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.
A
50.0\(\Omega\)
B
450.5 \(\Omega\)
C
195.0 \(\Omega\)
D
1950.1 \(\Omega\)
Verified Correct Answer: C.
Physical Principles & Numerical Solution
The impedance of a R-L circuit is given by:
Z = \sqrt(R\(^2\) + XL\(^2\)) ----------- eqn(1)
where Z is the impedance,
R is the resistance, and
XL is the inductive reactance
XL = 2\pifL ------------ eqn(2)
where f is the frequency
From eqn(2),
XL = 2 * \pi * 60 * 0.5
= 60\pi
Also, from eqn(1),
Z = \sqrt(50\(^2\) + (60\pi)\(^2\))
Z = 195.0\Omega
Source: WAEC / JAMB / NECO Classroom Archive (Record #59467)