for an RLC circuit with a resistance of 11 k*ohm, a capacitance of 2 uF, and an inductance of 24 H, what frequency is needed to minimize the impedance?
R = 11 K Ohm,
L = 24 H.
C = 2uF.
The impedance is min. when Xl = Xc:
6.28*F*L = 1/6.28*F*C.
6.28*F*24 = 1/(6.28*F*2*10^-6).
150.72F = 1/(12.56*F*10^-6).
Multiply both sides by F:
150.72F^2 = 1/(12.56*10^-6).
150.72F^2 = 0.07962*10^6.
150.72F^2 = 79620.
F^2 = 528.26.
F = 23 Hz.
To find the frequency that minimizes the impedance of an RLC (resistance, inductance, capacitance) circuit, we need to calculate the resonant frequency. The resonant frequency is the frequency at which the reactive components of the circuit cancel each other out, resulting in minimum impedance.
The formula to calculate the resonant frequency is:
f = 1 / (2π√(LC))
Where:
f = Resonant frequency
L = Inductance in Henrys
C = Capacitance in Farads
π = Pi (approximately 3.14159)
Given the values:
R = 11 kΩ (kilo-ohm), which is equivalent to 11,000 Ω
C = 2 uF (microfarad), which is equivalent to 2 x 10^(-6) F (Farad)
L = 24 H (Henry)
We can substitute the values into the formula:
f = 1 / (2π√(24 x 2 x 10^(-6)))
Now, let's calculate it step by step:
1. Calculate the square root of (24 x 2 x 10^(-6)):
√(24 x 2 x 10^(-6)) = √(48 x 10^(-6)) = √(48) x √(10^(-6)) = 6.928 x 10^(-3)
2. Multiply the result by 2π:
2π x (6.928 x 10^(-3)) = 0.0436 π
3. Finally, calculate the reciprocal to find the resonant frequency:
f = 1 / (0.0436 π) = 1 / 0.1373 = 7.28 Hz
Therefore, the resonant frequency for the given RLC circuit is approximately 7.28 Hz. At this frequency, the impedance will be minimized.