00:01
All right.
00:03
So for this problem, we have the arc series circuit.
00:08
And we know the capacitance, window inductance, and also window resistance.
00:13
So we also know the amplitude of the source is 15 watts.
00:18
And for part a, so we want to find out under what circumstances is the average power delivered to the circuit equal to half of vrms times irms.
00:30
Okay.
00:31
So because we have this.
00:32
Half in front of the vrms times rms.
00:36
So we know cosine 5, which is the power factor equal to half.
00:45
And power factor also equal to r over z.
00:52
So we can solve for z in this case should be equal to 150 homes.
01:03
All right.
01:07
And then we should solve for the angular frequencies to reach this impedance for the circuit.
01:14
We know the definition of z is square root r squared plus omega l minus one over omega c squared.
01:30
So this is z which equals to 150 in this problem.
01:36
And in this expression we are already know r which is 75 oms.
01:41
And the window l and the c also is right here.
01:45
So the only unknown variable is the omega.
01:48
So we can solve this omega by solving this equation.
01:52
And the values i found for omega, so there are two positive values...