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(C) For $DC$ circuits,inductor acts as a short circuit $(V_L = 0)$ and capacitor acts as an open circuit $(I = 0)$.For $AC$ circuits,$X_L = \omega L =

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$C) V_1 = 0, V_2 = V$

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(C) For $DC$ circuits,inductor acts as a short circuit $(V_L = 0)$ and capacitor acts as an open circuit $(I = 0)$.For $AC$ circuits,$X_L = \omega L = 2\pi f L$ and $X_C = 1/(\omega C) = 1/(2\pi f C)$.Circuit $(p)$: $DC$ source. Inductor $(V_1)$ is shorted,so $V_1 = 0$. Capacitor $(V_2)$ is open,so $I = 0$. Matches $C$.Circuit $(q)$: $DC$ source. Inductor $(V_1)$ is shorted,so $V_1 = 0$. Resistor $(V_2)$ has $V_2 = IR$. Matches $C, D$.Circuit $(r)$: $AC$ source. $V_1 = I X_L$ and $V_2 = IR$. $X_L = 2\pi(50)(6 	imes 10^{-3}) \approx 1.88 \ \Omega$. $R = 2 \ \Omega$. Since $R > X_L$,$V_2 > V_1$. Also $V_1 \propto I$ and $V_2 \propto I$. Matches $A, B, D$.Circuit $(s)$: $AC$ source. $V_1 = I X_L$ and $V_2 = I X_C$. $X_L \approx 1.88 \ \Omega$. $X_C = 1/(2\pi(50)(3 	imes 10^{-6})) \approx 1061 \ \Omega$. Since $X_C > X_L$,$V_2 > V_1$. Also $V_1 \propto I$ and $V_2 \propto I$. Matches $A, B, D$.Circuit $(t)$: $AC$ source. $V_1 = IR$ and $V_2 = I X_C$. $R = 1000 \ \Omega$,$X_C \approx 1061 \ \Omega$. $V_2 > V_1$. Also $V_1 \propto I$ and $V_2 \propto I$. Matches $A, B, D$.Summary:$A ightarrow (r, s, t)$$B ightarrow (r, s, t)$$C ightarrow (p, q)$$D ightarrow (q, r, s, t)$

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