A 220V, 50Hz A.C. supply is applied to series combination of a 2.5Ω, 6mH inductor and a 6.5μF capacitor. Calculate (i) reactance of inductor (ii) reactance of capacitor (iii) impedance of circuit.
Question
A 220V, 50Hz A.C. supply is applied to series combination of a 2.5Ω, 6mH inductor and a 6.5μF capacitor. Calculate (i) reactance of inductor (ii) reactance of capacitor (iii) impedance of circuit.
Solution
(i) Reactance of inductor (XL)
The formula for the reactance of an inductor is XL = 2πfL, where f is the frequency and L is the inductance.
Substituting the given values, we get:
XL = 2π * 50Hz * 6mH = 2 * 3.14 * 50 * 6 * 10^-3 = 1.884 Ohms
(ii) Reactance of capacitor (XC)
The formula for the reactance of a capacitor is XC = 1 / (2πfC), where f is the frequency and C is the capacitance.
Substituting the given values, we get:
XC = 1 / (2π * 50Hz * 6.5μF) = 1 / (2 * 3.14 * 50 * 6.5 * 10^-6) = 490.5 Ohms
(iii) Impedance of circuit (Z)
The formula for the impedance of a series circuit is Z = sqrt(R^2 + (XL - XC)^2), where R is the resistance, XL is the reactance of the inductor, and XC is the reactance of the capacitor.
Substituting the given values, we get:
Z = sqrt((2.5Ω)^2 + (1.884 Ohms - 490.5 Ohms)^2) = sqrt(6.25 + 238370.6) = 488.24 Ohms
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