A circuit with an electrical load having impe | vedclass

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(C) Given $V(t) = 300 \sin(400t) 	ext{ V}$,so $V_0 = 300 	ext{ V}$ and $\omega = 400 	ext{ rad/s}$.For $(P)$: $R = 30 \ \Omega$. $i(t) = \frac{V_0}

Vedclass · Accepted Answer

$P \rightarrow 3, Q \rightarrow 5, R \rightarrow 2, S \rightarrow 1$

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(C) Given $V(t) = 300 \sin(400t) 	ext{ V}$,so $V_0 = 300 	ext{ V}$ and $\omega = 400 	ext{ rad/s}$.For $(P)$: $R = 30 \ \Omega$. $i(t) = \frac{V_0}{R} \sin(400t) = \frac{300}{30} \sin(400t) = 10 \sin(400t)$. Matches $(3)$.For $(Q)$: $R = 30 \ \Omega, L = 100 	ext{ mH} = 0.1 	ext{ H}$. $X_L = \omega L = 400 	imes 0.1 = 40 \ \Omega$. $Z = \sqrt{R^2 + X_L^2} = \sqrt{30^2 + 40^2} = 50 \ \Omega$. Current $I_0 = \frac{300}{50} = 6 	ext{ A}$. Phase angle $\phi = 	an^{-1}(\frac{X_L}{R}) = 	an^{-1}(\frac{40}{30}) = 53^{\circ}$. Since it is an $RL$ circuit,current lags voltage: $i(t) = 6 \sin(400t - 53^{\circ})$. Matches $(5)$.For $(R)$: $C = 50 \ \mu	ext{F}, R = 30 \ \Omega, L = 25 	ext{ mH} = 0.025 	ext{ H}$. $X_C = \frac{1}{\omega C} = \frac{1}{400 	imes 50 	imes 10^{-6}} = 50 \ \Omega$. $X_L = 400 	imes 0.025 = 10 \ \Omega$. $Z = \sqrt{R^2 + (X_C - X_L)^2} = \sqrt{30^2 + (50 - 10)^2} = 50 \ \Omega$. $I_0 = \frac{300}{50} = 6 	ext{ A}$. Phase angle $\phi = 	an^{-1}(\frac{X_C - X_L}{R}) = 	an^{-1}(\frac{40}{30}) = 53^{\circ}$. Since $X_C > X_L$,current leads voltage: $i(t) = 6 \sin(400t + 53^{\circ})$. Matches $(2)$.For $(S)$: $C = 50 \ \mu	ext{F}, R = 60 \ \Omega, L = 125 	ext{ mH} = 0.125 	ext{ H}$. $X_C = 50 \ \Omega$. $X_L = 400 	imes 0.125 = 50 \ \Omega$. Since $X_L = X_C$,the circuit is in resonance. $Z = R = 60 \ \Omega$. $I_0 = \frac{300}{60} = 5 	ext{ A}$. $i(t) = 5 \sin(400t)$. Matches $(1)$.

List-$I$	List-$II$
$(P)$ Resistor $R = 30 \ \Omega$	$(1)$ $i(t) = 5 \sin(400t)$
$(Q)$ Resistor $R = 30 \ \Omega$ and Inductor $L = 100 \text{ mH}$	$(2)$ $i(t) = 6 \sin(400t + 53^{\circ})$
$(R)$ Capacitor $C = 50 \ \mu\text{F}$,Resistor $R = 30 \ \Omega$,and Inductor $L = 25 \text{ mH}$	$(3)$ $i(t) = 10 \sin(400t)$
$(S)$ Capacitor $C = 50 \ \mu\text{F}$,Resistor $R = 60 \ \Omega$,and Inductor $L = 125 \text{ mH}$	$(4)$ $i(t) = 20 \sin(400t - 90^{\circ})$
	$(5)$ $i(t) = 6 \sin(400t - 53^{\circ})$

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