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 In an electric circuit, an inductor of inductance 0.5 H and resistance 50\(\Omega\) is connected...

Physics
WAEC 2020

 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 \(\Omega\)
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Correct Answer: Option C
Explanation

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


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WAEC May/June 2024 - Practice for Objective & Theory - From 1988 till date, download app now - 99995
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WAEC offline past questions - with all answers and explanations in one app - Download for free
Post-UTME Past Questions - Original materials are available here - Download PDF for your school of choice + 1 year SMS alerts
WAEC May/June 2024 - Practice for Objective & Theory - From 1988 till date, download app now - 99995
Post-UTME Past Questions - Original materials are available here - Download PDF for your school of choice + 1 year SMS alerts
WAEC offline past questions - with all answers and explanations in one app - Download for free
WAEC Past Questions, Objective & Theory, Study 100% offline, Download app now - 24709