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Question

(A) The total capacitance $C$ of the wire is given by the product of capacitance per unit length and the total length:$C = 0.014 	imes 10^{-6} \, F/k

Vedclass · Accepted Answer

$0.35$

Vedclass · Answer

(A) The total capacitance $C$ of the wire is given by the product of capacitance per unit length and the total length:$C = 0.014 	imes 10^{-6} \, F/km 	imes 200 \, km = 2.8 	imes 10^{-6} \, F = 2.8 \, \mu F$.For the impedance of an $LCR$ circuit to be minimum,the circuit must be in resonance,which occurs when the inductive reactance equals the capacitive reactance:$X_L = X_C$$2\pi 
u L = \frac{1}{2\pi 
u C}$.Rearranging for the inductance $L$:$L = \frac{1}{4\pi^2 
u^2 C}$.Given frequency $
u = 5 \, kHz = 5 	imes 10^3 \, Hz$ and $C = 2.8 	imes 10^{-6} \, F$:$L = \frac{1}{4 	imes (3.14)^2 	imes (5 	imes 10^3)^2 	imes 2.8 	imes 10^{-6}}$.$L = \frac{1}{4 	imes 9.8596 	imes 25 	imes 10^6 	imes 2.8 	imes 10^{-6}}$.$L = \frac{1}{2760.688} \approx 0.000362 \, H$.Re-calculating with standard approximation $\pi^2 \approx 10$:$L = \frac{1}{4 	imes 10 	imes 25 	imes 10^6 	imes 2.8 	imes 10^{-6}} = \frac{1}{1000 	imes 2.8} = \frac{1}{2800} \approx 0.000357 \, H = 0.357 \, mH$.Rounding to the nearest provided option,the value is $0.35 \, mH$.

$A$ telephone wire of length $200\, km$ has a capacitance of $0.014\, \mu F$ 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? (in $mH$)

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