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| − | [[Category: | + | [[Category:ECE201]] |
| + | [[Category:ECE]] | ||
| + | [[Category:ECE201Spring2015Peleato]] | ||
| + | [[Category:circuits]] | ||
| + | [[Category:linear circuits]] | ||
| + | [[Category:problem solving]] | ||
| − | = | + | =Voltage Division Practice= |
| + | <center><font size= 4> | ||
| + | '''Practice question for [[ECE201]]: "Linear circuit analysis I" ''' | ||
| + | </font size> | ||
| + | By: Chinar Dhamija | ||
| + | Topic: Voltage Division | ||
| − | + | </center> | |
| + | ---- | ||
| + | ==Question== | ||
| + | Find V1. | ||
| + | [[File:ECE201_P2.png|700px|center]] | ||
| + | ---- | ||
| + | ---- | ||
| + | ===Answer === | ||
| + | Bottom of the circuit is considered to be ground.<br /> | ||
| + | First we need to find Vx. Vx can be found by voltage division.<br /> | ||
| + | Vx = 24(4 / (6 + 4)) | ||
| + | = 9.6 V | ||
| + | Now, find the voltage at the independent source. | ||
| + | = .75 * 9.6 | ||
| + | = 7.2 | ||
| + | Finally, solve for V1 by using voltage division again. | ||
| + | V1 = 7.2(3 / (3 + 2)) | ||
| + | '''= 4.32 V''' | ||
| + | ---- | ||
| + | ==Questions and comments== | ||
| + | If you have any questions, comments, etc. please post them below | ||
| + | *Comment 1 | ||
| + | **Answer to Comment 1 | ||
| + | *Comment 2 | ||
| + | **Answer to Comment 2 | ||
| + | ---- | ||
| + | [[2015 Spring ECE 201 Peleato|Back to 2015 Spring ECE 201 Peleato]] | ||
| − | + | [[ECE201|Back to ECE201]] | |
| − | [[ | + | |
Latest revision as of 15:31, 26 April 2015
Voltage Division Practice
Practice question for ECE201: "Linear circuit analysis I"
By: Chinar Dhamija
Topic: Voltage Division
Question
Find V1.
Answer
Bottom of the circuit is considered to be ground.
First we need to find Vx. Vx can be found by voltage division.
Vx = 24(4 / (6 + 4))
= 9.6 V
Now, find the voltage at the independent source.
= .75 * 9.6
= 7.2
Finally, solve for V1 by using voltage division again.
V1 = 7.2(3 / (3 + 2))
= 4.32 V
Questions and comments
If you have any questions, comments, etc. please post them below
- Comment 1
- Answer to Comment 1
- Comment 2
- Answer to Comment 2
