The question is not complete. We are supposed to find the average value of v_o.
Answer:
v_o,avg = 0.441V
Explanation:
Let t1 and t2 be the start and stop times of the output waveforms. Thus, from the diagram i attached, using similar triangles, we have;
3/(T/4) = 0.7/t1
So, 12/T = 0.7/t1
So, t1 = 0.7T/12
t1 = 0.0583 T
Also, from symmetry of triangles,
t2 = T/2 - t1
So, t2 = T/2 - 0.0583 T
t2 = 0.4417T
Average of voltage output is;
v_o,avg = (1/T) x Area under small triangle
v_o,avg = (1/T) x (3 - 0.7) x (T/4 - t1)
v_o,avg = (1/T) x (2.3) x (T/4 - 0.0583 T)
v_o,avg = (1/T) x 2.3 x 0.1917T
T will cancel out to give;
v_o,avg = 0.441V
Answer:
(a) the maximum and minimum temperatures for the cycle in K;
Maximum temprature = T1 = 600.05K
Minimum temprature = 300.03K
(b) the pressure and volume at the beginning of the isothermal expansion in bar and m3 respectively;
Pressure (P4) = 0.67842 bar
Volume (V4) = 1.2693
(c) the work and heat transfer for each of the four processes in kJ.
= -215.25 kJ
The answers are clearly explained in the below mentioned document.
Explanation:
Kindly download the below mentioned document, I have explained in quite detail in it. I hope it will help. Thanks.
Answer:
Engineering law and ethics are the moral codes associated with the field in which engineers are meant to adhere strictly to.
These ethics help in the safety of the engineers as a result of wearing protective clothing being a part of the ethics . Accuracy should also be taken important in research and daily activities. Corrupt practices and activities which poses risk to the environment is also strictly prohibited.