Question

A series R−L−C circuit has R=425Ω,R=1.25H and 3.50μF. It is connected to an a.c source with frequency of 60.0 Hz and ΔV

max=150 V. Determine the Inductivtive reactance.

Answer 1

The inductive reactance of an R-L-C series circuit connected to a 60Hz AC source with an inductor of 1.25H can be calculated using the formula X_L = 2πfL, which results in an inductive reactance of 471.24Ω.

Step-by-step explanation:

The student is asking how to calculate the inductive reactance of an R-L-C series circuit. The given values are: resistance R = 425Ω, inductance L = 1.25H, capacitance C = 3.50μF, and the circuit is connected to an alternating current (AC) source with a frequency of 60Hz.

Inductive reactance (XL) can be calculated using the formula XL = 2πfL, where f is the frequency of the AC source, and L is the inductance of the circuit. Plugging in the given values:

XL = 2 × π × 60Hz × 1.25H = 2 × 3.14159 × 60 × 1.25 = 471.24Ω.

Therefore, the inductive reactance of the circuit is 471.24Ω.