00:01
Capacitors placed in parallel with some device b is in figure 3018, part b.
00:06
To filter out stray high frequency signals, but to allow ordinary 60 hertz ac to pass through with low loss.
00:12
Suppose that circuit b in figure 30 -18b is a resistance r equals 490 oms connected to ground, and that c equals 0 .35 mu .f.
00:24
What percent of the incoming current will pass through c rather than r if it is 60, and then we'll do 60 ,000 hertz.
00:35
So starting with 60, to get the current to the capacitor, we'll use the formula i subc equals v sub rms over xc.
00:47
And if we first compute the reactants of the capacitor at 60 hertz, then xc equals 1 over wc, which is the same as 1 over 2 pi f times c.
01:02
And so if we substitute knowing that c equals 0 .35 times 10 to the negative 6 at 60 hertz, we'll have xc equals 1 over 2 pi times 60 hertz...