View source for Talk:EGR 224/Spring 2009

== Typographical / Other Errors ==
=== Lab 1 ===
* Below the circuit drawing on p. 2, the statement about the number of equations should be:
<source lang="text">
 There are four elements and three nodes, meaning a total of 4 element equations, 
 2 independent KCL equations, and (4-3+1)=2 independent KVL equations.
</source>
* In Section 1.3, the circuit's voltage source is labeled using the '''active sign convention''' not the passive sign convention as recommended in the lab manual.  The equations are correctly written for the circuit, however, and the active sign convention '''is''' allowed to be used for sources.

== Homework ==
=== Homework 1 ===
==== 1.14 ====
*You should assume Passive Sign Convention. [[User:DukeEgr93|DukeEgr93]] 16:21, 11 January 2009 (EST)
==== 1.20 ====
*This question regards current controlled voltage sources. How do you find the voltage across one? Consider problem 1.20 in A&S. The source is labeled 5I<sub>0</sub> with a current of 3 A passing through it. I thought the voltage is just 5*3=15 V, but using Kirchhoff's Voltage Law the voltage definitely should be 10 V. How do you utilize "5I<sub>0</sub>" to find the voltage? Thanks! [[User:Brs16|Brs16]] 12:24, 10 January 2009 (EST)
** For the circuit in 1.20, the current controlled voltage source (CCVS) has an element equation that says the voltage drop is equal to <math>5 I_0</math> - the key here is to notice that <math>I_0</math> is the current through the 28 V element at the top of the circuit; it is listed as being 2 A.  Given that, the CCVS has a voltage drop of 10 V, which can be verified with KVL. [[User:DukeEgr93|DukeEgr93]] 12:10, 11 January 2009 (EST)

=== Homework 2 ===
====3.10 ====
*Once you get the equations symbolically, you can solve and subs using Maple.  You probably do not want to do this by hand. [[User:DukeEgr93|DukeEgr93]] 20:38, 17 January 2009 (EST)

=== Homework 3 ===
==== 3.38 ====
* How do the mesh currents relate to the 4 A source?
** Assuming you have drawn the mesh currents clockwise, <math>i_1</math> measures the current flowing clockwise through mesh 1; its component through the source flows from top to bottom; alternately, <math>i_2</math> measures the current flowing clockwise through mesh 2 and so its component is going from bottom to top of that source.  All of which means <math>i_2</math> is in the same direction as the source and <math>i_1</math> is in the opposite direction, so <math>i_{source} = i_{2} - i_{1}</math> [[User:DukeEgr93|DukeEgr93]] 17:59, 29 January 2009 (EST)

=== Homework 6 ===
==== 14.102 ====
*Over which element are we taking the cutoff frequency?  For example, in part (a) the circuit is reduced to a simple RC circuit.  To find the transfer function, we need to have a V_output, right? {{unsigned|njg6}}
** Turns out "which element" will not matter...  If you reduce it to two elements, then the half-power frequency will be the same for each.  [[User:DukeEgr93|DukeEgr93]] 18:26, 28 February 2009 (EST)

=== Homework 8 ===
==== 16.60 ====
*What should the value of the capacitor above the voltage source be? The circuit diagram shows 0.5 ''Ohm'' F. I'm assuming it's supposed to be ''mu'' like the other capacitor? [[User:MoZ|MoZ]] 14:28, 8 April 2009 (EDT)
** <math>\mu</math>F it is!  [[User:DukeEgr93|DukeEgr93]] 15:15, 8 April 2009 (EDT)

== General Questions ==