In an electric circuit, an inductor of inductance 0.5 H and resistance 50\(\Omega\) is connected to an alternating current source of frequency 60 Hz. Calculate the impedance of the circuit.
50.0\(\Omega\)
450.5 \(\Omega\)
195.0 \(\Omega\)
1950.1 \(\Omega\)
Explanation
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The impedance (Z) of a circuit with inductance (L) and resistance (R) connected to an alternating current source of frequency (f) is given by the formula: Z = √(R² + XL²), where XL is the inductive reactance.
Given:
Inductance (L) = 0.5 H
Resistance (R) = 50 Ω
Frequency (f) = 60 Hz
First, calculate the inductive reactance (XL) using the formula: XL = 2πfL
XL = 2π × 60 Hz × 0.5 H
XL = 60π Ω
Now, substitute the values of XL and R into the formula for impedance:
Z = √(R² + XL²)
Z = √(50 Ω)² + (60π Ω)²
Z ≈ 195.0 Ω (rounded to three significant figures)
Therefore, the impedance of the circuit is approximately 195.0 Ω, which corresponds to option C.
