Lesson 2 - Source Transformations, Part 2 (Engineering Circuits)
Math and Science
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20 Comments
@eccleziassenat3262 6 years ago
Man we need the rest of it ahaahahaha.
@arisbautista4510 1 year ago
If you guys haven't solved it yet, the people that calculated here just solved for Req of 20||5, resulting to 4 ohms. With the 4 ohms, we then do source transformation with the 8A again, so 8A*4 ohms will result to 32v.
After that, perform mesh equivalent on the circuit, so by now we have 6v on the left side and 32v on the right side, both facing upwards, but we will assume the current is flowing clockwise.
The Mesh 1 equation will be:
34i1-30i2=6v
The Mesh 2 equation will be:
-30i1+50i2=-32v
Now when you use the elimination method, we can multiply 5 on the Mesh 1, and 3 on the Mesh 2 to eliminate the i2. Thus,
170i1-150i2=30v
-90i1+150i2=-96v
Now from here solve for i1
80i1=-66v
U will get i1=-0.875 which is really negative I don't know how it is positive with the others.
Hope this helps!
After that, perform mesh equivalent on the circuit, so by now we have 6v on the left side and 32v on the right side, both facing upwards, but we will assume the current is flowing clockwise.
The Mesh 1 equation will be:
34i1-30i2=6v
The Mesh 2 equation will be:
-30i1+50i2=-32v
Now when you use the elimination method, we can multiply 5 on the Mesh 1, and 3 on the Mesh 2 to eliminate the i2. Thus,
170i1-150i2=30v
-90i1+150i2=-96v
Now from here solve for i1
80i1=-66v
U will get i1=-0.875 which is really negative I don't know how it is positive with the others.
Hope this helps!
@estevangtz6041 6 years ago (edited)
Coming to comments for video solutions? Lol same .... after video paused through parallel series of resistors on right side then a source transformation on right side then did mesh equations. You can cancel out one variable and I got i1=0.825A . Looking at other comments that's what others got, but yea. Goodluck with your exam
@osamamahmoud7170 4 years ago
Hello , I wanna thank you I really learnt a lot from your great lessons sir , It's Osama Mahmoud From Egypt Here
@wltkrisha 7 years ago (edited)
i have a question.so at the first step you only considered the 5 ohm resister and 40v source to calculate the current source.so from that,you are finding the maximum current that can be provided from the 40v source? no need to consider the other components to calculate the current source?so basically i'm asking is that,for example,40v is connected with R1 and R2 resistors.all the components are in series.so to calculate how much current the voltage source produce,is it enough to consider R1 or R2 only?
@arghyabanerjee3044 5 years ago
Is it possible without source transformation and without applying kcl/kvl solve this equation? only using ohm's law does this equation solve.if possible then please make a video to solve only using ohm's law.
Lesson 2 - Source Transformations, Part 2 (Engineering Circuits)
Math and Science
111,009 views
This is just a few minutes of a complete course.
Get full lessons & more subjects at: http://www.MathTutorDVD.com.