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| − | + | By: [[ECE]] student Paul Wonnacott | |
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| + | Topic: Current Division | ||
| + | |||
| + | |||
| + | ---- | ||
| + | ==Question== | ||
| + | Determine the current I<sub>R</sub> in the figure below. | ||
| + | https://www.projectrhea.org/rhea/images/8/80/Current_Division.pdf | ||
| + | ---- | ||
| + | ===Answer === | ||
| + | First, you can ignore the 2 ohm resistor since all the current needs to go through there. Then, apply the current division formula to the other two resistors. Take the 3 ohm resistor and divide by the sum of the 3 and 6 ohm resistors, and multiply this quantity by the 12 amps from the source. The answer is 4 amps. | ||
| + | |||
| + | |||
| + | ---- | ||
| + | ==Questions and comments== | ||
| + | If you have any questions, comments, etc. please post them below | ||
| + | |||
| + | ---- | ||
| + | [[2015 Spring ECE 201 Peleato|Back to 2015 Spring ECE 201 Peleato]] | ||
| + | |||
| + | [[ECE201|Back to ECE201]] | ||
Revision as of 13:34, 29 April 2015
Contents
Paul_Wonnacott_Current_Division_ECE201S15
By: ECE student Paul Wonnacott
Topic: Current Division
Question
Determine the current IR in the figure below. https://www.projectrhea.org/rhea/images/8/80/Current_Division.pdf
Answer
First, you can ignore the 2 ohm resistor since all the current needs to go through there. Then, apply the current division formula to the other two resistors. Take the 3 ohm resistor and divide by the sum of the 3 and 6 ohm resistors, and multiply this quantity by the 12 amps from the source. The answer is 4 amps.
Questions and comments
If you have any questions, comments, etc. please post them below
Back to 2015 Spring ECE 201 Peleato
