Finding the impedance of circuits.

  • Thread starter Thread starter RobLikesBrunc
  • Start date Start date
  • Tags Tags
    Circuits Impedance
Click For Summary
The discussion revolves around understanding and calculating impedance in circuits, particularly for a junior circuits lab assignment. Participants confirm the correctness of the impedance problems and discuss the formula for calculating the phase of Z1, which is arctan(y/x). They emphasize the importance of considering the signs of x and y to determine the correct quadrant for the angle. A minor mistake in notation is noted, but it is quickly corrected. Overall, the conversation highlights the critical aspects of impedance calculations and phase plotting in circuit analysis.
RobLikesBrunc
Messages
29
Reaction score
0
I'm currently taking a junior circuits lab that expects familiarity with impedance (my intro E/M course never covered it), so I just want to make sure I did these two impedance problems correctly (the second one asks merely for a formula, so no calculation was done):

tFqniA0.png


t5RyQE3.png


Later in the assignment, I'm also asked the plot the phase of Z1 as a function of frequency. For that (letting Z1 = x + yi), the phase is just arctan(y/x), right?

Thanks.
 
Physics news on Phys.org
Seems all good to me.
 
RobLikesBrunc said:
Later in the assignment, I'm also asked the plot the phase of Z1 as a function of frequency. For that (letting Z1 = x + yi), the phase is just arctan(y/x), right?

Thanks.

Looks good to me also.

Caution: in computing arc tan (y/x), be sure to retain the signs of x and y to determine in which quadrant the angle belongs. Arc tan b/(-a) is not the same angle as arc tan (-b)/a.
 
rude man said:
Looks good to me also.

Caution: in computing arc tan (y/x), be sure to retain the signs of x and y to determine in which quadrant the angle belongs. Arc tan b/(-a) is not the same angle as arc tan (-b)/a.

There was actually a little mistake--on the last line I replaced the -i with -1 :P

Anyway, I was actually plotting the phase over 10^6 frequency range, so which quadrant to place it in is a non-issue. It looks pretty neat too (note the x-axis is logarithmic):

4RHtjSy.png
 
Whoops, messed up a sign. Should be this:

yD2t7Ia.png
 
RobLikesBrunc said:
Whoops, messed up a sign. Should be this:

yD2t7Ia.png

I was a about to draw your attention to that! Now it's OK.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 20 ·
Replies
20
Views
3K
Replies
5
Views
2K
Replies
9
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
2
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K