If the frequency of the ac. circuit illustrated above is 500/π Hz what would be the reactance in the circuit?
(Inductance (L) = 0.9H, Capacitance (C) = \(2 \times 10^{-6}\)
Attached Figure / Diagram
A
0.0009Ω/π
B
100Ω
C
1030Ω
D
400Ω
E
2500Ω
Verified Correct Answer: D.
Physical Principles & Numerical Solution
Given Data: Frequency (F) = \(\frac{500}{\pi}\) , Inductance (L) = 0.9H, Capacitance (C) = \(2 \times 10^{-6}\)
Total circuit reactance = Inductive reactance ( X\(_L\) ) - Capacitive reactance ( X\(_C\) )
when ( X\(_L\) ) > ( X\(_C\) )
Inductive reactance ( X\(_L\) ) = 2\piFL = 2 \(\times\) \pi \(\times\) \(\frac{500}{\pi}\) \(\times\) 0.9 = 900\Omega
Capacitive reactance ( X\(_C\) ) = \(\frac{1}{2\piFC}\) = \(\frac{1}{2 \times \pi \times 500/\pi \times 2 \times 10^{-6}}\)
= \(\frac{1}{2 \times 10^{-3}}\) = \(\frac{1}{0.002}\)
= 500\Omega
Total circuit reactance = ( X\(_L\) ) - ( X\(_C\) ) = (900 - 500)\Omega
=400\Omega
Source: WAEC / JAMB / NECO Classroom Archive (Record #24151)