Question 1: Palette 1/94

DPT MCQ

Find the magnetic field due to a small element $d\vec{l} = (3\hat{i} + 4\hat{j})\text{ mm}$ having current $10\text{ A}$ through it at a point whose position vector is $(5\hat{i} - 12\hat{j})\text{ m}$ from this element.

A

$-\frac{56}{2197} \times 10^{-9} \hat{k}\text{ T}$

B

$\frac{56}{2197} \times 10^{-9} \hat{k}\text{ T}$

C

$-\frac{28}{2197} \times 10^{-9} \hat{k}\text{ T}$

D

$-\frac{56}{169} \times 10^{-9} \hat{k}\text{ T}$

Question 2: Palette 2/94

DPT MCQ

From the image given below, find the magnetic field $B_P$ at point P due to a straight wire carrying current $i$ at a perpendicular distance $r$, where the angles subtended by the upper and lower ends of the wire at point P are $37^\circ$ and $53^\circ$ respectively.

A

$\frac{7 K i}{5 r} (-\hat{k})$

B

$\frac{1.4 K i}{r} (\hat{k})$

C

$\frac{K i}{5 r} (-\hat{k})$

D

$\frac{12 K i}{5 r} (-\hat{k})$

Question 3: Palette 3/94

DPT MCQ

From the image gven below, find the magnetic field $B_P$ at point P due to a straight wire carrying current $i$ at a perpendicular distance $r$, where the angles subtended by both ends of the wire at point P are $45^\circ$ and $45^\circ$ respectively.

A

$\frac{K i}{\sqrt{2} r} \otimes$

B

$\frac{2 K i}{r} \otimes$

C

$\frac{\sqrt{2} K i}{r} \otimes$

D

$\frac{K i}{2\sqrt{2} r} \otimes$

Question 4: Palette 4/94

DPT MCQ

From the image gven below, find the magnetic field $B_P$ at point P due to a current-carrying wire of current $i$ where one end subtends an angle $\theta_1 = 53^\circ$ and the other end is directly opposite to point P ($\theta_2 = 0^\circ$) at a perpendicular distance $r$.

A

$\frac{3 Ki}{5 r} \otimes$

B

$\frac{4 Ki}{5 r} \otimes$

C

$\frac{Ki}{r} \otimes$

D

$\frac{2 Ki}{5 r} \otimes$

Question 5: Palette 5/94

DPT MCQ

From the image gven below, find the magnetic field $B_P$ at point P due to a semi-infinite straight wire carrying current $i$ at a perpendicular distance $r$, where one end extends to infinity ($\theta_1 = 90^\circ$) and the other end is directly opposite to point P ($\theta_2 = 0^\circ$).

A

$\frac{2 K i}{r} \otimes$

B

$\frac{K i}{2 r} \otimes$

C

$\frac{K i}{r} \otimes$

D

$\frac{\mu_0 i}{2\pi r} \otimes$

Question 6: Palette 6/94

DPT MCQ

From the image gven below, find the magnetic field $B_P$ at point P due to a semi-infinite straight wire carrying current $i$ at a perpendicular distance $r$, where one end extends to infinity ($\theta_1 = 90^\circ$) and the other end subtends an angle $\theta_2 = 37^\circ$.

A

$\frac{3 K i}{5 r} (-\hat{k})$

B

$\frac{8 K i}{5 r} (-\hat{k})$

C

$\frac{4 K i}{5 r} (-\hat{k})$

D

$\frac{2 K i}{5 r} (-\hat{k})$

Question 7: Palette 7/94

DPT MCQ

From the image gven below, find the magnetic field $B_P$ at point P due to an infinitely long straight wire carrying current $i$ at a perpendicular distance $r$, where both ends extend to infinity ($\theta_1 = 90^\circ$ and $\theta_2 = 90^\circ$).

A

$\frac{K i}{r} (-\hat{k})$

B

$\frac{2 K i}{r} (-\hat{k})$

C

$\frac{4 K i}{r} (-\hat{k})$

D

$\frac{K i}{2 r} (-\hat{k})$

Question 8: Palette 8/94

DPT MCQ

From the image gven below, find the magnetic field at points P and Q which lie on the axis (extended line) of a straight wire carrying current $i$.

