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JEE Advanced (Matrix Match & Integer Answer): Electromagnetic Induction & Alternating | Chapter-wise Tests for JEE Main & Advanced PDF Download

Match the Following

DIRECTIONS : Each question contains statements given in two columns, which have to be matched.

The statements in Column-I are labelled A, B, C and D, while the statements in Column-II are labelled p, q, r and s. Any given statement in Column-I can have correct matching with ONE OR MORE statement(s) in Column-II. The appropriate bubbles corresponding to the answers to these questions have to be darkened as illustrated in the following example :

If the correct matches are A-p, s and t; B-q and r; C-p and q; and D-s then the correct darkening of bubbles will look like the given.

JEE Advanced (Matrix Match & Integer Answer): Electromagnetic Induction & Alternating | Chapter-wise Tests for JEE Main & Advanced

Q.1. You are given many resistances, capacitors and inductors. These are connected to a variable DC voltage source (the first two circuits) or an AC voltage source of 50 Hz frequency (the next three circuits) in different ways as shown in Column II. When a current I (steady state for DC or rms for AC) flows through the circuit, the corresponding voltage V1 and V2 , (indicated in circuits) are related as shown in Column I. Match the two

 Column I
Column II
(A) I ≠ 0,V1 is
proportional  to I
JEE Advanced (Matrix Match & Integer Answer): Electromagnetic Induction & Alternating | Chapter-wise Tests for JEE Main & Advanced
(B) I ≠ 0, V2>V1
JEE Advanced (Matrix Match & Integer Answer): Electromagnetic Induction & Alternating | Chapter-wise Tests for JEE Main & Advanced
(C) V1 = 0,V2 = V
JEE Advanced (Matrix Match & Integer Answer): Electromagnetic Induction & Alternating | Chapter-wise Tests for JEE Main & Advanced
(D) I ≠ 0,V2 is
proportional to I
JEE Advanced (Matrix Match & Integer Answer): Electromagnetic Induction & Alternating | Chapter-wise Tests for JEE Main & Advanced

JEE Advanced (Matrix Match & Integer Answer): Electromagnetic Induction & Alternating | Chapter-wise Tests for JEE Main & Advanced

Ans. A-r, s, t; B-q, r, s, t; C-p, q; D-q, r, s, t

Solution. The following are the important concepts which are applied in the given situation. 

(i) For DC circuit, in steady state, the current I through the capacitor is zero. In case of L-C circuit, the potential difference across the inductor is zero and that across the capacitor is equal to the applied potential difference.

In case of L-R circuit, the potential difference across inductor is zero across resistor is equal to the applied voltage.
(ii) For AC circuit in steady state, Irms curr ent flows through the capacitor, inductor and resistor. The potential difference across resistor, inductor and capacitor is proportional to I.
(iii) For DC circuit, for changing current, the potential difference across inductor, capacitor or resistor is proportional to the current.


Integer Value Correct Type

Q.1. A series R-C combination is connected to an AC voltage of angular frequency ω = 500 radian/s. If the impedance of the R-C circuit is JEE Advanced (Matrix Match & Integer Answer): Electromagnetic Induction & Alternating | Chapter-wise Tests for JEE Main & Advanced

the time constant (in millisecond) of the circuit is    (2011)

Ans. 4

Solution. Time constant = RC

JEE Advanced (Matrix Match & Integer Answer): Electromagnetic Induction & Alternating | Chapter-wise Tests for JEE Main & Advanced

JEE Advanced (Matrix Match & Integer Answer): Electromagnetic Induction & Alternating | Chapter-wise Tests for JEE Main & Advanced

Q.2. A circular wire loop of radius R is placed in the x-y plane centered at the origin O. A square loop of side a(a<<R) having two turns is placed with its centre at JEE Advanced (Matrix Match & Integer Answer): Electromagnetic Induction & Alternating | Chapter-wise Tests for JEE Main & Advanced along the axis of the circular wire loop, as shown in figure. The plane of the square loop makes an angle of 45° with respect to the z-axis. If the mutual inductance between the loops is givenJEE Advanced (Matrix Match & Integer Answer): Electromagnetic Induction & Alternating | Chapter-wise Tests for JEE Main & Advanced then the value of p is     (2012)

Ans. 7

Solution. The magnetic field due to current carrying wire at the location of square loop is

JEE Advanced (Matrix Match & Integer Answer): Electromagnetic Induction & Alternating | Chapter-wise Tests for JEE Main & Advanced

The mutual induction

JEE Advanced (Matrix Match & Integer Answer): Electromagnetic Induction & Alternating | Chapter-wise Tests for JEE Main & Advanced

JEE Advanced (Matrix Match & Integer Answer): Electromagnetic Induction & Alternating | Chapter-wise Tests for JEE Main & Advanced

Q.3. Two inductors L1 (inductance 1 mH, internal resistance 3 Ω) and L2 (inductance 2 mH, internal resistance 4 Ω), and a resistor R (resistance 12 Ω) are all connected in parallel across a 5 V battery. The circuit is switched on at time t = 0.

The ratio of the maximum to the minimum current (Imax / I min) drawn from the battery is      (JEE Adv. 2016)

Ans. 8

Solution. 

JEE Advanced (Matrix Match & Integer Answer): Electromagnetic Induction & Alternating | Chapter-wise Tests for JEE Main & Advanced

JEE Advanced (Matrix Match & Integer Answer): Electromagnetic Induction & Alternating | Chapter-wise Tests for JEE Main & Advanced

JEE Advanced (Matrix Match & Integer Answer): Electromagnetic Induction & Alternating | Chapter-wise Tests for JEE Main & Advanced

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