I hope this is a step in the right direction,
1. For the first positive cycle, Vc1 = Vi and Vc2 does not charge is there is a short circuit due to the left ideal diode
2. For the first negative cycle, Vc2 = Vi + Vc1 = 2Vi. As the polarity of the output terminals reversed, the output should be...
Homework Statement
Both diodes are ideal.
Answers:
(a) Vi
(b) -Vi
(c) -2Vi
(d) none of the above
Homework Equations
Q=CV
The Attempt at a Solution
1. For t=0 to Pi:
Cap1 and Cap2 are in series and hence Q1=Q2, therefore Vc1 = Vc2.
From this we get Vi = Vc1 + Vc2, hence Vc1 =Vc2 = Vi/2.
2...
Question
http://puu.sh/52zAa.png
Attempt
http://puu.sh/52AVq.png
I've attempted to use Riemann sums and use the integral to prove the inequality, not sure if this was the right approach to start with as I am now stuck and don't see what to do next.
For part (b), I know that if (2√n...
Homework Statement
http://puu.sh/2mrWM
I'm practicing for my upcoming exam on discrete mathematics (we're not really too far in yet), and I cannot no matter how hard I try prove the equations. I expand tan(k-1) and then multiply by tan(k) and always end up at a dead end, it is driving me...
I'm aware of the the point (1,-16),
So as you already know that b=-1 and c=-3, sub in the point to get:
-16 = a(1)(1+1)(1-3)^d
= 2a(-2)^d
Two varibles makes this unsolvable but using guess and check you'll find 3 is the only correct solution (which just makes the whole question...
Hey guys, I was looking at an exam I did last year and tried to solve a question, which at the time I couldn't do.
Unfortunately I'm running into the same problem I had during the exam, so hear me out on this one
Question:
The graph below has equation y =ax(x-b)(x+c)^d. Write down the values...