A

$B_P = 0, B_Q = 0$

B

$B_P = \frac{K i}{r}, B_Q = \frac{K i}{r}$

C

$B_P = \frac{2 K i}{r}, B_Q = 0$

D

$B_P = 0, B_Q = \frac{2 K i}{r}$

Question 9: Palette 9/94

DPT MCQ

From the image gven below, find the magnetic field vector $\vec{B}_P$ at point P $(0,0)$ due to the wire arrangement consisting of a horizontal current-carrying wire lying on the x-axis and a vertical semi-infinite wire carrying current $i$ extending upwards to infinity from $(-a, 0)$.

A

$\frac{2 K i}{a} (-\hat{k})$

B

$\frac{K i}{a} (-\hat{k})$

C

$\frac{K i}{2 a} (-\hat{k})$

D

$0$

Question 10: Palette 10/94

DPT MCQ

From the image gven below, find the total magnetic field vector $\vec{B}_A$ at point P situated between two infinitely long parallel straight wires carrying currents $i$ in opposite directions at perpendicular distances $r_1$ and $r_2$ respectively from point P.

A

$2 K i \left[\frac{1}{r_1} - \frac{1}{r_2}\right] (-\hat{k})$

B

$2 K i \left[\frac{1}{r_1} + \frac{1}{r_2}\right] (-\hat{k})$

C

$K i \left[\frac{1}{r_1} + \frac{1}{r_2}\right] (-\hat{k})$

D

$2 K i \left[\frac{1}{r_1} + \frac{1}{r_2}\right] (\hat{k})$

Question 11: Palette 11/94

DPT MCQ

From the image gven below, find the net magnetic field vector $\vec{B}_P$ at point P situated between two infinitely long parallel straight wires carrying currents $i_1$ and $i_2$ both in the upward direction at perpendicular distances $r_1$ and $r_2$ respectively from point P.

A

$\frac{2 K i_1}{r_1} (-\hat{k}) + \frac{2 K i_2}{r_2} (-\hat{k})$

B

$\frac{2 K i_1}{r_1} (+\hat{k}) + \frac{2 K i_2}{r_2} (-\hat{k})$

C

$\frac{2 K i_1}{r_1} (-\hat{k}) + \frac{2 K i_2}{r_2} (+\hat{k})$

D

$\frac{K i_1}{r_1} (-\hat{k}) + \frac{K i_2}{r_2} (+\hat{k})$

Question 12: Palette 12/94

DPT MCQ

From the image gven below, find the net magnetic field $B_P$ at point P situated midway between two long parallel straight wires, each carrying equal current $i$ in the same direction, separated by a distance $2r$ (such that each wire is at a distance $r$ from point P).

A

$\frac{4 K i}{r}$

B

$\frac{2 K i}{r}$

C

$0$

D

$\frac{K i}{r}$

Question 13: Palette 13/94

DPT MCQ

From the image gven below, find the total magnitude of the magnetic field $B_P$ at point P situated midway between two infinitely long parallel straight wires carrying equal current $i$ in opposite directions, where each wire is at a perpendicular distance $r$ from point P.

A

$\frac{2 K i}{r}$

B

$0$

C

$\frac{4 K i}{r}$

D

$\frac{K i}{r}$

Question 14: Palette 14/94

DPT MCQ

From the image gven below, find the net magnetic field $B_P$ at point P situated below two semi-infinite parallel straight wires, each extending upwards to infinity, carrying current $i$ in opposite directions at equal horizontal distances $r$ from point P, where point P lies on the line passing through their lower ends.

A

$\frac{2 K i}{r} \otimes$

B

$0$

C

$\frac{K i}{r} \otimes$

D

$\frac{K i}{2 r} \odot$

Question 15: Palette 15/94

DPT MCQ

From the image gven below, find the total magnetic field $B_P$ at the center P of a square loop of side length $L$ carrying current $i$.

A

$\frac{4\sqrt{2} K i}{L} \otimes$

B

$\frac{8\sqrt{2} K i}{L} \otimes$

C

$\frac{2\sqrt{2} K i}{L} \otimes$

D

$\frac{16 K i}{L} \otimes$

Question 16: Palette 16/94

DPT MCQ

From the image gven below, find the total magnetic field $B_P$ at the center P of an equilateral triangular loop of side length $l$ carrying current $i$.

A

$\frac{18 K i}{l} \otimes$

B

$\frac{9 K i}{l} \otimes$

C

$\frac{6\sqrt{3} K i}{l} \otimes$

D

$\frac{3\sqrt{3} K i}{l} \otimes$

Question 17: Palette 17/94

DPT MCQ

From the image gven below, find the net magnetic field $B_P$ at point P with coordinates $(3, 4)$ due to two infinitely long perpendicular wires lying along the coordinate axes, carrying currents $12\text{ A}$ along the y-axis (upward) and $4\text{ A}$ along the x-axis (leftward).

