00:01
For this problem on the topic of alternating current, we have an ac generator with an emf, that's epsilon m, sine omega -t, with the emf amplitude being 25 volts, and the driving angular frequency 377 radiance per second.
00:14
It is connected to a 12 .7 -henry inductor, and we want to find the maximum value of the current, the emf of the generator at this maximum current, and when the emf of the generator is minus 12 .5 volts and increasing in magnitude we want to again find the current.
00:34
Now the circuit consists of one generator across one inductor and so we have epsilon the emf to be the voltage across the inductor vl and so the current amplitude i is epsilon m divided by the inductive reactance xl which is epsilon divided by the driving angular frequency omega d times the inductance l putting in our putting in our values this is 25 volts divided by 377 radians per second times an inductance of 12 .7 henries which gives us the current amplitude of 5 .22 times 10 to the minus 3 ampheres for part b when the current is at a maximum limits derivative is 0, and so from equation 35, we get the emf, epsilon, l to be 0 at that instant.
01:42
In another way, since epsilon and i have a 90 -degree phase difference, then epsilon must be 0 when the current is equal to i, and the fact that the phase angle 5's pi over 2 radiance can be used in the following question.
02:05
Now we'll consider equation 28 with emf epsilon equal to the emf amplitude minus epsilon m divided by two...