Series Circuts

**VonNemo19** · January 30th 2011, 06:18 PM

An electromotive force

$E(t)=\left\{\begin{array}{cc}120,&\mbox{ if }<br /> 0\leq{t}\leq20\\0, & \mbox{ if } t>20\end{array}\right.$

is applied to an LR series circut in which the inductance is 2o henries and the resistance is 2 ohms. Find the current $i(t)$ is $i(0)=0$ .

Now, I would totally know what to do here guys if the voltage wasn't piecewise. How do I proceed?

**dwsmith** · January 30th 2011, 06:28 PM

Originally Posted by VonNemo19

An electromotive force

$E(t)=\left\{\begin{array}{cc}120,&\mbox{ if }<br /> 0\leq{t}\leq20\\0, & \mbox{ if } t>20\end{array}\right.$

is applied to an LR series circut in which the inductance is 2o henries and the resistance is 2 ohms. Find the current $i(t)$ is $i(0)=0$ .

Now, I would totally know what to do here guys if the voltage wasn't piecewise. How do I proceed?

DE

$\displaystyle L\frac{di}{dt}+Ri=E(t)$

L = 20 and R = 2. For $0\leq t\leq 20$ , then $\displaystyle 20\frac{di}{dt}+2i=120\Rightarrow\frac{di}{dt}+\fr ac{1}{10}i=6$

Solve like a DE

For $t\leq 20$

$\displaystyle 20\frac{di}{dt}+2i=0\Rightarrow\frac{di}{i}=-\frac{dt}{10}$

Solve as a DE.

**VonNemo19** · January 30th 2011, 06:36 PM

Originally Posted by dwsmith

DE

$\displaystyle L\frac{di}{dt}+Ri=E(t)$

L = 20 and R = 2. For $0\leq t\leq 20$ , then $\displaystyle 20\frac{di}{dt}+2i=120\Rightarrow\frac{di}{dt}+\fr ac{1}{10}i=6$

Solve like a DE

For $t\leq 20$

$\displaystyle 20\frac{di}{dt}+2i=0\Rightarrow\frac{di}{i}=-\frac{dt}{10}$

Solve as a DE.

So, I get

$i(t)=\left\{\begin{array}{cc}60(1-e^{-t/10}),&\mbox{ if }<br /> 0\leq{t}\leq20\\60(e^2-1)e^{-t/10}, & \mbox{ if } t>20\end{array}\right.$

Right?

**dwsmith** · January 30th 2011, 06:38 PM

Originally Posted by VonNemo19

So, I get

$i(t)=\left\{\begin{array}{cc}60(1-e^{-t/10}),&\mbox{ if }<br /> 0\leq{t}\leq20\\60(e^2-1)e^{-t/10}, & \mbox{ if } t>20\end{array}\right.$

Right?

Correct.

**dwsmith** · January 30th 2011, 06:46 PM

Ok so we have done 25 and 33. Is 35 next?

**VonNemo19** · January 30th 2011, 06:49 PM

Well, dwsmith...Thank you so very much for your help tonight. I think I'll ace this exam that I have Tuesday and I attribute part of my future success to you. I'll ace it because my teacher doesn't even know the material that well - it's obvious in how she responds to some of the tougher questions during class. So, since she doesn't know it that well, she's limited in the types of problems that she can assign because she has to be able not only to grade them, but articulate her reasons in a clear and professional manner as to why she may have marked a particular problem incorrect. Therefore, I think that she will stick to the very basic notions within differential equations and leave a more advanced treatment to someone who knows what they're talking about.

Anyways, thanks again,
VonNemo.

**dwsmith** · January 30th 2011, 06:55 PM

Originally Posted by VonNemo19

Well, dwsmith...Thank you so very much for your help tonight. I think I'll ace this exam that I have Tuesday and I attribute part of my future success to you. I'll ace it because my teacher doesn't even know the material that well - it's obvious in how she responds to some of the tougher questions during class. So, since she doesn't know it that well, she's limited in the types of problems that she can assign because she has to be able not only to grade them, but articulate her reasons in a clear and professional manner as to why she may have marked a particular problem incorrect. Therefore, I think that she will stick to the very basic notions within differential equations and leave a more advanced treatment to someone who knows what they're talking about.

Anyways, thanks again,
VonNemo.

Just be open and honest with her and tell her how you feel. It is a high risk high reward strategy that will 99.9999% backfire i.e. don't try it.

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