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 = true.
Explanation:
The statement made in the Question or problem above is TRUE/correct. The term known as " catch basin" is an important and essential part of the drainage system that is should be built in a community(whether housing community, industrial community or commercial community).
The catch basin is very important because it helps in making sure that they act as sieve to get all the dirt such as leaves from water running on the surface(runoffs).
Answer:
a, c
Explanation:
As the problem statement tells you, the phasor technique cannot be used when the frequencies are different. The frequencies are different when the coefficients of t are different. The different ones are highlighted.
a. 45 sin(2500t – 50°) + 20 cos(1500t +20°)
b. 25 cos(50t + 160°) + 15 cos(50t +70°)
c. 100 cos(500t +40°) + 50 sin(500t – 120°) – 120 cos(500t + 60°) -100 sin(10,000t +90°) + 40 sin(10, 100t – 80°) + 80 cos(10,000t)
d. 75 cos(8t+40°) + 75 sin(8t+10°) – 75 cos(8t + 160°)
Answer:
See the attached file for the answers.
Explanation:
Find attached of the explanation
Yes she did that’s the answers
