so even though this is a part summer, part difference amplifier, the method we used will apply? anyway, thanks a lot for your help berkeman, this problem was giving me nightmares last night..
what confuses me is that why can't the current i_1= 1/2k, i_2=.25/3k, i_3=.75/5k for the currents entering v_i+?
also would the current into the node with the two 4k resistors be 2/4k + 0-v_i-/4k?
ok so from what i understand from what you said, here is what i have done:
.067/2000 + 2/4000 + -2/4000 + ((2-v_o)/50000) = 1/2000 + 0.25/3000 + 0.75/5000
then i solve for v_o?
does that look correct?
IDEAL CHARACTERISTICS:
A_ol=inf
R_i=inf
R_o=0
i_+=0
i_-=0
v_in=v_+ - v_-
v_out=A_ol(v_+ - v_-)
v_+=v_-
but i don't see how any of them would make this problem easier
the "1",".25",".75",".67", and "2" are all voltages. are you saying that i should right nodal equations for the inputs? i didnt quite understand what you meant...
hey guys, i have a HW problem of a combination of a summing and difference op amp. I got the first few steps of the problem, but i don't know how to continue from where i am. I am having a hard time solving for v_+. any help is greatly appreciatd, maybe if you guys can not solve the problem for...
Im trying to get the Fourier transform of x(t)=u(t)-u(t-1)
from what i know the FT of u(t) is pi*delta(omega)+1/jw
so for the u(t-1) would we have to use the time shifting property of Fourier transforms so that it becomes pi*delta(omega)+1/jw*(exp(-jw_o)??
im given a a 20-mH inductor and a 30 ohm resistor in parallel. Z_in is 25 degrees. and I am asked to find the frequency omega in rad/s here's what i try to do:
25=(jw.02*30)/(jw.02+30), and solve for w, but i don't get the right answer.
what is it that I am doing wrong?
Im having trouble comparing sinusoidal waves and their phases.
As a sample problem I was given
v_1=120cos(120*pi*t - 40deg)
and i_1=2.5cos(120*pi*t +20deg)
and I was asked to find the angle by which i_1 lags v_1.
I have no clue on how to go through with this problem, I don't even know...
you may be right about the load resistor lying betwen the two points. But even if i don't remove the load resistor, and use the same technique to find V_AB, then id still have the same equations because there would still be no current through the 10k resistor.
i have an attempted solution to a HW problem, and I want to know if it is right or wrong. Could you guys please help me out. I have attached the image as a pdf
ok here's the URL of the work
[img=http://www.freeimagehosting.net/uploads/th.d669771024.jpg]
http://www.freeimagehosting.net/d669771024.jpg
[img=http://www.freeimagehosting.net/uploads/d669771024.jpg]
theres a bunch of links to the image, this is my first time using this imageshack thing...
hello. I was reading through the chapter, and i came across a practice problem. The given answer for the problem is 79.2V. I attemted to solve it, but i don't get the same answer. ICould you guys please help me out and tell me where i am making a mistake? Thanx in advance...
its ok, I am just going to wait until the professor goes over the answer in class today. i mean that's the circuit the book has given us, so that's what I've been trying to work on. thanks for your help tho
i keep getting stuck in this problem, i don't see what I am doing wrong.
the problem says to determine which sources absorb positive power, then show that algebraic sum of the five absorbed power values is zero.
i have attached a picture of the circuit, and here is what i have done:
Starting...
Can anyone recommend a good book for me to review general math such as fractions, decimals, percenttages, and division etc.. those are my weakest points in math, and I would like to strenghten them a bit.
we know that
y'= \sum_{n=0}^{\infty} n a_{n} (x-1)^{n-1}
y"= \sum_{n=0}^{\infty} n (n-1) a_{n} (x-1)^{n-2}
when we compute xy', doesn't it become just:
\sum_{n=0}^{\infty} n a_{n} (x-1)^{n} ?
i don't see how \sum_{n=0}^{\infty} n a_{n} (x-1)^{n} and
- \sum_{n=0}^{\infty} (n+1) a_{n+1}...
i don't understand how you got the -(x-1)y'. does the term xy' change to (x-1)y' when the x in the y=SUM an(x-1)^n changes also? I am sorry, but is there a way you can show you got the 2nd and 3rd summation in your solution in more detail please?
