JEE Advanced 2025 — Paper 1 — PHY-IV solution

JEE Advanced2025Paper 1PHY-IV

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 $V (t) = 300 sin (400 t)$ V, where $t$ is time in s. 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 $\to$ 3; Q $\to$ 5; R $\to$ 2; S $\to$ 1
B
P $\to$ 1; Q $\to$ 5; R $\to$ 2; S $\to$ 3
C
P $\to$ 3; Q $\to$ 4; R $\to$ 2; S $\to$ 1
D
P $\to$ 1; Q $\to$ 4; R $\to$ 2; S $\to$ 5

Solution

Source: $V_{0} = 300$ V, $ω = 400$ rad/s.

(P) Pure resistor, $Z = 30 Ω$ . Peak current $i_{0} = 300/30 = 10$ A, in phase. $i (t) = 10 sin (400 t)$ . Matches (3).

(Q) $R = 30 Ω$ , $X_{L} = ω L = 400 \times 0.1 = 40 Ω$ . $Z = 3 0^{2} + 4 0^{2} = 50 Ω$ ; current lags voltage by $tan^{- 1} (40/30) = 5 3^{\circ}$ . Peak current $= 6$ A; $i (t) = 6 sin (400 t - 5 3^{\circ})$ . Matches (5).

(R) $X_{C} = \frac{1}{ω C} = \frac{1}{400 \times 50 \times 1 0 ^{- 6}} = 50 Ω$ ; $X_{L} = 400 \times 25 \times 1 0^{- 3} = 10 Ω$ ; net reactance $X_{C} - X_{L} = 40 Ω$ (capacitive). $Z = 3 0^{2} + 4 0^{2} = 50 Ω$ , current leads by $tan^{- 1} (40/30) = 5 3^{\circ}$ . Peak current $= 6$ A; matches (2).

(S) $X_{C} = 50 Ω$ , $X_{L} = 400 \times 125 \times 1 0^{- 3} = 50 Ω$ . Resonance: $X_{L} = X_{C}$ , so $Z = R = 60 Ω$ . Peak current $= 5$ A, in phase. Matches (1).

$∴$ P $\to$ 3, Q $\to$ 5, R $\to$ 2, S $\to$ 1. Option (A).

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