In circuit $(P)$, under steady state, capacitor will act like an infinite impedance and inductor will act like a zero impedance. Thus, $I=0, V_1=0$, and $V_2=V$.

In circuit $(Q)$, inductor will act as zero resistance in steady state giving us $I=V / R=V / 2, V_1=0$, and $V_2=V$.

In circuit $(R)$, the inductive reactance $X_L$ and impedance $Z$ are

$$ \begin{aligned} X_L & =\omega L=2 \pi \nu L=1.88 \Omega, \quad \text { and } \\\\ Z & =\sqrt{X_L^2+R^2}=2.75 \Omega \end{aligned} $$

Thus, $I=V / Z \neq 0, V_1=X_L I=1.88 I, V_2=R I=2 I$, and $V_2>V_1$.

In circuit (S), inductive reactance $X_L$, capacitive reactance $X_C$, and impedance $Z$ are

$$ \begin{aligned} X_L & =1.88 \Omega \\\\ X_C & =1 /(\omega C)=1061 \Omega, \quad \text { and } \\\\ Z & =X_C-X_L=1059 \Omega \end{aligned} $$

Thus the current in the circuit $I=V / Z \neq 0, V_1=$ $X_L I=1.88 I, V_2=X_C I=1061 I$, and $V_2>V_1$.

In circuit $(T), X_C, R$, and $Z$ are

$$ \begin{aligned} X_C & =1 /(\omega C)=1061 \Omega \\\\ R & =1000 \Omega, \quad \text { and } \\\\ Z & =\sqrt{R^2+X_C^2}=1458 \Omega \end{aligned} $$

Thus, $I=V / Z \neq 0, V_1=R I=1000 I, V_2=X_C I=$ $1061 I$, and $V_2 > V_1$.