Difference between revisions of "Talk:EGR 224/Spring 2009"
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| + | == Typographical / Other Errors == | ||
| + | === Lab 1 === | ||
| + | * 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 1 == | == Homework 1 == | ||
=== 1.14 === | === 1.14 === | ||
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*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) | *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) | ** 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) | ||
| + | == General Questions == | ||
Revision as of 17:42, 13 January 2009
Contents
Typographical / Other Errors
Lab 1
- 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 1
1.14
- You should assume Passive Sign Convention. 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 5I0 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 "5I0" to find the voltage? Thanks! 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 \(5 I_0\) - the key here is to notice that \(I_0\) 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. DukeEgr93 12:10, 11 January 2009 (EST)
