WAEC - Physics (2020 - No. 36)
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 \(\Omega\)
Wyjaśnienie
The impedance of a R-L circuit is given by:
Z = √(R\(^2\) + XL\(^2\)) ----------- eqn(1)
where Z is the impedance,
R is the resistance, and
XL is the inductive reactance
XL = 2πfL ------------ eqn(2)
where f is the frequency
From eqn(2),
XL = 2 * π * 60 * 0.5
= 60π
Also, from eqn(1),
Z = √(50\(^2\) + (60π)\(^2\))
Z = 195.0Ω