In a series AC circuit, R=20(ohms), Xl=10(ohms), and Xc=25(ohms), where the frequency is 400 Hz. Find the resonant frequency of the circuit.
To find the resonant frequency of a series AC circuit, we need to find the frequency at which the inductive reactance (Xl) equals the capacitive reactance (Xc).
In a series AC circuit, the total impedance (Z) is given by the formula:
Z = sqrt(R^2 + (Xl - Xc)^2)
Given that R = 20 ohms, Xl = 10 ohms, and Xc = 25 ohms, we can substitute these values into the above formula:
Z = sqrt(20^2 + (10 - 25)^2)
Z = sqrt(400 + (-15)^2)
Z = sqrt(400 + 225)
Z = sqrt(625)
Z = 25 ohms
At resonance, the reactance of the inductor and the reactance of the capacitor cancel each other out, resulting in the total impedance (Z) being equal to the resistance (R).
Since the total impedance (Z) is equal to the resistance (R) at resonance, we can set Z equal to R and solve for the resonant frequency (f):
Z = R
sqrt(R^2 + (Xl - Xc)^2) = R
sqrt(20^2 + (10 - 25)^2) = 20
sqrt(400 + (-15)^2) = 20
sqrt(400 + 225) = 20
sqrt(625) = 20
25 = 20
At resonance, the impedance is equal to the resistance, therefore, the resonant frequency of the circuit is 400 Hz.