ?? And finally, multiply the 2.7 back in?

This isn't coming out to one of my answer choices. Where have I gone wrong in my setup?

Thanks!

 

[djeitnstine]

e^x = 1 +(-0.1) + \frac{(-0.1)^2}{2} Substituting you get 2.7 (1- \left( 1 +(-0.1) + \frac{(-0.1)^2}{2} \right)) Watch your minus signs.

 

[jbunniii]

Homework Statement

A resistor and inductor are connected in series to a battery. The current in the circuit ( in A ) is given by I = 2.7(1-e^-0.1), where t is the time since the circuit was closed. By using the series for e^x, find the first three terms of the expansion of this function.

Do you mean

I(t) = 2.7(1 - e^{-0.1t})? (There's no "t" in your formula.)

Would I substitute -0.1 for x in this expansion like so: e^x = 1 - 0.1 - 0.1²/2! ? And then place this expansion into the original function 2.7(1-(1 - 0.1 - 0.1²/2!)) ??

This isn't coming out to one of my answer choices. Where have I gone wrong in my setup?

Thanks!

I see a couple of problems. First,

(-0.1)^2 = 0.1^2, not -0.1^2

Second, notice that the "1" terms cancel, leaving you with only two terms. The problem asks for three, so you probably need to include the 0.1^3 term as well.

 

[CaptGoodvibes]

It kills me when I try to get clever and my signs get skewed.

So, now I've added a couple steps to insure I'm sign-correct and also added the third term to account for the 1's canceling.

Here is the setup:
I = 2.7 (1- \left( 1 +(-0.1) + \frac{(-0.1)^2}{2} + \frac{(-0.1)^3}{6}\right))

Clear the 1's and multiply by 2.7...

2.7(-0.1) + \frac{2.7(-0.1)^2}{2} + \frac{2.7(-0.1)^3}{6}

Pull out the fractions to please the prof:

2.7(-0.1) + \frac{1}{2}2.7(-0.1)^2+ \frac{1}{6}2.7(-0.1)^3

The numbers come out almost right. And the missing 't' from the problem shows in the answer key so I've arbitrarily added the t,t^2, and t^3

The answer I got is:
I(t) = -0.27t + 0.0135t^2 - 0.00045t^3 but to match the answer, I'd have to multiply by -1 to get the signs right. I'm close but there must be another tidbit I've overlooked. Or there's a typo in the answer key...

I wonder if this is like a falling object where the negative sign is removed(or multiplied by -1) in relation to time.

 

[jbunniii]

You only applied the - sign to the first term inside the inner parentheses. It should have been applied to all of them! (Distributive law.)

 

[CaptGoodvibes]

You only applied the - sign to the first term inside the inner parentheses. It should have been applied to all of them! (Distributive law.)

Ahhhhh... it's not 1-(...) it's 1-1(...) Dangit! I'm so caught up in the new stuff I forgot the details.

Thanks so much!