A

$6 K \odot$

B

$10 K \odot$

C

$6 K \otimes$

D

$14 K \odot$

Question 18: Palette 18/94

DPT MCQ

From the image gven below, find the total magnetic field vector at point O due to the given current-carrying wire configuration carrying current $i$.

A

$\frac{2 K i}{r} (-\hat{k})$

B

$\frac{K i}{r} (-\hat{k})$

C

$\frac{4 K i}{r} (-\hat{k})$

D

$0$

Question 19: Palette 19/94

DPT MCQ

From the image gven below, find the magnetic field $B_A$ at point A due to a straight wire carrying current $i$ where both ends P and Q lie on the same side of the perpendicular drawn from point A to the line containing the wire (extended line), subtending angles $53^\circ$ and $37^\circ$ with the perpendicular line of length $r$.

A

$\frac{K i}{5 r}$

B

$\frac{7 K i}{5 r}$

C

$\frac{2 K i}{5 r}$

D

$\frac{K i}{r}$

Question 20: Palette 20/94

DPT MCQ

From the image gven below, find the magnetic field $B_O$ at the center O for the given current configurations: (a) a complete circular loop of radius $R$ carrying current $i$, and (b) a semi-circular arc of radius $R$ carrying current $i$.

A

(a) $B_O = \frac{\mu_0 i}{2 R}$, (b) $B_O = \frac{K i}{R} \pi$

B

(a) $B_O = \frac{\mu_0 i}{4 R}$, (b) $B_O = \frac{2 K i}{R} \pi$

C

(a) $B_O = \frac{2 K i}{R}$, (b) $B_O = \frac{K i}{2 R} \pi$

D

(a) $B_O = \frac{K i}{R} \pi$, (b) $B_O = \frac{\mu_0 i}{2 R}$

Question 21: Palette 21/94

DPT MCQ

From the image gven below, find the magnetic field $B_O$ at the center O for the given current-carrying circular arcs: (a) a quarter-circular arc subtending an angle of $90^\circ$ at center O, and (b) a circular arc subtending an angle of $60^\circ$ at center O, both having radius $R$ and carrying current $i$.

A

(a) $B_O = \frac{K i}{R} \left(\frac{\pi}{2}\right)$, (b) $B_O = \frac{K i}{R} \left(\frac{\pi}{3}\right)$

B

(a) $B_O = \frac{K i}{R} \pi$, (b) $B_O = \frac{K i}{R} \left(\frac{\pi}{2}\right)$

C

(a) $B_O = \frac{2 K i}{R} \left(\frac{\pi}{2}\right)$, (b) $B_O = \frac{K i}{R} \left(\frac{\pi}{6}\right)$

D

(a) $B_O = \frac{K i}{R} \left(\frac{\pi}{4}\right)$, (b) $B_O = \frac{K i}{R} \left(\frac{\pi}{3}\right)$

Question 22: Palette 22/94

DPT MCQ

From the image gven below, find the net magnetic field vector $\vec{B}_O$ at center O due to the closed loop consisting of two concentric circular arcs of radii $R_1$ and $R_2$ subtending an angle of $60^\circ$ at O, and two radial straight wire segments carrying current $i$.

A

$K i \frac{\pi}{3} \left[ \frac{1}{R_1} (-\hat{k}) + \frac{1}{R_2} (+\hat{k}) \right]$

B

$K i \frac{\pi}{6} \left[ \frac{1}{R_1} (-\hat{k}) + \frac{1}{R_2} (+\hat{k}) \right]$

C

$K i \frac{\pi}{3} \left[ \frac{1}{R_1} (+\hat{k}) + \frac{1}{R_2} (-\hat{k}) \right]$

D

$K i \frac{\pi}{3} \left[ \frac{1}{R_1} (+\hat{k}) + \frac{1}{R_2} (+\hat{k}) \right]$

Question 23: Palette 23/94

DPT MCQ

From the image gven below, find the total magnetic field vector $\vec{B}_O$ at point O due to the wire configuration consisting of a semi-circular arc of radius $R$ and two horizontal straight wire segments extending along the line passing through point O, carrying a current $i$.

