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Given : $L =5.0 H , C =80 \mu F=80 \times 10^{-6} F$$R =40 \Omega, E _{ rms }=230 V$
(a) Frequency f of the source which will cause reso-nance in the circuit
$f=f_r=\frac{1}{2 \pi \sqrt{ LC }}$
$f=f_r=\frac{1 \times 7}{2 \times 22 \sqrt{5.0 \times 80 \times 10^{-6}}}$
$=\frac{7}{2 \times 22 \times 20 \times 10^{-3}}$
$=\frac{7000}{880}=\frac{175}{22} Hz$
Hence, $\quad \omega_r=2 \pi f_r$
$=2 \times \frac{22}{7} \times \frac{175}{22}=50 rad s ^{-1}$
(b) Let the impedance at resonance be $Z$.
$Z=\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}$
At resonance, $\omega=\omega_r$
and$\omega L=\frac{1}{\omega C}$
$\therefore \quad Z=R=40 \Omega$
$I_{ rms }=\frac{E_{ rms }}{Z}=\frac{E_{ rms }}{R}=\frac{230}{40}=5.75 A$
Amplitude of current at resonant frequency
$I_{rms}=\frac{I_0}{\sqrt{2}}$
$\therefore \quad I _0= I _{ rms } \sqrt{2}$
$=5.75 \times \sqrt{2} A$
$=8.13 A=8.1 A$.
(c) Potential drop across the ends of resistor
$V_R=I_{r m s} R$
$=5.75 \times 40=230 V$
Potential drop across the ends of L
$V_{r m s}=I_{r m s} X_L$
$= I _{ rms } \times \omega_{,} L$
$=5.75 \times 50 \times 5.0$
$V _{ rms }=1437.5 V$
In the same way, potential drop across the ends of capacitor
$= I _{ rms } \times X _{ C }$
$= I _{ rms } \times \frac{1}{\omega_r C }$"
$=5.75 \times \frac{1}{50 \times 80 \times 10^{-6}}$
$V _{ rms }=1437.5 V$
Potential drop across the ends of C = Potential difference across the endsL. $E _{ L }$ of and $E _{ C }$ are equal and opposite.
Hence magnitude of potential drop across the combination of LC combination is zero at resonant frequency.
(a) Frequency f of the source which will cause reso-nance in the circuit
$f=f_r=\frac{1}{2 \pi \sqrt{ LC }}$
$f=f_r=\frac{1 \times 7}{2 \times 22 \sqrt{5.0 \times 80 \times 10^{-6}}}$
$=\frac{7}{2 \times 22 \times 20 \times 10^{-3}}$
$=\frac{7000}{880}=\frac{175}{22} Hz$
Hence, $\quad \omega_r=2 \pi f_r$
$=2 \times \frac{22}{7} \times \frac{175}{22}=50 rad s ^{-1}$
(b) Let the impedance at resonance be $Z$.
$Z=\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}$
At resonance, $\omega=\omega_r$
and$\omega L=\frac{1}{\omega C}$
$\therefore \quad Z=R=40 \Omega$
$I_{ rms }=\frac{E_{ rms }}{Z}=\frac{E_{ rms }}{R}=\frac{230}{40}=5.75 A$
Amplitude of current at resonant frequency
$I_{rms}=\frac{I_0}{\sqrt{2}}$
$\therefore \quad I _0= I _{ rms } \sqrt{2}$
$=5.75 \times \sqrt{2} A$
$=8.13 A=8.1 A$.
(c) Potential drop across the ends of resistor
$V_R=I_{r m s} R$
$=5.75 \times 40=230 V$
Potential drop across the ends of L
$V_{r m s}=I_{r m s} X_L$
$= I _{ rms } \times \omega_{,} L$
$=5.75 \times 50 \times 5.0$
$V _{ rms }=1437.5 V$
In the same way, potential drop across the ends of capacitor
$= I _{ rms } \times X _{ C }$
$= I _{ rms } \times \frac{1}{\omega_r C }$"
$=5.75 \times \frac{1}{50 \times 80 \times 10^{-6}}$
$V _{ rms }=1437.5 V$
Potential drop across the ends of C = Potential difference across the endsL. $E _{ L }$ of and $E _{ C }$ are equal and opposite.
Hence magnitude of potential drop across the combination of LC combination is zero at resonant frequency.
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