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A filter consist of resistance = 10k Ohm. and capacitance = 220 pF. Find out the cut of frequency.72.34 KHz7.234 KHz7.234 MHz72.34 Hz

Question

A filter consist of resistance = 10k Ohm. and capacitance = 220 pF. Find out the cut of frequency.72.34 KHz7.234 KHz7.234 MHz72.34 Hz

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Solution

The cut-off frequency for a RC (resistor-capacitor) filter can be calculated using the formula:

f_c = 1 / (2πRC)

where: f_c is the cut-off frequency, R is the resistance (in ohms), C is the capacitance (in farads), and π is a constant (approximately 3.14159).

Given that R = 10k Ohm = 10,000 Ohm and C = 220 pF = 220 x 10^-12 Farads, we can substitute these values into the formula:

f_c = 1 / (2π * 10,000 * 220 x 10^-12)

Solving this equation gives:

f_c = 1 / (2 * 3.14159 * 10,000 * 220 x 10^-12) f_c = 1 / (6.28318 * 10,000 * 220 x 10^-12) f_c = 1 / (1.3823 x 10^-6) f_c = 723.4 x 10^3 Hz

Therefore, the cut-off frequency is approximately 723.4 KHz.

This problem has been solved

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