$A$ 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 $mH$)

  • A
    $0.35$
  • B
    $35$
  • C
    $3.5$
  • D
    $0$

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$A$ series $LCR$ circuit is connected to a $45 \sin (\omega t) \text{ V}$ source. The resonant angular frequency of the circuit is $10^5 \text{ rad s}^{-1}$ and current amplitude at resonance is $I_0$. When the angular frequency of the source is $\omega = 8 \times 10^4 \text{ rad s}^{-1}$, the current amplitude in the circuit is $0.05 I_0$. If $L = 50 \text{ mH}$, match each entry in List-$I$ with an appropriate value from List-$II$ and choose the correct option.
List-$I$List-$II$
$(P)$ $I_0$ in $\text{mA}$$(1)$ $44.4$
$(Q)$ The quality factor of the circuit$(2)$ $18$
$(R)$ The bandwidth of the circuit in $\text{rad s}^{-1}$$(3)$ $400$
$(S)$ The peak power dissipated at resonance in $\text{Watt}$$(4)$ $2250$
$(5)$ $500$

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