The following series L-C-R circuit,when drive | vedclass

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(C) The impedance of an $L-C-R$ series circuit is given by $Z = \sqrt{(X_L - X_C)^2 + R^2}$,where $X_L = \omega L$ and $X_C = \frac{1}{\omega C}$.Give

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series $R$-$C$ circuit

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(C) The impedance of an $L-C-R$ series circuit is given by $Z = \sqrt{(X_L - X_C)^2 + R^2}$,where $X_L = \omega L$ and $X_C = \frac{1}{\omega C}$.Given: $\omega = 70 	imes 10^3 	ext{ rad/s}$,$L = 100 	imes 10^{-6} 	ext{ H}$,and $C = 1 	imes 10^{-6} 	ext{ F}$.Calculating inductive reactance: $X_L = \omega L = (70 	imes 10^3) 	imes (100 	imes 10^{-6}) = 7 	ext{ } \Omega$.Calculating capacitive reactance: $X_C = \frac{1}{\omega C} = \frac{1}{(70 	imes 10^3) 	imes (1 	imes 10^{-6})} = \frac{1}{70 	imes 10^{-3}} = \frac{1000}{70} \approx 14.29 	ext{ } \Omega$.Since $X_C > X_L$,the net reactance is capacitive $(X_C - X_L > 0)$.Therefore,the circuit behaves like a series $R-C$ circuit.

The following series $L-C-R$ circuit,when driven by an emf source of angular frequency $70 \text{ krad/s}$,the circuit effectively behaves like:

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