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: [itex]V=L\frac{di}{dt}[/itex]
And together with the resistor: (b) [itex]V=L\frac{di}{dt}+iR[/itex]
[tex]\Rightarrow\mbox{ (c) }\frac{di}{dt}=\frac{V}{L}-\frac{R}{L}i[/tex]

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:

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

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

[tex]\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)[/tex]

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

[tex]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)[/tex]

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

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

 

[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 [tex] \LaTeX [/tex] 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.

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

Is just not right. Let's analyze it by looking at the extreme conditions where [itex] t=0 [/itex] and [itex] t = \infty [/tex].

When [itex] t \rightarrow \infty [/tex], the exponential term approaches 0, leaving

[tex]
V_{ac}(\infty)=\frac{V_{dc}R}{\left(R_{0}+R \right) },
[/tex]

which makes sense. The inductor becomes a short circuit in steady state, and we're just left with a voltage divider. So far the answer makes sense. But now let's look at what happens when [itex] t = 0 [/itex]. Here the exponential term becomes 1, leaving

[tex]
V_{ac}(0)=\frac{2V_{dc}R}{\left(R_{0}+R\right)},
[/tex]

which is just silly. At the instant of time [itex] t = 0 [/itex], 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,

[tex]
V_{ac}(0)=V_{dc}.
[/tex]

////////////////////////////////
// Finding [itex] di/dt [/itex]
////////////////////////////////

I can't give you the actual answer to this, because according to the rules of the forum, I can't just do your work for you. But allow me to point you in the right direction.

Note that,

[tex] \frac{d}{dt}(e^{\alpha t}) = \alpha e^{\alpha t} [/tex]

Similarly, as another example,

[tex] \frac{d}{dt}\left( A(1-e^{-\alpha t}) \right) = \alpha Ae^{\alpha t} [/tex]

So if you have an expression for i , take its derivative with respect to t, and that becomes [itex] di/dt [/itex].

//////////////////////////////
// Correcting the book's
// mistake.
/////////////////////////////

Normally, I couldn't or wouldn't give the final answer, per the forum rules. But since this situation is correcting the mistake in a textbook, I feel it's justified. Moderators: please let me know if I am out-of-line here. I will retract this book's-correction if you feel it is not justified.

I haven't thoroughly double checked my answer, but I am somewhat confident that it is correct. I believe the book should have given you the answer of:

[tex] V_{ac} = \frac{V_{dc}}{R + R_0} \left(R + R_0 e^{-(R + R_0)t/L} \right) [/tex]

I'm guessing that the authors of the book got confused and thought that the R₀ in the second term's was R, and then factored the equation to give the answer that they gave you. But that's not right.

I'm somewhat confident that my corrected answer is correct. At time [itex] t=0 [/itex], the exponential becomes 1. Simple algebra shows that [itex] V_{ac}(0) = V_{dc} [/itex]. Similarly at [itex] t = \infty [/itex] 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.