A

$\frac{K i}{R} \pi \otimes$

B

$\frac{2 K i}{R} \pi \otimes$

C

$\frac{K i}{2 R} \pi \odot$

D

$0$

Question 24: Palette 24/94

DPT MCQ

From the image gven below, find the total magnetic field vector $\vec{B}_O$ at origin O due to the current-carrying wire configuration consisting of a horizontal straight wire segment along the x-axis, a quarter-circular arc of radius $R$, and a vertical straight wire segment along the y-axis extending to infinity, carrying current $i$.

A

$\frac{K i}{2 R} \pi \odot$

B

$\frac{K i}{R} \frac{\pi}{2} \odot$

C

$\frac{2 K i}{R} \pi \otimes$

D

$0$

Question 25: Palette 25/94

DPT MCQ

From the image gven below, find the net magnetic field $B_O$ at the center O due to two concentric semi-circular arcs of radii $R$ and $2R$ carrying current $i$ in opposite directions as shown.

A

$\frac{\mu_0 i}{8 R} \odot$

B

$\frac{\mu_0 i}{4 R} \odot$

C

$\frac{\mu_0 i}{8 R} \otimes$

D

$\frac{\mu_0 i}{16 R} \odot$

Question 26: Palette 26/94

DPT MCQ

From the image gven below, find the total magnetic field $B_O$ at point O due to a wire configuration consisting of a circular arc of radius $R$ subtending an angle of $60^\circ$ carrying current $i$, along with two radial straight wire segments extending from point O.

A

$\frac{K i}{R} \frac{\pi}{3} \otimes$

B

$\frac{K i}{R} \frac{\pi}{6} \otimes$

C

$\frac{2 K i}{R} \frac{\pi}{3} \otimes$

D

$0$

Question 27: Palette 27/94

DPT MCQ

From the image gven below, find the total magnetic field $B_O$ at the center O due to a major circular arc wire segment of radius $R$ carrying current $i$ that subtends an angle $\theta = \frac{3\pi}{2}$ radians at center O.

A

$\frac{\mu_0 i}{4\pi R} \frac{3\pi}{2}$

B

$\frac{\mu_0 i}{2\pi R} \frac{3\pi}{2}$

C

$\frac{\mu_0 i}{4\pi R} \frac{\pi}{2}$

D

$\frac{\mu_0 i}{8\pi R} \pi$

Question 28: Palette 28/94

DPT MCQ

From the image gven below, find the total magnetic field vector $\vec{B}_O$ at point O due to the current-carrying wire configuration consisting of a circular arc of radius $R$ subtending an angle of $\frac{3\pi}{2}$ radians at center O, and two semi-infinite straight wires extending along lines passing through point O, carrying current $i$.

A

$\frac{K i}{R} \frac{3\pi}{2} \odot$

B

$\frac{K i}{R} \pi \otimes$

C

$\frac{2 K i}{R} \frac{3\pi}{2} \otimes$

D

$0$

Question 29: Palette 29/94

DPT MCQ

From the image gven below, find the total magnetic field vector $\vec{B}_O$ at point O due to the current-carrying wire configuration consisting of a circular arc of radius $R$ subtending an angle of $\frac{3\pi}{2}$ radians at center O, a semi-infinite straight wire whose line of action passes through O, and a semi-infinite straight wire running parallel to the horizontal at a distance $R$ from O with $\theta_1 = 90^\circ$ and $\theta_2 = 0^\circ$, carrying current $i$.

A

$\frac{K i}{R} \left( \frac{3\pi}{2} + 1 \right) \odot$

B

$\frac{K i}{R} \left( \frac{3\pi}{2} - 1 \right) \otimes$

C

$\frac{K i}{R} \left( \frac{\pi}{2} + 1 \right) \odot$

D

$\frac{2 K i}{R} \left( \frac{3\pi}{2} + 1 \right) \otimes$

Question 30: Palette 30/94

DPT MCQ

From the image gven below, find the total magnetic field $B_O$ at the center O of a square loop of side length $L$ carrying current $i$ where current enters and leaves through long diagonal leads.

A

$\frac{3\sqrt{2} K i}{L/2} \otimes$

B

$\frac{\sqrt{2} K i}{L/2} \otimes$

C

$\frac{2\sqrt{2} K i}{L/2} \otimes$

D

$0$

Question 31: Palette 31/94

DPT MCQ

From the image gven below, find the total magnetic field vector $\vec{B}_O$ at point O due to the current-carrying wire configuration consisting of two semi-infinite straight wires and a semi-circular arc of radius $R$ in the xy-plane carrying current $i$.

