Match AC and DC Circuit Voltage and Current Relationships

To determine the correct matching between Column I and Column II, we analyze the behavior of each circuit in Column II:

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Analysis of Circuit (p)

• Components: Inductor ($L = 6\text{ mH}$) and Capacitor ($C = 3\ \mu\text{F}$) connected in series to a variable DC voltage source $V$.
• In steady-state DC, the capacitor acts as an open circuit (blocks DC), so the current through the circuit is:
  $I = 0$
• Since no current flows, the voltage drop across the inductor is:
  $V_1 = 0$
• The entire source voltage appears across the capacitor:
  $V_2 = V$
• Therefore, circuit (p) satisfies C ($V_1 = 0, V_2 = V$).

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Analysis of Circuit (q)

• Components: Inductor ($L = 6\text{ mH}$) and Resistor ($R = 2\ \Omega$) connected in series to a variable DC voltage source $V$.
• In steady-state DC, the inductor acts as an ideal short circuit (zero resistance):
  $V_1 = L \frac{dI}{dt} = 0$
• The steady-state current is:
  $I = \frac{V}{R} = \frac{V}{2} \neq 0$
• The voltage across the resistor is:
  $V_2 = I R = V$
• Thus:
  • $V_1 = 0$ and $V_2 = V$, which matches C.
  • Since $I \neq 0$ and $V_2 = V > 0 = V_1$, we have $V_2 > V_1$, which matches B.
  • Since $V_2 = I R$, $V_2$ is proportional to $I$, which matches D.
• Therefore, circuit (q) matches B, C, D.

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Analysis of Circuit (r)

• Components: Inductor ($L = 6\text{ mH}$) and Resistor ($R = 2\ \Omega$) connected to an AC source of frequency $f = 50\text{ Hz}$.
• Angular frequency:
  $\omega = 2\pi f = 2\pi \times 50 = 100\pi \approx 314.16\text{ rad/s}$
• Reactance of the inductor:
  $X_L = \omega L = 100\pi \times (6 \times 10^{-3}) = 0.6\pi \approx 1.88\ \Omega$
• Resistance:
  $R = 2\ \Omega$
• Since $R > X_L$:
  $V_2 = I_{\text{rms}} R > V_1 = I_{\text{rms}} X_L$
• Here, $I \neq 0$, and:
  • $V_1 = I X_L \propto I$ (matches A)
  • $V_2 > V_1$ (matches B)
  • $V_2 = I R \propto I$ (matches D)
• Therefore, circuit (r) matches A, B, D.

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Analysis of Circuit (s)

• Components: Inductor ($L = 6\text{ mH}$) and Capacitor ($C = 3\ \mu\text{F}$) connected to an AC source of frequency $f = 50\text{ Hz}$.
• Reactance of the inductor:
  $X_L = \omega L \approx 1.88\ \Omega$
• Reactance of the capacitor:
  $X_C = \frac{1}{\omega C} = \frac{1}{100\pi \times (3 \times 10^{-6})} = \frac{10^4}{3\pi} \approx 1061\ \Omega$
• Clearly, $X_C \gg X_L$, which implies:
  $V_2 = I_{\text{rms}} X_C > V_1 = I_{\text{rms}} X_L$
• Here, $I \neq 0$, and:
  • $V_1 = I X_L \propto I$ (matches A)
  • $V_2 > V_1$ (matches B)
  • $V_2 = I X_C \propto I$ (matches D)
• Therefore, circuit (s) matches A, B, D.

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Analysis of Circuit (t)

• Components: Resistor ($R = 1\text{ k}\Omega = 1000\ \Omega$) and Capacitor ($C = 3\ \mu\text{F}$) connected to an AC source of frequency $f = 50\text{ Hz}$.
• Resistance:
  $R = 1000\ \Omega$
• Reactance of the capacitor:
  $X_C = \frac{10^4}{3\pi} \approx 1061\ \Omega$
• Since $X_C > R$:
  $V_2 = I_{\text{rms}} X_C > V_1 = I_{\text{rms}} R$
• Here, $I \neq 0$, and:
  • $V_1 = I R \propto I$ (matches A)
  • $V_2 > V_1$ (matches B)
  • $V_2 = I X_C \propto I$ (matches D)
• Therefore, circuit (t) matches A, B, D.

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Conclusion and Final Matching

• A $\rightarrow$ r, s, t
• B $\rightarrow$ q, r, s, t
• C $\rightarrow$ p, q
• D $\rightarrow$ q, r, s, t