Help with a Current question 3

[dhambly1]

Dear All

Im having trouble with a question involving current flow.

I = I1 + I2 in the form R sin(wt+ a) and state amplitude and phase angle.
I1 = 0.02 sin wt and I2 = 0.032 cos (wt- pi/3)

Could someone point me in the right direct please as its been many years since ive done any of this.

thanks

 

[cuky2000]

The easy analitical way that I can suggest is to use the trigonometric method or a phasor notation with the max. amplitude values without converting to RMS :

1-Convert the cosine funtion to a sine function using a supplementary. Angle2= pi/2-pi/3=pi/6 =30 deg.

I2= 0.032 cos (wt- pi/3)= 0.032*sin(wt+pi/6).

2- Calculate the max. amplitud of both sine functions.

A2 = 0.032*cos(pi/3)+ 0.032*sin(pi/6)*j
A1 = 0.02*cos(0) + 0.02*sin(0 )*j _________________________________________________ A1+A2 = 0.048 + 0.016*j

|A1+A2| = SQRT(0.048^2+0.016^2)]=0.5032.

3.- Determine the phase shift angle between both sine functions:
Angle = arctg(0.016/0.048)= 18.54 deg.~0.1*pi

I1+I2 = |A1+A2|*sin(wt+Angle)=0.5032*sin(wt+18.54deg).
I1+I2 ~ 0.5*sin(wt+0.1pi).

NOTE: If you have an utility program such as MathCad, MathLab or perhaps in Excel, the result will be even easier using specific values or provide a plot of the functions in any range.

 

[230842]

This is the complete and detailed development allowin to get values of amplitude C and angle c of general expresion:

A.sin(wt+a) + B.sin(wt+b) = C.sin(wt+c)

First, devlope binomials of sines:

A.sin(wt+a) + B.sin(wt+b) = A(sin wt.cos a + cos wt.sin a) + B(sin wt.cos b + cos wt.sin b)

Grouping term in sin and cos:

= (A.cos a + B.cos b).sin wt + (A.sin a + B.sin b).cos wt = C.sin(wt+c) = C.cos c.sin wt + C.sin c. cos wt

Let's make now:

(A.cos a + B.cos b) = C.cos c and
(A.sin a + B.sin b) = C.sin c

Dividing both equations we get angle:

tan c = (A.sin a + B.sin b)/(A.cos a + B.cos b)

And adding both equations, but squared, and taking into account that te sum of squares of sin and cos equals 1:

C^2 = A^2 + B^2 + 2.AB.(cos a cos b + sin a sin b)

So, finnally the amplitude becomes:

C = sqrt[A^2 + B^2 + 2.A.B.cos(a-b)]
Julian

 

[stevenal]

It all depends on the tools available. Does your calculator or software do complex calculations in polar form? Simply use a trig identity to convert the cos term to a sin term. Create phasors out of both values, add, and convert answer back to the time domain.

Maybe your tool will not add complex values in polar form. Then it probably has a polar to rectagular conversion utilty. Add in rectangular form and convert back to polar.

If your tool does not do complex calculations at all, you are stuck with the methods given above. The first one is easier, but the magnitude seems to be off by an order of magnitude.

Here is what you I get:

0.02+0.032*e^(jPI/6)=0.050*e^j0.32

In the time domain:

0.050*sin(wt+0.32)