Im given y"-xy'-y=0 at x0=1.
The problem asks for recurrene relation, and the first four terms in each of two linearly independant solutions, and the general term in each solution.
Whats thrwoing me off is the x0=1. I tried doing y= SUM an(x-1)^n, but when i differenetiate and plug in, i get...
Im given this problem:
Ge forms a substitutional solid solution with Si. find the weight% of Ge, that must be added to Si to yield an alloy that contains 2.43x10^21 Ge atoms per cubic centimetr. The densities of pure Ge and Si are 5.32 and 2.33 g/cm^3, respectively.
then I have a formula...
for example I am given a triple integral for the function w=z
outermost integral is:
the integral from -2 to 2, dx
middle integral is:
the integral from -sqrt(4-x^2) to sqrt(4-x^2), dy
and inner most integral is:
the integral from x^+y^2 to 4
they say to convert this to spherical...
im having trouble determining the angles of phi in spherical coordinates when asked to convert a triple integral into spherical, and find the limits of the phi integral. can anybody point out any hints/tips/tricks how this may be done??Please...i have an exam tomorrow and I am tryn to prepare...
thanx for tryn Benny, but unfortanately the answer you gave was incorrect. the way the book does it is like this:
inner integral= sqrt(1-x^2)dy,0,x
outer integals limits are 0-1, dx
and the entire integral multiplied by two.
the final answer is 2/3
the equation they gave for the xy...
i have to setup a doble integral to find the volume of the solid bounded by the graphs of the equation.
x^2+z^2=1, and y^2+z^2=1
z=sqrt(1-x^2)
z=sqrt(1-y^2)
then substituting in z=sqrt(1-y^2) into x^2+z^2=1, i got y=x.
so when i setup a double integral
for the dy i get integral...
Hey guys i keep getting stuck with this:
integral of ( ln(2x+1)dx)
im supposed to use by parts
heres what i have done
u=ln(2x+1)
du=2/2x+1
dv=dx
v=x
then i apply the formula uv-vdu
and i end up with another integral I am supposed to use by parts for:
integral of(2x/2x+1 dx)
heres...
A 1500 kg car traveling at 20m/s skids to a halt. What is the change in the thermal energy of the car and the road surface?
Here is how i try to approach it but i get a wrong answer:
First I solve for All dissipstive forces which is the frictional force in this problem.
I said the...
200 ml of a 6 M HCl solution is added to 400ml of a
1.50 M HCl solution. find the concentration of the
mixture.
here is what I think should be done:
(6M)(V concentration)=(.8L)(1.50M)
can i add the two volumes together?
Is this correct?
a block of mass m is resting on a 20degree slope. The block has coefficients of friction mu_s =0.80 and mu_k =0.50 with the surface. It is connected via a massless string over a massless, frictionless pulley to a hanging block of mass 2.0 kg. (sorry didnt know of a way to distribute a picture)...
2Al+3Cl2=Al2Cl6
a)which reactant is limiting if 2.70g Al and 4.05g Cl3 is mixed.
b)what mass of Al2Cl3 can be produced.
for part a i found the limiting reactant to be Cl2. I don't understand what I am supposed to do for the second part. Should i write another equation as Al2Cl3 as the...
ok. the first part of the problem asked me to find the tension in the string. to do that i used Newtons second law where Tcos(theta)=mg
to find theta, i looksed at the picture, and since the radius(opposite side of the angle) and the hypotenuse were given, i applied sin, which was...
A conical pendulum is formed by attaching a 0.100kg ball to a 1.00m-long string, then allowing the mass to move in a horizontal circle of radius 40.0cm.
What is the ball's angular velocity, in rpm?
from using sin, i obtained 23.57 degress. I am not sure how i would convert that to rpm