JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Alternating Current

Match List - I with List - II :

List - I
(a)

ω L > \frac{1}{ω C}

(b)

ω L = \frac{1}{ω C}

(c)

ω L < \frac{1}{ω C}

(d) Resonant frequency

List - II
(i) Current is in phase with emf
(ii) Current lags behind the applied emf
(iii) Maximum current occurs
(iv) Current leads the emf

Choose the correct answer from the options given below :

(A)(a)-(iv); (b)-(iii); (c)-(ii); (d)-(i)
(B)(a) - (ii); (b) - (i); (c) - (iv); (d) - (iii)✓ Correct
(C)(a)-(iii); (b)-(i); (c)-(iv); (d) - (ii)
(D)(a) -(ii); (b) -(i); (c) - (iii); (d) -(iv)

Answer: (B)

Solution

In a series LCR circuit the net reactance is

X = ω L - \frac{1}{ω C}

, and its sign decides the phase of the current relative to the applied emf.
For

ω L > \frac{1}{ω C}

the circuit is inductive, so the current lags behind the applied emf.
For

ω L = \frac{1}{ω C}

the two reactances cancel, the impedance is purely resistive and the current is in phase with the emf.
For

ω L < \frac{1}{ω C}

the circuit is capacitive, so the current leads the emf.
The resonant frequency is the one at which

X = 0

; the impedance

Z = R

is then at its smallest, so maximum current occurs.
The matching is (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii).

JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Single correct option

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JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Alternating Current

Solution

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