A telephone wire of length 200 km has a capacitance of 0.014 μF km^(-1). If it carries an AC frequency of 5 kHz, what should be the value of an…

Question

A telephone wire of length 200 km has a capacitance of 0.014 μF km^(-1). If it carries an AC frequency of 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?

MARKS Content Team · Accepted Answer

Correct answer: (A) 0.35 mH. Capacitance of wire C=0.014×10^(-6)×200 =2.8×10^(-6) F=2.8 μF For impedance of the circuit to be minimum X_L=X_C ⇒2πfL=1/(2πfC) L=1/(4π^2f^2C) =1/(43.14^2×5×10^32×2.8×10^(-6)) =0.35×10^(-3 )H=0.35 mH

A telephone wire of length $200 km$ has a capacitance of $0.014 μ F$ ${km}^{- 1}$ . If it carries an AC frequency of $5 kHz,$ what should be the value of an inductor required to be connected in series so that the impedance of the circuit is minimum?

Solution

More Alternating Current Questions

A telephone wire of length 200 km has a capacitance of 0.014 μF km-1. If it carries an AC frequency of 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?

Solution

More Alternating Current Questions

A telephone wire of length $200 km$ has a capacitance of $0.014 μ F$ ${km}^{- 1}$ . If it carries an AC frequency of $5 kHz,$ what should be the value of an inductor required to be connected in series so that the impedance of the circuit is minimum?