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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


2 Answers 2


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

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

  • \$\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\$ Dec 28, 2016 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\$ Dec 28, 2016 at 17:18

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 =


>> A=156

A =


>> C=1e-6

C =


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

ans =



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