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machuBest ResponseYou've already chosen the best response.5
Just done Lab 12..Homework 12 later. R1=10 R2=2490 C1=0.0000001 C2=0.0000001
 one year ago

Jose.oliveira2Best ResponseYou've already chosen the best response.6
H12P1 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
 one year ago

machuBest ResponseYou've already chosen the best response.5
HWP3  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
 one year ago

machuBest ResponseYou've already chosen the best response.5
Remaining answers HWP3 part 6 : R3= R1 part 7 : R4= R2 part 8: c3=c1 part 9 : c4= c2 part 9 :Midband gain = (R4/R3)
 one year ago

machuBest ResponseYou've already chosen the best response.5
some one asked me the value of pi. its always 3.14 buddy!
 one year ago

machuBest ResponseYou've already chosen the best response.5
H12 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
 one year ago

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

TaimoorBest ResponseYou've already chosen the best response.0
what is the value of your vout machu ?
 one year ago

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

machuBest ResponseYou've already chosen the best response.5
I'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
 one year ago

ppichBest ResponseYou've already chosen the best response.0
Thank you for H12P1 part c). After calculate my answer is 0.00883218771969
 one year ago

flashsahooBest ResponseYou've already chosen the best response.0
m nt getng rite ans frm this equatio...plz help mee out
 one year ago

ppichBest ResponseYou've already chosen the best response.0
Write 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.
 one year ago

QaisersattiBest ResponseYou've already chosen the best response.0
HW12 QNo 2 Part 3 4 5 and r any help please
 one year ago

ash__aryaBest ResponseYou've already chosen the best response.2
h12p2 part c,d..help
 one year ago

Jose.oliveira2Best ResponseYou've already chosen the best response.6
H12P2 a) vin*(R1+R2)/R2 b) vz*(R1+R2)/R2 c) 700 500 500 d) 0.15
 one year ago

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

Jose.oliveira2Best ResponseYou've already chosen the best response.6
try 4.92/R R = first R (a)
 one year ago

srikar4135Best ResponseYou've already chosen the best response.0
For 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.
 one year ago

srikar4135Best ResponseYou've already chosen the best response.0
Our 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.
 one year ago

aparna.bBest ResponseYou've already chosen the best response.0
i need answers for H12p2
 one year ago

offroader100Best ResponseYou've already chosen the best response.0
H12P1: 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=????
 one year ago
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