Question

Matlab

Answer 1

@UnkleRhaukus

Answer 2

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Answer 3

**math lab

Answer 4

% defining input variables for RLC circuit

f_volt = 60; % measured in hertz
W_volt = 120; % measured in volts rms
R = 100; % measured in ohms


L_o = 0.005:0.005:0.245; % matrix L 2D matrix incr. from left to right
C_o = 5e-6:10e-10:9.5e-5; % matrix C 2D matrix incr. from top to bottom

% current output for RLC circuit (I) in RLC circuit (note: j = sqrt(-1))
I = 120. / ((100 + j*2*pi(f_volt).*L_o - (j / 2*pi*(f_volt).*C_o))


% total power ouput in RLC circuit
%

Answer 5

My main concern is if the L_o and C_o are setup correctly?

Answer 6

so I guess my matrices don't agree in dimension so...

Answer 7

[x-new, y-new] = meshgrid(x,y) ?

Answer 8

ahhh okay. Why wouldn't we use meshgrid?

Answer 9

if you transpose cap the dimensions don't work

Answer 10

nvm i will figure it out, night!

Answer 11

it's not night time

Answer 12

yes it is.

Answer 13

were?

Answer 14

Did you mean:
>> I = 120. / ((100 + j*2*pi(f_volt)*L_o - ((j) / (2*pi*(f_volt)*C_o))))
Error using pi
Too many input arguments.

Answer 15

slacker

Answer 16

% defining input variables for RLC circuit

f_volt = 60; % measured in hertz
W_volt = 120; % measured in volts rms
R = 100; % measured in ohms

L_o = 0.005:0.005:0.245; % matrix L 2D matrix incr. from left to right
C_o = 5e-6:10e-10:9.5e-5; % matrix C 2D matrix incr. from top to bottom

[L_o, C_o] = meshgrid(L_o,C_o);

% current output for RLC circuit (I) in RLC circuit (note: j = sqrt(-1))
I = 120 / ((100 + j*2.*pi(f_volt)*L_o - (j / 2.*pi*(f_volt)*C_o))


% power observed by resistor
P_R = 100*abs(I.^2);

@UnkleRhaukus

Answer 17

@reemii

Answer 18

So for some reason my current equation isn't running smoothly

Answer 19

Look at the assignment above...

Answer 20

there's a missing parenthesis in `120 / ((100 + j*2.*pi(f_volt)*L_o - (j / 2.*pi*(f_volt)*C_o))`

Answer 21

`I = 120 / ((100 + j*2.*pi(f_volt)*L_o - (j / 2.*pi*(f_volt)*C_o))`
on that line

Answer 22

@reemii I don't know what the problem is...

Answer 23

my editor says that there the parenthesis starting just after `120 /` is not closed.

Answer 24

>> I = 120. / ((100 + j*2*pi*f_volt*L_o) - (j. / 2*pi*f_volt*C_o))
I = 120. / ((100 + j*2*pi*f_volt*L_o) - (j. / 2*pi*f_volt*C_o))
↑
Error: Unexpected MATLAB operator.

Answer 25

I closed it and now it says the / is not valid operator

Answer 26

I = 120. / (100 + j*2.*pi*f_volt*L_o - (j. / (2.*pi*f_volt*C_o)))

Answer 27

do i have to put a . right after the numerator when I am dividing by a matrix?

Answer 28

I don't know. What is the matrix there?

Also, I don't understand the meaning of the `j.` in the denominator . Why the dot?

Answer 29

http://www.mathworks.com/matlabcentral/newsreader/view_thread/74362?requestedDomain=www.mathworks.com

Answer 30

L_o = 0.005:0.005:0.245; % matrix L 2D matrix incr. from left to right
C_o = 5e-6:10e-10:9.5e-5; % matrix C 2D matrix incr. from top to bottom

Answer 31

j is an imaginary number instead of using i. Do I need to use i? Even if I use i the code shows a problem

Answer 32

>> lab03
Error using /
Matrix dimensions must agree.

Error in lab03 (line 18)
I = (120)/(100 + 1i*2*pi*60*L_o - ((1i)/(2*pi*60*C_o)));

Answer 33

`I = 120 / ((100 + j*2.*pi(f_volt)*L_o - (j / 2.*pi*(f_volt)*C_o)) ` without `j.`

`I = 120 / ((100 + j*2.*pi(f_volt)*L_o - (j. / 2.*pi*(f_volt)*C_o))` with `j.`

Try the first one. Also, are you sure you want `2.*pi*(f_volt)*C_o)` and not `(2.*pi*(f_volt)*C_o))` ?

Answer 34

^as you can see above i made all of those corrections

Answer 35

`120. /` is not `120 ./` . You wrote `120. /`, did you try `120 ./` ?

Answer 36

okay well that worked.

Answer 37

so when I run the code I just get an array of calculations that go on forever how can i stop it ?

Answer 38

oh nvm it stopped

Answer 39

% power obsorbed by resistor
P_R = 100*abs(I.^2);

% stores max power inside P_R matrix
max_PR = max(P_R);

so how do I go about solving this part?
Use your results to report which specific value for L and value for C resulted in max PR. Store
these two results in the variables max L and max C, respectively. The find function should be
instrumental here

Answer 40

% power obsorbed by resistor
P_R = 100*abs(I.^2);

% stores max power inside P_R matrix
max_PR = max(P_R);

Max_L = find(max(L_o))

???