Art Of Electronics Second Edition - Math Question - Complex Number Current Calculation

Question

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

Answer 1

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

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

Answer 2

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

>>