A series LCR circuit of R=5 , Omega, L=20 , t | vedclass

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(B) At resonance,the inductive reactance $X_L$ is equal to the capacitive reactance $X_C$,so the impedance $Z$ of the circuit is equal to the resistan

Vedclass · Accepted Answer

$125$

Vedclass · Answer

(B) At resonance,the inductive reactance $X_L$ is equal to the capacitive reactance $X_C$,so the impedance $Z$ of the circuit is equal to the resistance $R$.$Z = R = 5 \, \Omega$.The root mean square current $I_{	ext{rms}}$ is given by $I_{	ext{rms}} = \frac{V}{Z} = \frac{V}{R}$.The power dissipated at resonance is $P = I_{	ext{rms}}^2 R = \left(\frac{V}{R}ight)^2 R = \frac{V^2}{R}$.Substituting the given values: $P = \frac{250 	imes 250}{5} = \frac{62500}{5} = 12500 \, 	ext{W}$.Expressing this in the form $..... 	imes 10^2 \, 	ext{W}$,we get $125 	imes 10^2 \, 	ext{W}$.Thus,the correct option is $B$.

$A$ series $LCR$ circuit of $R=5 \, \Omega, L=20 \, \text{mH}$ and $C=0.5 \, \mu \text{F}$ is connected across an $AC$ supply of $250 \, \text{V}$,having variable frequency. The power dissipated at resonance condition is $..... \times 10^{2} \, \text{W}$.

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