In the RC high-pass filter shown in the figure, R = 13.1 kΩ and C = 0.0365 μF. What is the 3.00-dB frequency of this circuit (where dB means basically the same for electric current as it did for sound in Chapter 16)? That is, at what frequency does the ratio of output voltage to input voltage satisfy 20log(VoutVin)=−320log𝑉out𝑉in=-3 ? Hz
Question
In the RC high-pass filter shown in the figure, R = 13.1 kΩ and C = 0.0365 μF. What is the 3.00-dB frequency of this circuit (where dB means basically the same for electric current as it did for sound in Chapter 16)? That is, at what frequency does the ratio of output voltage to input voltage satisfy 20log(VoutVin)=−320log𝑉out𝑉in=-3 ? Hz
Solution
The 3 dB frequency, also known as the cut-off frequency, of a high-pass RC filter can be calculated using the formula:
f_c = 1 / (2πRC)
where:
- f_c is the cut-off frequency
- R is the resistance
- C is the capacitance
- π is a mathematical constant whose approximate value is 3.14159
Given that R = 13.1 kΩ = 13.1 x 10^3 Ω and C = 0.0365 μF = 0.0365 x 10^-6 F, we can substitute these values into the formula:
f_c = 1 / (2π * 13.1 x 10^3 Ω * 0.0365 x 10^-6 F)
Solving this will give us the cut-off frequency in Hz.
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