The RC/IC circuit frequency variation is defined as the frequency of signals that can flow through the circuit. It is also known as the characteristic frequency. An RC circuit is one that has a resistor and a capacitor connected in series. It suppresses frequencies less than f while allowing signals with frequencies greater than f to flow freely. However, signals with frequencies close to f are nonetheless partially transferred. Depending on the design, the RC filter can be used to filter out low or high frequencies. An integrated circuit (IC) is a semiconductor that contains hundreds of millions of small capacitors, resistors, and transistors.
Formula
The formula for frequency variation is given by the reciprocal of twice the product of pi, resistance, and capacitance of the circuit. It is represented by the symbol f. Its standard unit of measurement is hertz or per second (Hz or s-1), and its dimensional formula is given by [M0L0T-1].
f = 1/(2πRC)
Where,
- f is the frequency variation,
- π is a constant with the value of 3.14,
- R is the resistance,
- C is the capacitance of the circuit.
Sample Problems
Problem 1: Calculate the frequency variation for a circuit with a resistance of 2 ohms and capacitance of 3 F.
Solution:
We have,
R = 2
C = 3
Using the formula we have,
f = 1/(2πRC)
= 1 / (2 × 3.14 × 2 × 3)
= 0.0265 Hz
Problem 2: Calculate the frequency variation for a circuit with a resistance of 4 ohms and capacitance of 5 F.
Solution:
We have,
R = 4
C = 5
Using the formula we have,
f = 1/(2πRC)
= 1 / (2 × 3.14 × 4 × 5)
= 0.007958 Hz
Problem 3: Calculate the frequency variation for a circuit with a resistance of 2.5 ohms and capacitance of 6 F.
Solution:
We have,
R = 2.5
C = 6
Using the formula we have,
f = 1/(2πRC)
= 1 / (2 × 3.14 × 2.5 × 6)
= 0.01061 Hz
Problem 4: Calculate the resistance of a circuit with a frequency variation of 0.2 Hz and capacitance of 1 F.
Solution:
We have,
f = 0.2
C = 1
Using the formula we have,
f = 1/(2πRC)
=> R = 1/(2πfC)
= 1 / (2 × 3.14 × 0.2 × 1)
= 0.7958 ohms
Problem 5: Calculate the resistance of a circuit with a frequency variation of 0.06 Hz and capacitance of 3.5 F.
Solution:
We have,
f = 0.06
C = 3.5
Using the formula we have,
f = 1/(2πRC)
=> R = 1/(2πfC)
= 1 / (2 × 3.14 × 0.06 × 3.5)
= 0.758 ohms
Problem 6: Calculate the capacitance of a circuit with a frequency variation of 0.03 Hz and resistance of 2 ohms.
Solution:
We have,
f = 0.03
R = 2
Using the formula we have,
f = 1/(2πRC)
=> C = 1/(2πfR)
= 1 / (2 × 3.14 × 0.03 × 2)
= 2.653 F
Problem 7: Calculate the capacitance of a circuit with a frequency variation of 0.8 Hz and resistance of 1 ohm.
Solution:
We have,
f = 0.8
R = 1
Using the formula we have,
f = 1/(2πRC)
=> C = 1/(2πfR)
= 1 / (2 × 3.14 × 0.8 × 1)
= 0.197 F