This is a matching-type question where we need to relate voltage conditions (Column I) to different circuits (Column II). We will analyze each circuit and determine which conditions it can satisfy.

Analysis of Circuits in Column II

1. Circuit (P): R-L circuit with a DC source

• In a DC circuit at steady state, an inductor (L) acts as a short circuit (a wire with zero resistance).
• The voltage across the inductor is $V_2 = V_L = L \frac{dI}{dt} = 0$, as the current $I$ is constant.
• The entire source voltage $V$ drops across the resistor (R). So, $V_1 = V_R = V$.
• This corresponds to the condition (B) $V_2 = 0, V_1 = V$.
• Also, Kirchhoff's Voltage Law (KVL) for DC gives $V_1 + V_2 = V$. Substituting the values, we get $V + 0 = V$, which is true. So, it also matches condition (C) $V_1 + V_2 = V$.

2. Circuit (Q): R-C circuit with a DC source

• In a DC circuit at steady state, a capacitor (C) acts as an open circuit. It gets fully charged, and no current flows through the circuit.
• The steady-state current is $I = 0$.
• The voltage across the resistor is $V_1 = V_R = I \cdot R = 0$.
• The entire source voltage appears across the capacitor. So, $V_2 = V_C = V$.
• This corresponds to the condition (A) $V_1 = 0, V_2 = V$.
• KVL gives $V_1 + V_2 = V$. Substituting the values, we get $0 + V = V$, which is true. So, it also matches condition (C) $V_1 + V_2 = V$.
• The condition (D) $V_2 > V_1$ is also satisfied, as $V > 0$ (assuming a non-zero source voltage).

3. AC Circuits (R, S, T) For the AC circuits, the voltages $V_1$ and $V_2$ are RMS values. The problem states we are given many components, which implies we can choose their values to satisfy certain conditions.

• Circuit (R): AC, R-L series

  • $V_1 = I R$, $V_2 = I X_L$. The source voltage is $V = \sqrt{V_1^2 + V_2^2}$.
  • (A) $V_1=0, V_2=V$: Possible if we choose $R=0$. Then $V_1=0$ and $V = \sqrt{0^2 + V_2^2} \Rightarrow V=V_2$.
  • (B) $V_2=0, V_1=V$: Possible if we choose $L=0$ ($X_L=0$). Then $V_2=0$ and $V = \sqrt{V_1^2 + 0^2} \Rightarrow V=V_1$.
  • (D) $V_2 > V_1$: Possible if we choose components such that $X_L > R$.

• Circuit (S): AC, R-C series

  • $V_1 = I R$, $V_2 = I X_C$. The source voltage is $V = \sqrt{V_1^2 + V_2^2}$.
  • (A) $V_1=0, V_2=V$: Possible if we choose $R=0$.
  • (B) $V_2=0, V_1=V$: Possible if we choose $C=\infty$ ($X_C=0$).
  • (D) $V_2 > V_1$: Possible if we choose components such that $X_C > R$.

• Circuit (T): AC, L-C series

  • $V_1 = I X_L$, $V_2 = I X_C$. The voltages are $180^\circ$ out of phase. The source voltage is $V = |V_1 - V_2|$.
  • (A) $V_1=0, V_2=V$: Possible if we choose $L=0$. Then $V_1=0$ and $V = |0 - V_2| \Rightarrow V=V_2$.
  • (B) $V_2=0, V_1=V$: Possible if we choose $C=\infty$. Then $V_2=0$ and $V = |V_1 - 0| \Rightarrow V=V_1$.
  • (D) $V_2 > V_1$: Possible if we choose components such that $X_C > X_L$.

Matching Column I to Column II

• (A) $V_1 = 0, V_2 = V$: This matches (Q) from DC analysis, and (R), (S), (T) from AC analysis (by choosing a component value to be zero). Thus, (A) → (Q), (R), (S), (T).

• (B) $V_2 = 0, V_1 = V$: This matches (P) from DC analysis, and (R), (S), (T) from AC analysis (by choosing a component value to be zero or infinite). Thus, (B) → (P), (R), (S), (T).

• (C) $V_1 + V_2 = V$:

  • For DC circuits (P) and (Q), this is a direct application of KVL, so it holds true.
  • For AC circuits (R), (S), and (T), the RMS voltages do not add arithmetically unless they are in phase. The voltages across R-L and R-C are $90^\circ$ out of phase, and across L-C are $180^\circ$ out of phase. So, in general, $V_1+V_2 \neq V$. The equality holds only in the limiting cases already covered by (A) and (B). Given that (P) and (Q) satisfy this for any component values, they are the primary matches. Thus, (C) → (P), (Q).

• (D) $V_2 > V_1$:

  • For (P), $0 > V$, which is false.
  • For (Q), $V > 0$, which is true.
  • For (R), (S), (T), this is possible by choosing appropriate component values. Thus, (D) → (Q), (R), (S), (T).

Evaluating the Options

Let's summarize our findings:

• (A) → (Q), (R), (S), (T)
• (B) → (P), (R), (S), (T)
• (C) → (P), (Q)
• (D) → (Q), (R), (S), (T)

Now we compare this with the given options: Option A: (A)→(R), (S), (T); (B)→(Q), (R), (S), (T); (C)→(P), (Q); (D)→(Q), (R), (S), (T)

• (C)→(P), (Q): Matches our derivation.
• (D)→(Q), (R), (S), (T): Matches our derivation.
• (A)→(R), (S), (T): This differs from our derivation; it excludes (Q). This could be based on the interpretation that the phrase "When a current I ... flows" implies $I \neq 0$. In the DC steady state of circuit (Q), $I=0$.
• (B)→(Q), (R), (S), (T): This is incorrect. Our analysis shows (B) matches (P), not (Q). In fact, (Q) matches (A). This part of the option seems to contain a typo, likely swapping P and Q.

Despite the clear error in the mapping for (B), Option A has the correct mappings for (C) and (D), which are the most definitive parts of the analysis. The discrepancy in (A) is based on a subtle interpretation, and the error in (B) is very likely a typo in the question or options (a common issue). Among the given choices, Option A is the most plausible intended answer.