On Mon, 29 Jan 2007 kwooley@purdue.edu wrote:

>
> On question one of assignment 2, I am confused as to how the current
> will flow through this circuit, and thus how to draw the loops for
> Kirchoff's law.
>
> From your drawing in class, it seems there will be two loops, one with a
> capacitor and inductor with the given current, and another with the capacitor
> and resistor.  From the handout it seems current may be drawn through the
> Vout wire, or does this not form a loop because, as you said this is not the
> same "ground" as the others? Also, could you explain why a wire lead that
> connects to ground forms a loop (an inductor)?  Thanks,


Date: Mon, 29 Jan 2007 14:24:34 -0500 (EST)
From: Timothy Matthew Jones <jones105@physics.purdue.edu>
To: kwooley@purdue.edu
Cc: mjones@physics.purdue.edu
Subject: Re: 536 hw question

Remember that you can connect all the gound symbols together.  The result
is that the current drawn in the loop flows up through the inductor to get
back to the voltage source on the integrated circuit.  In practice, what
happens is the charge on the capacitor has to flow back through the device
to get to ground.  Indeed, there are two loops in the circuit as it is
drawn, but I have given you the form of the current in the loop that has
the inductor in it.

The wire lead connected to ground forms an inductor because of the (small)
magnetic field that the loop induces when current flows through it.  If
the lead is very short then the inductance is very small, but if the lead
is longer, or the current being carried on the loop changes very rapidly,
then the effects of the inductance is larger.

------------------------------------------------------------------------

On Thu, 1 Feb 2007 mgoodwin@purdue.edu wrote:

> I am confused about how to go about solving #3.  Should we simply try to
> maximize the voltage across RL?  In class you gave the equation V2=MVR2/
> (R1L2+R2L1), but I am confused how to determine what L1/L2 ratio will
> maximize the power across RL, and I really have no idea how to relate the
> number of turns to the power.  I have looked over the notes and read the
> book, but things just aren't clicking.  Any help would be appreciated.

------------------------------------------------------------------------

From: Timothy Matthew Jones <jones105@physics.purdue.edu>
Date: Fri, 2 Feb 2007 08:55:01 -0500 (EST)
To: mgoodwin@purdue.edu
Cc: mjones@physics.purdue.edu
Subject: Re: homework #2

Hi Mike,

First, you need to calculate the power dissipated by the load resistor,
RL.  This will just be I^2 RL where you calculate the current in the
second loop using Kirchoff's laws, etc...

You can express this power entirely in terms of the ratio L1/L2.  If you
call this ratio J, then to maximize the power you need to find the value
of J for which dP/dJ = 0.

Recall that L is proportional to the square of the number of turns on an
inductor.  That's what we worked out for the ideal solenoid and other
inductors are basically the same except for different geometric factors.
Basically, the magnetic field is proportional to N and the induced
electric field is proportional to N so the electric field induced by a
current is proportional to N^2.


Let me know if you still have qustions...