Difference between revisions of "Talk:EGR 224/Spring 2009"
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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) | + | == Homework 1 == |
| + | *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) | ||
Revision as of 17:10, 11 January 2009
Homework 1
- 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)
