An ac circuit consists of a resistance of 5 ohms aninductance of 0.1H and a capacitance of 100μF allconnected in series. Determine for this circuits.a)Totalreactance b) impedance c) admittance d) susceptance ande) conductance . The supply frequency is 60 Hz.
Question
An ac circuit consists of a resistance of 5 ohms aninductance of 0.1H and a capacitance of 100μF allconnected in series. Determine for this circuits.a)Totalreactance b) impedance c) admittance d) susceptance ande) conductance . The supply frequency is 60 Hz.
Solution
a) To determine the total reactance of the circuit, we need to calculate the reactance of the inductor and the reactance of the capacitor separately.
The reactance of an inductor (XL) can be calculated using the formula XL = 2πfL, where f is the frequency and L is the inductance. Substituting the given values, we have XL = 2π(60)(0.1) = 37.7 ohms.
The reactance of a capacitor (XC) can be calculated using the formula XC = 1/(2πfC), where f is the frequency and C is the capacitance. Substituting the given values, we have XC = 1/(2π(60)(100x10^-6)) = 26.5 ohms.
The total reactance (XT) is the algebraic sum of the reactance of the inductor and the reactance of the capacitor. Therefore, XT = XL - XC = 37.7 - 26.5 = 11.2 ohms.
b) The impedance (Z) of the circuit can be calculated using the formula Z = √(R^2 + X^2), where R is the resistance and X is the reactance. Substituting the given values, we have Z = √(5^2 + 11.2^2) = √(25 + 125.44) = √150.44 = 12.26 ohms.
c) The admittance (Y) of the circuit can be calculated as the reciprocal of the impedance. Therefore, Y = 1/Z = 1/12.26 = 0.0816 S.
d) The susceptance (B) of the circuit can be calculated as the imaginary part of the admittance. Since the circuit is purely resistive, the susceptance is zero.
e) The conductance (G) of the circuit can be calculated as the real part of the admittance. Since the circuit is purely resistive, the conductance is equal to the reciprocal of the resistance. Therefore, G = 1/5 = 0.2 S.
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