A telephone wire of length 200 km has a capacitance
of 0.014 uF per km. If it carries an AC of frequency
5 kHz, what should be the value of an inductor
required to be connected in series so that the
impedance of the circuit is minimum?
(a) 0.35 mH (b) 35 mH
(c) 3.5 mH
(d) Zero
Answers
Answered by
5
Answer:
L = 0.35 mH
Explanation:
XL = 2πfL
Xc = -1/2πfC
C = 0.014 μF * 200 = 2.8 μF = 2.8 * 10⁻⁶ F
f = 5 kHz = 5* 10³ Hz
Z = XL + Xc
impedance of the circuit is minimum if Z = 0
=> XL + Xc = 0
=> XL = -Xc
=> 2πfL = 1/2πfC
=> L = 1/(2πf)²C
=> L = 1/(2 * π * 5 * 10³)²(2.8 * 10⁻⁶)
=> L = 0.35 mH
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