RL Circuit Refresher: Solve Total Current

Click For Summary
The discussion revolves around calculating the total current in a parallel RL circuit. The original poster attempted to use the formula for total current but encountered discrepancies in their results. It was pointed out that admittances in parallel should be added, and a mistake in the placement of the imaginary unit 'j' was identified, which affected the calculations. The conversation highlights the importance of careful algebraic manipulation in circuit analysis. Overall, the thread emphasizes the need for accuracy in applying formulas for electrical circuits.
jendrix
Messages
120
Reaction score
4
Hi, I have just been refreshing my theory and was going over a parallel RL circuit, like this one.

vFu2isf.png


The site I was reading had this for the total current

wSOg5fX.png


I tried some values as a test but it doesn't give me the correct answer, for total current. eg if I set
V=10V
R=2.5ohm
jXl=5ohm

Then IR=10/2.5 =4A

IL=10/5=2A

IT=√42 +22 =4.47A

However when I try the formula from the second answer I get 10/(2.5-2j) which following the maths gives me 2.44+1.95j=3.12A

Am I going wrong somewhere or is it the formula?Thanks
 

Attachments

  • vFu2isf.png
    vFu2isf.png
    32.7 KB · Views: 881
  • wSOg5fX.png
    wSOg5fX.png
    8 KB · Views: 878
Engineering news on Phys.org
jendrix said:
However when I try the formula from the second answer I get 10/(2.5-2j) which following the maths gives me 2.44+1.95j=3.12A

Admittances in parallel add. That should be your approach.

upload_2018-8-8_21-15-30.png

A 'Site', you said ? If it's on the internet it must be right, eh ? See my signature.....

see if that works any better.

old jim
 

Attachments

  • upload_2018-8-8_21-15-30.png
    upload_2018-8-8_21-15-30.png
    6 KB · Views: 420
  • Like
Likes jendrix and Delta2
Oops ----- make that it equals Vin X ( 1/R -j1/ωL )

i put my j in wrong place . :cry:
 
  • Like
Likes Delta2
jim hardy said:
Oops ----- make that it equals Vin X ( 1/R -j1/ωL )

i put my j in wrong place . :cry:

Seems we can't even trust Old Jim when he posts at the internet with a computer, cause we can never trust a computer with anything important :-p

Ok seriously speaking that was a minor mistake, I didn't notice it either, I just noticed the huge mistake in the algebra in the OP.
 
  • Like
Likes jim hardy
i do that all the time. I'm just clumsy.

Thanks @Delta2
 
  • Like
Likes jendrix
jim hardy said:
i do that all the time. I'm just clumsy.

Thanks @Delta2

Thanks Jim for clearing that up
 
  • Like
Likes jim hardy and Delta2
I am trying to understand how transferring electric from the powerplant to my house is more effective using high voltage. The suggested explanation that the current is equal to the power supply divided by the voltage, and hence higher voltage leads to lower current and as a result to a lower power loss on the conductives is very confusing me. I know that the current is determined by the voltage and the resistance, and not by a power capability - which defines a limit to the allowable...

Similar threads

Replies
12
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
14
Views
4K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
12K
  • · Replies 1 ·
Replies
1
Views
1K