For a series LCR circuit with R=100,Omega,L=0 | vedclass

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(D) Given: $R=100\,\Omega$,$L=0.5\,mH = 0.5 	imes 10^{-3}\,H$,$C=0.1\,pF = 0.1 	imes 10^{-12}\,F$,$f=50\,Hz$.Angular frequency $\omega = 2\pi f = 2\

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$\approx 90^{\circ}$,predominantly capacitive circuit

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(D) Given: $R=100\,\Omega$,$L=0.5\,mH = 0.5 	imes 10^{-3}\,H$,$C=0.1\,pF = 0.1 	imes 10^{-12}\,F$,$f=50\,Hz$.Angular frequency $\omega = 2\pi f = 2\pi 	imes 50 = 100\pi\,rad/s$.Inductive reactance $X_L = \omega L = 100\pi 	imes 0.5 	imes 10^{-3} = 0.05\pi\,\Omega \approx 0.157\,\Omega$.Capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{100\pi 	imes 0.1 	imes 10^{-12}} = \frac{10^{11}}{10\pi} = \frac{10^{10}}{\pi}\,\Omega \approx 3.18 	imes 10^9\,\Omega$.Since $X_C \gg X_L$ and $X_C \gg R$,the circuit is predominantly capacitive.The phase angle $\phi$ is given by $	an \phi = \frac{X_L - X_C}{R}$.Since $X_C$ is extremely large compared to $R$ and $X_L$,$	an \phi \approx -\infty$,which implies $\phi \approx -90^{\circ}$.The negative sign indicates that the current leads the voltage by approximately $90^{\circ}$,confirming the circuit is predominantly capacitive.

For a series $LCR$ circuit with $R=100\,\Omega$,$L=0.5\,mH$ and $C=0.1\,pF$ connected across $220\,V-50\,Hz$ $AC$ supply,the phase angle between current and supplied voltage and the nature of the circuit is:

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