A telephone wire of length 200 km has a capacitance of 0.014μF per km. If it carries an ac of frequency 2.5 kHz, what should be the value of an inductor required to be connected in series so that the impedance of the circuit is minimum
Capacitance of wire
C=0.014× 10−6× 200=2.8× 10−6F=2.8μF
For impedance of the circuit to be minimum XL=Xc⇒ 2π vL=12π vC
⇒ L=14π2v2C=14(3.14)2× (2.5× 103)2× 2.8× 10−6
=1.4× 10−3H = 1.4mH