Kaplan problem 32 Afternoon exam
- Thread starter eng787
- Start date
Help Support Professional Engineer & PE Exam Forum:
cableguy
Has never sniffed a stink bug
Looks like they excluded the region from pi to 2pi in their integration, but used 2pi as the period.
So it looks like they did 1/(2pi) integral from alpha to pi of the function (2pi) sin theta d-theta
That gives us -cos(pi)-(-cos alpha) which = 1 + cos alpha
Since the thyristor cuts off at pi (zero crossing), I can see why they did it that way. It's a lot like NCEES Sample Exam 115. Imagine Sample Exam 115 but with half wave rectification. That's what it's like. The Kaplan voltage function is V(t) = 2*pi*sin(wt)
So it looks like they did 1/(2pi) integral from alpha to pi of the function (2pi) sin theta d-theta
That gives us -cos(pi)-(-cos alpha) which = 1 + cos alpha
Since the thyristor cuts off at pi (zero crossing), I can see why they did it that way. It's a lot like NCEES Sample Exam 115. Imagine Sample Exam 115 but with half wave rectification. That's what it's like. The Kaplan voltage function is V(t) = 2*pi*sin(wt)
Last edited by a moderator:
You can also solve this problem by realizing that the maimum conduction and thefore voltage would occur when the diode is conducting for the full half cycle as a half wave rectifier.
For a half wave rectifier the you can realize that the Vave of Vdc voltage is given by Vave=(Vm)(1/pi) where Vm is the peak magnitude of the AC voltage. This is given in books such as Camara's which has tables for the different Vave vales of different rectifiers.
So taking the peak of the AC sine wave to be 2pi as given in the problem we get Vave=(2pi)(1/pi)=2
For a half wave rectifier the you can realize that the Vave of Vdc voltage is given by Vave=(Vm)(1/pi) where Vm is the peak magnitude of the AC voltage. This is given in books such as Camara's which has tables for the different Vave vales of different rectifiers.
So taking the peak of the AC sine wave to be 2pi as given in the problem we get Vave=(2pi)(1/pi)=2
I believe I have version 8. Anyway the table that I am referring to in section 27-7 the AC Circuit Fundamentals section. Table 27.2 on page 27-6 gives "Characteristics of Alternating Waveforms" In the table they give ratios for full wave, half wave, square, etc.... waveforms which from the ratios you can calculate the desired Vave value.
In my previous post I stated that this table gave the Vave/Vm ratio as being 1/pi. Similar ratios are given for other waveforms. You can then easily sove for Vave in these cases.
Note however that this only applies to where the voltage is conduting during the full time half cycle. For thyristors which can be turned on any any point and wont necessarily conduct for the full half cycle and you need to either integrate between turn on and turn off time or I believe you can multiply the given equaion in the table times the cos of the angle at which the gate turns on and the device starts conducting.
