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anonymous
 3 years ago
homework 12 and lab 12 anyone....??? pls hlep
anonymous
 3 years ago
homework 12 and lab 12 anyone....??? pls hlep

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Just done Lab 12..Homework 12 later. R1=10 R2=2490 C1=0.0000001 C2=0.0000001

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0H12P1 A) Just V/i b) first find R2 = VS / i second the anwser is R = R2  R1 (R1 is the awnser in a) c) I'm Trying LAB12 R1 (in Ohms): 1k C1 (in Farads): 1.59n R2 (in Ohms): 100k C2 (in Farads): 1.59n

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0HWP3  first 5 filter: 10Hz and 19kHz. signals around 1kHz, to be amplified by about a factor of 2. 0.68μF and a 390pF capacitor. resistors between 100Ω and 100kΩ. RS=20kΩ. 1) First find C1=0.68*10^6 = 6.8e7 (answer of part 4) Then R1*C1=1/(2*pi*f) R1*C1=1/(2*pi*10) so R1= 23417 2)First find C2=390*10^12 = 3.9e10 (answer of part 5) R2*C2=1/(2*pi*19*10^3) R2= 21489 3) Rs = Rf = 20kΩ = 2000 Ω

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Remaining answers HWP3 part 6 : R3= R1 part 7 : R4= R2 part 8: c3=c1 part 9 : c4= c2 part 9 :Midband gain = (R4/R3)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0some one asked me the value of pi. its always 3.14 buddy!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0H12 P1 c) Q: v=5v A=100, load resistance =500 Solve two equations to find value of vx vx/R=(vOut–vx)/500,with R=answer from Part1. vx=5–vOut/100 Solving two equations we get Vx Current = Vx / R

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hi mach.. i followed your method, since my variables are the same, but it gives me wrong answer!! can you check PLZ...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what is the value of your vout machu ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Write the answer for A=100, load 500 many errors when calculate this, because answer is in to within 0.1% of the exact answer

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'll be more clear with H12 P1 part c) my values are v=5v A=100, load resistance =500 First Solve two equations to find value of vx equation 1 : vx/R=(vOut–vx)/500,with R=answer from Part1. equation 2: vx=5–(vOut/100) Solving two equations we get Vx equation 1 :vout =1.5vx substitute the value of vout in equation 2 and u get vx Final Answer : Current = Vx/R

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thank you for H12P1 part c). After calculate my answer is 0.00883218771969

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0m nt getng rite ans frm this equatio...plz help mee out

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Write your R1, v, A and load to calculate if problem in HP12P1 c). For H12P2 first two is a) vin*(R1+R2)/R2 b) vz*(R1+R2)/R2 for other 4 question I have too many work to calculate and do many error after 23 may be have time to give you answers.

Qaisersatti
 3 years ago
Best ResponseYou've already chosen the best response.0HW12 QNo 2 Part 3 4 5 and r any help please

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0H12P2 a) vin*(R1+R2)/R2 b) vz*(R1+R2)/R2 c) 700 500 500 d) 0.15

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0H12P1: CURRENT SOURCE if A=100 v=5V, i=5mA ±VS=20V 500Ω within 0.1% of the exact What is ans of c=????

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0try 4.92/R R = first R (a)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0For example, suppose that we need to put precisely i=8mA through a load that presents a resistance but the value of the resistance is not well controlled. Suppose also that we have a precision voltage source of v=5V, but this voltage source is pretty wimpy: it cannot supply more than 0.1mA without smoke coming out of it. With the circuit shown above we can choose a resistor of resistance R that can accomplish this goal. First, assuming that the op amp is ideal (A=∞) give the value of the resistance R, in Ohms, that we need.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Our specification is that we want there to be two break frequencies for this filter: 12Hz and 18kHz. We also want midband frequencies, such as signals around 1kHz, to be amplified by about a factor of 2. We have available a 1.2μF and a 390pF capacitor. We also have a variety of resistors between 100Ω and 100kΩ. However, our op amp really likes RS=20kΩ. Your job is to fill in the remaining details of this design.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i need answers for H12p2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0H12P1: CURRENT SOURCE if A=100 v=5V, i=9mA ±VS=17V 500Ω within 0.1% of the exact. I got: a)555.555555556 b)1333.33333333 c) ??????? What is ans of c=????

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0any one did lab 12 please help how to calculate R1 C1 R2 C2
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