anonymous
  • anonymous
hw10 part 3
MIT 6.002 Circuits and Electronics, Spring 2007
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
H10P3 a) (((j*w*L+R2)*(1/(j*w*C))/(j*w*L+R2+(1/(j*w*C))))/((j*w*L + R2)*(1/(j*w*C))/(j*w*L+R2+(1/(j*w*C)))+R1)*(R2/(j*w*L + R2))) b) ((j * w * L + R2) * (1 / (j * w * C))) / (j * w * L + R2 + (1 / (j * w * C ))) c) (sqrt (R1/R2 -1)) / (w * R1) d) ((sqrt (R1/R2 -1)) * R2) / w e) Change values in C (with w = 2*pi*f) be careful, because the f is normaly in Khz (*10^3) f) Change values in D (with w = 2*pi*f) be careful, because the f is normaly in Khz (*10^3)
anonymous
  • anonymous
part e answer is: 1101.175 part f: 55.0587
anonymous
  • anonymous
correct is part e anwser is: 889.4pF part f answer is: 44.5uH

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anonymous
  • anonymous
Lmatch = sqrt(R1R2 – R2^2)/w and Cmatch = R1R2/(w*sqrt(R1R2 – R2^2))
anonymous
  • anonymous
becarful there are diferent values for someone....
anonymous
  • anonymous
@dubiduba we might be having different values for R1 and R2 it all depends on the previous 2 formulas of Lmatch and Cmatch.
anonymous
  • anonymous
Cmatch = 1/(w*sqrt(R1R2 – R2^2))
anonymous
  • anonymous
Lmatch = sqrt(R1R2 – R2^2)/w
anonymous
  • anonymous
Part-1 ?????
anonymous
  • anonymous
confused with the last 2
anonymous
  • anonymous
w??
anonymous
  • anonymous
w = 2*pi*f see the question C in Pico F in Micro
anonymous
  • anonymous
Finally got the h9p1 completely......guys those who are strugling with h9p1 part e they need to use the formula.with sin term as 1.
anonymous
  • anonymous
H10P3 e &f ans please
anonymous
  • anonymous
giv me ur question pls
anonymous
  • anonymous
w = 2*pi*f Cmatch = 1/(w*sqrt(R1R2 – R2^2)) Lmatch = sqrt(R1R2 – R2^2)/w

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