A conduction coil and 2 resistors

[Karol]

Homework Statement

a conduction coil with 2 resistors are connected, according to the drawing. write the expression for the potential difference V_ac over the coil after t seconds after the switch is closed.

Homework Equations

The current at any time: (a) i=I(1-e^-Rt/L)

The electro motive force (votage) at a pure coil: V=L\frac{di}{dt}
And together with the resistor: (b) V=L\frac{di}{dt}+iR
\Rightarrow\mbox{ (c) }\frac{di}{dt}=\frac{V}{L}-\frac{R}{L}i

The voltage is the sum of the one on the coil plus the one on the resistor R₀:
(d) V_ac=V_ab+V_bc

The Attempt at a Solution

The current in the hole circuit is:

\mbox{(e) }i=\frac{V}{R+R_{0}}\left(1-e^{-(R+R_{0})t/L}\right)

I use the last aquation (e) and insert into (c), instead of the current i:

\mbox{(f) }\frac{di}{dt}=\frac{V}{L}-\frac{(R+R_{0})}{L}\frac{V}{(R+R_{0})}}\left(1-e^{-(R+R_{0})t/L}\right)=\frac{V}{L}\left(1+e^{-(R+R_{0})t/L}\right)

Finally, to get the required voltage on the coil+resistor R, i use, again, (b) and insert into (d) equations (e)+(f):

V_{ac}=V_{bc}+V_{ab}=V\left(1+e^{-(R+R_{0})t/L}\right)+\frac{VR}{(R+R_{0})}}\left(1-e^{-(R+R_{0})t/L}\right)

This does not lead to the required result, according to my book, Sears-Zemansky, 1965:

V_{dc}=\frac{V_{dc}R}{\left(R_{0}+R\right)}\left(1+e^{-(R+R_{0})t/L}\right)

 

[collinsmark]

Hello Karol,

Which are you supposed to find the potential for? The problem statement says V_ac. But the problem statement also says, "potential difference ... over the coil," which sort of implies V_bc. But then the book's answer gives V_dc.

By the way, I might see a problem with your derivation of di/dt, but I'd like to clear up the above first.

[Edit: Second, if the answer is supposed to be V_ac, I think your book's answer is wrong. I can explain in more detail later. Everybody makes mistakes, even textbook authors. It wouldn't be the first time a mistake has occurred in a textbook. But if you'd like, redo your derivation of di/dt, and you can compare your final V_ac answer against what I came up with (rather than finding di/dt by substituting it into anther equation, just take the derivative of i).]

[Another edit: Nice \LaTeX usage btw!]

 

[Karol]

Thanks, i worked hard, also with trial and error, with Latex.
The problem is finding V_ac, the coil's voltage plus the resistor R attached to it.
Please tell me your answer, and then say how do you find, directly, the derivative instead of inserting one formula into the other.
Again-Thanks.

 

[collinsmark]

Let me start by explaining the error in the book's final answer.

<br /> V_{ac}=\frac{V_{dc}R}{\left(R_{0}+R\right)}\left(1 +e^{-(R+R_{0})t/L}\right)<br />

Is just not right. Let's analyze it by looking at the extreme conditions where t=0 and t = \infty [/tex]. <br /> <br /> When t \rightarrow \infty [/tex], the exponential term approaches 0, leaving &lt;br /&gt; &lt;br /&gt; &amp;lt;br /&amp;gt; V_{ac}(\infty)=\frac{V_{dc}R}{\left(R_{0}+R \right) },&amp;lt;br /&amp;gt;&lt;br /&gt; &lt;br /&gt; which makes sense. The inductor becomes a short circuit in steady state, and we&amp;#039;re just left with a voltage divider. So far the answer makes sense. But now let&amp;#039;s look at what happens when t = 0. Here the exponential term becomes 1, leaving&lt;br /&gt; &lt;br /&gt; &amp;lt;br /&amp;gt; V_{ac}(0)=\frac{2V_{dc}R}{\left(R_{0}+R\right)},&amp;lt;br /&amp;gt;&lt;br /&gt; &lt;br /&gt; which is just silly. At the instant of time t = 0, no current flows through the circuit. That is because it is impossible to change the current through a inductor instantaneously (well, ignoring infinite voltages anyway). Since there is no current going through the resistors, their respective voltage drops are 0. This means correctly, &lt;br /&gt; &lt;br /&gt; &amp;lt;br /&amp;gt; V_{ac}(0)=V_{dc}.&amp;lt;br /&amp;gt;&lt;br /&gt; &lt;br /&gt; ////////////////////////////////&lt;br /&gt; // Finding di/dt&lt;br /&gt; ////////////////////////////////&lt;br /&gt; &lt;br /&gt; I can&amp;#039;t give you the actual answer to this, because according to the rules of the forum, I can&amp;#039;t just do your work for you. But allow me to point you in the right direction. &lt;br /&gt; &lt;br /&gt; Note that,&lt;br /&gt; &lt;br /&gt; \frac{d}{dt}(e^{\alpha t}) = \alpha e^{\alpha t}&lt;br /&gt; &lt;br /&gt; Similarly, as another example, &lt;br /&gt; &lt;br /&gt; \frac{d}{dt}\left( A(1-e^{-\alpha t}) \right) = \alpha Ae^{\alpha t}&lt;br /&gt; &lt;br /&gt; So if you have an expression for i , take its derivative with respect to &lt;i&gt;t&lt;/i&gt;, and that becomes di/dt.&lt;br /&gt; &lt;br /&gt; //////////////////////////////&lt;br /&gt; // Correcting the book&amp;#039;s&lt;br /&gt; // mistake.&lt;br /&gt; /////////////////////////////&lt;br /&gt; &lt;br /&gt; Normally, I couldn&amp;#039;t or wouldn&amp;#039;t give the final answer, per the forum rules. But since this situation is correcting the mistake in a textbook, I feel it&amp;#039;s justified. Moderators: please let me know if I am out-of-line here. I will retract this book&amp;#039;s-correction if you feel it is not justified. &lt;br /&gt; &lt;br /&gt; I haven&amp;#039;t thoroughly double checked my answer, but I am somewhat confident that it is correct. I believe the book &lt;i&gt;should&lt;/i&gt; have given you the answer of:&lt;br /&gt; &lt;br /&gt; V_{ac} = \frac{V_{dc}}{R + R_0} \left(R + R_0 e^{-(R + R_0)t/L} \right)&lt;br /&gt; &lt;br /&gt; I&amp;#039;m guessing that the authors of the book got confused and thought that the &lt;i&gt;R&lt;sub&gt;0&lt;/sub&gt;&lt;/i&gt; in the second term&amp;#039;s was &lt;i&gt;R&lt;/i&gt;, and then factored the equation to give the answer that they gave you. But that&amp;#039;s not right. &lt;br /&gt; &lt;br /&gt; I&amp;#039;m somewhat confident that my corrected answer is correct. At time t=0, the exponential becomes 1. Simple algebra shows that V_{ac}(0) = V_{dc}. Similarly at t = \infty we are left with the simple voltage divider, which is what we expect.

 

[Karol]

Thanks a lot for your reply, it taught me something.
I wish more people were acting like you, generally.
Thanks again-Karol.