You need a filter circuit to eliminate a 70-Hz signal and lower frequencies. The

Question

You need a filter circuit to eliminate a 70-Hz signal and lower frequencies. The load you are trying to drive has a resistance of 50 ?. You decide to design a low-pass filter that has a cutoff frequency of 400 Hz . (Figure 1)

Part A

If you use the circuit shown in the figure, what capacitance C should you use?

Part B

You need a filter circuit to eliminate a 70-Hz signal and lower frequencies. The load you are trying to drive has a resistance of 50 ?. You decide to design a low-pass filter that has a cutoff frequency of 400 Hz . (Figure 1)

Part A

If you use the circuit shown in the figure, what capacitance C should you use?

Part B

What fraction of any 70-Hz signal potential difference passes through this circuit?

Explanation / Answer

A)

At cut off frequancy, Vout/Vin must equal 1/sqrt(2)
Use:
Vout/Vin = R / sqrt (XC^2+ R^2)
1/sqrt(2) = 50/ sqrt (XC^2 + 50^2)
sqrt (XC^2 + 50^2) = 70.711
XC^2 + 50^2 = 5000
Xc= 50 ohm

Now use:
XC = 1/(2*pi*f*C)
50 = 1/ (2*pi*400*C)
C=7.96*10^-6 F
Answer: 7.96*10^-6 F

B)
at 70 Hz:
XC = 1/(2*pi*f*C)
= 1/ (2*pi*70*7.96*10^-6 )
= 35.9 ohm

Now use:
Vout/Vin = R / sqrt (XC^2+ R^2)
= 50/ sqrt (35.9^2 + 50^2)
=0.81
Answer: 0.81

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