A series LCR circuit is connected to a 45 sin | vedclass

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(B) Given: $V_0 = 45 	ext{ V}$, $L = 50 	imes 10^{-3} 	ext{ H}$, $\omega_0 = 10^5 	ext{ rad/s}$.At resonance, $\omega_0 = 1/\sqrt{LC} \Rightarrow

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$P \rightarrow 3, Q \rightarrow 1, R \rightarrow 4, S \rightarrow 2$

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(B) Given: $V_0 = 45 	ext{ V}$, $L = 50 	imes 10^{-3} 	ext{ H}$, $\omega_0 = 10^5 	ext{ rad/s}$.At resonance, $\omega_0 = 1/\sqrt{LC} \Rightarrow C = 1/(L \omega_0^2) = 1/(50 	imes 10^{-3} 	imes 10^{10}) = 2 	imes 10^{-9} 	ext{ F}$.At $\omega = 8 	imes 10^4 	ext{ rad/s}$, $I = 0.05 I_0 = I_0/20$. Since $I = V_0/Z$ and $I_0 = V_0/R$, we have $Z = 20R$.$X_L = \omega L = 8 	imes 10^4 	imes 50 	imes 10^{-3} = 4000 	ext{ } \Omega$.$X_C = 1/(\omega C) = 1/(8 	imes 10^4 	imes 2 	imes 10^{-9}) = 6250 	ext{ } \Omega$.$Z^2 = R^2 + (X_C - X_L)^2 \Rightarrow (20R)^2 = R^2 + (6250 - 4000)^2$.$399 R^2 = (2250)^2 \Rightarrow R = 2250 / \sqrt{399} \approx 2250 / 19.97 \approx 112.6 	ext{ } \Omega$.Using $R \approx 112.5 	ext{ } \Omega$ (for calculation simplicity), $I_0 = V_0/R = 45 / 112.5 = 0.4 	ext{ A} = 400 	ext{ mA}$. Thus $P ightarrow 3$.Quality factor $Q = (1/R) \sqrt{L/C} = (1/112.5) \sqrt{50 	imes 10^{-3} / 2 	imes 10^{-9}} = (1/112.5) 	imes 5000 \approx 44.4$. Thus $Q ightarrow 1$.Bandwidth $\Delta \omega = \omega_0 / Q = 10^5 / 44.4 \approx 2250 	ext{ rad/s}$. Thus $R ightarrow 4$.Peak power $P_{max} = I_0^2 R = (0.4)^2 	imes 112.5 = 0.16 	imes 112.5 = 18 	ext{ W}$. Thus $S ightarrow 2$.Correct match: $P ightarrow 3, Q ightarrow 1, R ightarrow 4, S ightarrow 2$.

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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