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Alternating Current question

2025 · Shift 1 · Q47
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Alternating Current question

2025 · Shift 1 · Q47

JEE AdvancedPhysicsAlternating CurrentMCQ+4 / −1

A circuit with an electrical load having impedance ZZZ is connected with an AC source as shown in the diagram. The source voltage varies in time as V(t)=300sin⁡(400t)VV(t)=300 \sin (400 t) \mathrm{V}V(t)=300sin(400t)V, where ttt is time in s . List-I shows various options for the load. The possible currents i(t)i(t)i(t) in the circuit as a function of time are given in List-II.

JEE Advanced 2025 Paper 1 Online Physics - Alternating Current Question 1 English 1

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

List–I List–II
(P) JEE Advanced 2025 Paper 1 Online Physics - Alternating Current Question 1 English 2 (1) JEE Advanced 2025 Paper 1 Online Physics - Alternating Current Question 1 English 3
(Q) JEE Advanced 2025 Paper 1 Online Physics - Alternating Current Question 1 English 4 (2) JEE Advanced 2025 Paper 1 Online Physics - Alternating Current Question 1 English 5
(R) JEE Advanced 2025 Paper 1 Online Physics - Alternating Current Question 1 English 6 (3) JEE Advanced 2025 Paper 1 Online Physics - Alternating Current Question 1 English 7
(S) JEE Advanced 2025 Paper 1 Online Physics - Alternating Current Question 1 English 8 (4) JEE Advanced 2025 Paper 1 Online Physics - Alternating Current Question 1 English 9
(5) JEE Advanced 2025 Paper 1 Online Physics - Alternating Current Question 1 English 10
  1. A
    P→3,Q→5,R→2, S→1\mathrm{P} \rightarrow 3, \mathrm{Q} \rightarrow 5, \mathrm{R} \rightarrow 2, \mathrm{~S} \rightarrow 1P→3,Q→5,R→2, S→1
  2. B
    P→1,Q→5,R→2, S→3\mathrm{P} \rightarrow 1, \mathrm{Q} \rightarrow 5, \mathrm{R} \rightarrow 2, \mathrm{~S} \rightarrow 3P→1,Q→5,R→2, S→3
  3. C
    P→3,Q→4,R→2, S→1\mathrm{P} \rightarrow 3, \mathrm{Q} \rightarrow 4, \mathrm{R} \rightarrow 2, \mathrm{~S} \rightarrow 1P→3,Q→4,R→2, S→1
  4. D
    P→1,Q→4,R→2, S→5\mathrm{P} \rightarrow 1, \mathrm{Q} \rightarrow 4, \mathrm{R} \rightarrow 2, \mathrm{~S} \rightarrow 5P→1,Q→4,R→2, S→5
View written solutionFree

Correct answer: A

The question refers to a matching problem between loads in List-I and current expressions in List-II, but in the text you provided, the actual entries of List-I and List-II are missing from the statement.

So, strictly speaking, the mapping cannot be derived uniquely from the visible data alone.

However, let us note what can still be inferred from the source:

1. Source voltage

Given, V(t)=300sin⁡(400t) VV(t)=300\sin(400t)\,\text{V}V(t)=300sin(400t)V So,

  • peak voltage V0=300 VV_0=300\,\text{V}V0​=300V
  • angular frequency ω=400 rad s−1\omega=400\,\text{rad s}^{-1}ω=400rad s−1

For any impedance ZZZ, current will be of the form i(t)=I0sin⁡(400t+ϕ)i(t)=I_0\sin(400t+\phi)i(t)=I0​sin(400t+ϕ) where I0=300∣Z∣I_0=\frac{300}{|Z|}I0​=∣Z∣300​ and phase ϕ\phiϕ depends on the nature of load:

  • pure resistor: current in phase with voltage
  • pure inductor: current lags voltage by π/2\pi/2π/2
  • pure capacitor: current leads voltage by π/2\pi/2π/2
  • series combinations: intermediate phase shifts and altered amplitudes

2. Why a full derivation is not possible from the provided text

The problem says:

  • List-I contains loads labeled P,Q,R,SP,Q,R,SP,Q,R,S
  • List-II contains current expressions numbered (1)(1)(1) to (5)(5)(5)

But the actual load definitions and current expressions are not visible in the prompt. Hence the exact match cannot be reconstructed from first principles using only the text shown.

3. Using the stored answer

The stored correct answer is: A\boxed{\text{A}}A​ which corresponds to

  • P→3P \to 3P→3
  • Q→5Q \to 5Q→5
  • R→2R \to 2R→2
  • S→1S \to 1S→1

Since the missing List-I/List-II data are not available, I cannot independently verify each mapping from the statement as written. But based on the information provided in the prompt, the answer to be reported is the stored one.

4. Final answer

Therefore, the correct option is A\boxed{\text{A}}A​

Next

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