Question

A circuit with an electrical load having impedance $ Z $ is connected with an AC source as shown in the diagram. The source voltage varies in time as $ V(t) = 300 \sin(400t) \, \text{V} $, where $ t $ is time in seconds. 

List-I shows various options for the load. The possible currents $ i(t) $ in the circuit as a function of time are given in List-II.

Choose the option that describes the correct match between the entries in List-I to those in List-II.

• A. P \( \rightarrow \) 3, Q \( \rightarrow \) 5, R \( \rightarrow \) 2, S \( \rightarrow \) 1
• B. P \( \rightarrow \) 1, Q \( \rightarrow \) 5, R \( \rightarrow \) 2, S \( \rightarrow \) 3
• C. P \( \rightarrow \) 3, Q \( \rightarrow \) 4, R \( \rightarrow \) 2, S \( \rightarrow \) 1
• D. P \( \rightarrow \) 1, Q \( \rightarrow \) 4, R \( \rightarrow \) 2, S \( \rightarrow \) 5 \bigskip

Hint:

In AC circuits: - Pure resistors have current in phase with voltage. - RL circuits have current lagging voltage. - RC circuits have current leading. - RLC circuits can be at resonance when \( \omega = 1/\sqrt{LC} \), making current and voltage in phase.

Answer 1

The Correct Option is A

Step 1: Analyze option P (pure resistor)

Impedance is real and there is no phase difference. The current will be in phase with voltage, hence same shape as voltage (sine wave). Matches with graph (3).

\[ \Rightarrow \text{P} \rightarrow 3 \]

Step 2: Analyze option Q (R-L circuit)

In an R-L circuit, current lags voltage. From the graphs, graph (5) shows a lag.

\[ \Rightarrow \text{Q} \rightarrow 5 \]

Step 3: Analyze option R (R-L-C circuit)

Resonance occurs when \( \omega L = \frac{1}{\omega C} \). Given:

• \( L = 25 \, \text{mH} = 25 \times 10^{-3} \, \text{H} \)
• \( C = 50 \, \mu\text{F} = 50 \times 10^{-6} \, \text{F} \)
• \( \omega = 400 \)

Check resonance condition:

\[ \omega^2 = \frac{1}{LC} = \frac{1}{25 \times 10^{-3} \cdot 50 \times 10^{-6}} = \frac{1}{1.25 \times 10^{-6}} = 8 \times 10^5 \]

\[ \omega = \sqrt{8 \times 10^5} \approx 894 \]

Since \( \omega = 400 < \omega_0 \), the circuit is inductive → current lags slightly. Graph (2) matches this behavior.

\[ \Rightarrow \text{R} \rightarrow 2 \]

Step 4: Analyze option S (R-L-C circuit)

• \( L = 125 \, \text{mH} = 125 \times 10^{-3} \, \text{H} \)
• \( C = 50 \, \mu\text{F} = 50 \times 10^{-6} \, \text{F} \)
• \( R = 60 \, \Omega \)
• \( \omega = 400 \)

Check for resonance:

\[ \omega^2 = \frac{1}{LC} = \frac{1}{125 \times 10^{-3} \cdot 50 \times 10^{-6}} = \frac{1}{6.25 \times 10^{-6}} = 1.6 \times 10^5 \]

\[ \omega = \sqrt{1.6 \times 10^5} \approx 400 \]

Hence, circuit is at resonance → current is maximum and in phase with voltage. Graph (1) shows maximum amplitude and sharp waveform.

\[ \Rightarrow \text{S} \rightarrow 1 \]