00:01
All right.
00:04
So for this problem, we have the lrc series circuit and we have the parameter for lrc.
00:13
We know the rm's current and the frequency.
00:18
So for part a, what is the p.
00:23
P.
00:23
Fault factor for the circuit? so the power factor is cosine 5 equal to r over z.
00:35
And z here is the impedance.
00:39
So to calculate z, it's r squared plus omega l minus one over omega c.
00:55
So this is the definition for the impedance.
01:00
Here, omega equal to two, five.
01:05
Because omega is the angular frequency, and f, which is 4 .00 in this problem.
01:14
And the unit plug is omega here.
01:18
You already know the r, you know, l, you know c, so we can find out the value for z.
01:24
The value i find is 3, 24, 5, homes.
01:39
So now that we have the z, we also have r given the problem, so we can find out causing pi.
01:46
Pi is 0 .6967...