Solving I2 in polar form

  • Thread starter noname1
  • Start date
  • #1
134
0
I got this equation

5<0° = -14.14<-45° + 2.24<116.6° I2

And i solved I2 this way

I2 = -14.14<-45° / 2.24<116.6°

I2 = 8<150.29°

I want to know complex numbers is the same way as normal math or not

thanks in advance
 

Answers and Replies

  • #2
berkeman
Mentor
59,419
9,539
I got this equation

5<0° = -14.14<-45° + 2.24<116.6° I2

And i solved I2 this way

I2 = -14.14<-45° / 2.24<116.6°

I2 = 8<150.29°

I want to know complex numbers is the same way as normal math or not

thanks in advance

What is I2? Is it something special, or just a term multiplying the 2nd number on the righthand side (RHS) of the equation?

If the latter, I think it would be more traditional to write it as:

5<0° = -14.14<-45° + I2 * 2.24<116.6°
 
  • #3
134
0
i2 is being multiplied, its a current in the second mesh but i just wanted to be sure if it is that way its calculated
 
  • #4
berkeman
Mentor
59,419
9,539
i2 is being multiplied, its a current in the second mesh but i just wanted to be sure if it is that way its calculated

In that case, I don't think you've solved for I2 correctly. For one thing, the LHS term disappeared?

To add the two terms on the RHS, you would convert from polar to rectangular form, and add the x and y components. Then represent the LHS in rectangular components (that part is easy), and solve for I2.
 
  • #5
134
0
In that case, I don't think you've solved for I2 correctly. For one thing, the LHS term disappeared?

To add the two terms on the RHS, you would convert from polar to rectangular form, and add the x and y components. Then represent the LHS in rectangular components (that part is easy), and solve for I2.


what do you mean by lhs and rhs? i am kind of lost
 
  • #6
berkeman
Mentor
59,419
9,539
what do you mean by lhs and rhs? i am kind of lost

I defined the term RHS in my post #2.
 
  • #7
134
0
I defined the term RHS in my post #2.

this is what i did, i put i2 term on left side and than the rest on the rhs and since i want I2 i divided both equations by 2.24<116.6°

5<0° = -14.14<-45° + I2 * 2.24<116.6°

5<0° + 14.14<-45° = I2 * 2.24<116.6°

(5<0° + 14.14<-45°) / 2.24<116.6° = I2

I2 = 8<150.29°
 
  • #8
berkeman
Mentor
59,419
9,539
this is what i did, i put i2 term on left side and than the rest on the rhs and since i want I2 i divided both equations by 2.24<116.6°

5<0° = -14.14<-45° + I2 * 2.24<116.6°

5<0° + 14.14<-45° = I2 * 2.24<116.6°

(5<0° + 14.14<-45°) / 2.24<116.6° = I2

I2 = 8<150.29°

How are you adding terms in polar form?
 
  • #10
134
0
How are you adding terms in polar form?

this part

(5<0° + 14.14<-45°) / 2.24<116.6° = I2

i have my ti-89 calculating it but i just want to know if it is correct if i can calculate polar form like a normal equation

example

3x -5 = 10
3x = 15
x=5
 
  • #11
berkeman
Mentor
59,419
9,539
this part

(5<0° + 14.14<-45°) / 2.24<116.6° = I2

i have my ti-89 calculating it but i just want to know if it is correct if i can calculate polar form like a normal equation

example

3x -5 = 10
3x = 15
x=5

LOL. Your calculator is doing the conversions for you (hopefully). You might want to double check the answer the good old fashioned way, by hand, just to be sure you understand the process.
 

Related Threads on Solving I2 in polar form

  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
1
Views
3K
Replies
1
Views
2K
  • Last Post
Replies
7
Views
5K
Replies
9
Views
2K
  • Last Post
Replies
17
Views
2K
Replies
1
Views
1K
Replies
0
Views
3K
Replies
2
Views
2K
  • Last Post
Replies
1
Views
1K
Top