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JEE Advanced · 2025 · 18 May 2025 · Paper 1 · Alternating Current

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.

List-I

(P) (see structure)
(Q) (see structure)
(R) (see structure)
(S) (see structure)

List-II

(1) (see structure)
(2) (see structure)
(3) (see structure)
(4) (see structure)
(5) (see structure)

Structure P — P

Structure Q — Q

Structure R — R

Structure S — S

Structure 1 — 1

Structure 2 — 2

Structure 3 — 3

Structure 4 — 4

Structure 5 — 5

(A)P→3, Q→5, R→2, S→1✓ Correct
(B)P→1, Q→5, R→2, S→3
(C)P→3, Q→4, R→2, S→1
(D)P→1, Q→4, R→2, S→5

Answer: (A)

Solution

Given:

V (t) = 300 sin (400 t)

V, so

V_{0} = 300

V,

ω = 400

rad/s.

(P) Pure resistor

R = 30Ω

:

i = \frac{V}{R} = 10 sin (400 t)

A, in phase → (3)

(Q)

R = 30Ω

,

L = 100

mH:

$X_{L} = ω L = 400 \times 0.1 = 40Ω$
$Z = 3 0^{2} + 4 0^{2} = 50Ω$
$i = 6 sin (400 t - 53°)$ A (current lags) → (5)

(R)

R = 30Ω

,

L = 25

mH,

C = 50 μ

F:

$X_{L} = 400 \times 0.025 = 10Ω$
$X_{C} = \frac{1 0 ^{6}}{400 \times 50} = 50Ω$
$Z = 3 0^{2} + (50 - 10)^{2} = 50Ω$
$i = 6 sin (400 t + 53°)$ A (current leads) → (2)

(S)

R = 60Ω

,

L = 125

mH,

C = 50 μ

F:

$X_{L} = 400 \times 0.125 = 50Ω$
$X_{C} = 50Ω$
Resonance: $Z = R = 60Ω$
$i = 5 sin (400 t)$ A, in phase → (1)

Therefore: P→3, Q→5, R→2, S→1

JEE Advanced · 2025 · 18 May 2025 · Paper 1 · Single correct option

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