0
\$\begingroup\$

enter image description here

How does the author obtain the expression I is approximately equal to .059sin(wt)

In the beginning of the book, 30 pages back, the author stated that V was equal to Asin(wt) (if you ignore phase), why is V now equal to Acos(wt)?

More importantly though, How does I = V/Z bring in sin(wt) when V nor Z had sin in it? Is sin(wt) equal to j where there is a 90 degree phase shift only?

Thanks for any help. This is a real great site and hopefully I will get through Chapter 1 of this book. Which should be downhill from there (I think).

I also don't calculate .059. I get .029 at best assuming w=2*3.14*f

\$\endgroup\$
3
\$\begingroup\$

First, the author mentions that, "The phase of the voltage is arbitrary," so he's defining the source voltage waveform as a cosine wave. The current function I(t) will be referenced to this function.

Next, since \$V(t) = Acos(\omega t)\$ and \$Z = -j/\omega C\$, then

$$I(t) = V/Z = A \omega Ccos(\omega t) / -j = j A \omega C cos(\omega t) $$

Since a multiplication by \$ j \$ represents a 90 degree phase shift, we yield

$$I(t) = A \omega C sin(\omega t) $$

which is what the book describes as well.

Now for the 0.059 coefficient: just check your math. You're correct in \$ \omega = 2 \pi f\$ where \$f = 60 Hz \$, thus

$$ A \omega C = 156 (2 \pi 60) 1e ^{-6} = 0.059$$

\$\endgroup\$
  • \$\begingroup\$ To me, the line [a multiplication by j represents a 90 degree phase shift] is key. As I understand it, that easily turns cos into sin. If it was a 180 degree phase shift, can multiply by j^2 or -1. How about a 135 degree phase shift? Thanks \$\endgroup\$ – Jeffrey Edward Messikian Dec 28 '16 at 17:14
  • \$\begingroup\$ You can use trigonometry to figure out a 135 degree shift by drawing your Cartesian coordinate system with j on the y-axis and 1 on the x-axis (hint: it'll be a negative real component with a positive imaginary one). Unfortunately your answer won't be as clean with a phase shift not equal to n*90 degrees where n is an integer. \$\endgroup\$ – calcium3000 Dec 28 '16 at 17:18
0
\$\begingroup\$

It's pretty crappy notation. In one half of the equation, they make f=60Hz, but don't make that clear inside of the argument for sin. As for the 0.059:

>> omega=2*pi*60

omega =

  376.9911

>> A=156

A =

   156  

>> C=1e-6

C =

   1.0000e-06

>> abs(j*omega*A*C)

ans =

    0.0588

>> 
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.