A

$\frac{K i}{R} [-2\hat{j} - \pi\hat{k}]$

B

$\frac{K i}{R} [2\hat{j} + \pi\hat{k}]$

C

$\frac{K i}{R} [-\hat{j} - \pi\hat{k}]$

D

$\frac{2 K i}{R} [-\hat{j} - \pi\hat{k}]$

Question 32: Palette 32/94

DPT MCQ

From the image gven below, find the total magnetic field $B_O$ at the center O of a square loop of side length $L$ and uniform resistance per unit length, when a total current $i$ enters at one corner and leaves at an adjacent corner.

A

$\frac{2\sqrt{2} K i}{L}$

B

$0$

C

$\frac{\sqrt{2} K i}{2 L}$

D

$\frac{3\sqrt{2} K i}{2 L}$

Question 33: Palette 33/94

DPT MCQ

From the image gven below, find the total magnetic field vector $\vec{B}_O$ at the center O of a circular loop of radius $R$ where current $i$ enters at one point, splits equally into two semi-circular paths carrying current $i/2$ each, and exits at the diametrically opposite point.

A

$0$

B

$\frac{K i}{R} \pi \otimes$

C

$\frac{K i}{2 R} \pi \odot$

D

$\frac{2 K i}{R} \pi$

Question 34: Palette 34/94

DPT MCQ

From the image gven below, find the total magnetic field vector $\vec{B}_O$ at the center O due to the current-carrying wire system consisting of a complete circular loop of radius $R$ carrying current $i$ and an infinitely long straight wire tangent to the bottom of the loop carrying current $i$ in the opposite direction.

A

$\left( \frac{2 K i}{R} - \frac{\mu_0 i}{2\pi R} \right) \hat{k}$

B

$\left( \frac{K i}{R} + \frac{\mu_0 i}{2\pi R} \right) \hat{k}$

C

$\left( \frac{2 K i}{R} + \frac{\mu_0 i}{\pi R} \right) \hat{k}$

D

$0$

Question 35: Palette 35/94

DPT MCQ

A coil of area $A = 0.5\text{ m}^2$ is situated in a uniform magnetic field $B = 4.0\text{ wb/m}^2$ and area vector makes an angle of $60^\circ$ with respect to the magnetic field from the image given below. The value of the magnetic flux through the area $A$ would be equal to

A

$2\text{ weber}$

B

$1\text{ weber}$

C

$3\text{ weber}$

D

$\frac{3}{2}\text{ weber}$

Question 36: Palette 36/94

DPT MCQ

A coil of $N$ turns and area $A$ is rotated at the rate of $n$ rotations per second in a magnetic field of intensity $B$, the magnitude of the maximum magnetic flux will be from the image given below

A

$NnAB$

B

$NAB$

C

$2\pi nNAB$

D

$nAB$

Question 37: Palette 37/94

DPT MCQ

A square coil of $10^{-2}\text{ m}^2$ area is placed perpendicular to a uniform magnetic field of intensity $10^3\text{ wb/m}^2$. The magnetic flux through the coil is

A

$10^5\text{ weber}$

B

$10\text{ weber}$

C

$10^{-5}\text{ weber}$

D

$100\text{ weber}$

Question 38: Palette 38/94

DPT MCQ

Consider the following figure, a uniform magnetic field of $0.2\text{ T}$ is directed along the positive x-axis from the image given below. What is the magnetic flux through top surface of the figure

A

$1.0\text{ m-wb}$

B

Zero

C

$0.8\text{ m-wb}$

D

$-1.8\text{ m-wb}$

Question 39: Palette 39/94

DPT MCQ

A coil of area $100\text{ cm}^2$ has $500$ turns. Magnetic field of $0.1\text{ weber/metre}^2$ is perpendicular to the coil from the image given below. The field is reduced to zero in $0.1\text{ sec}$. The induced emf in the coil is

A

$1\text{ V}$

B

$50\text{ V}$

C

$5\text{ V}$

D

Zero

Question 40: Palette 40/94

DPT MCQ

A coil has $1000$ turns and $500\text{ cm}^2$ as it is area. The plane of the coil is placed at right angles to a magnetic induction field of $2 \times 10^{-5}\text{ wb/m}^2$. The coil is rotated through $180^\circ$ in $0.2\text{ sec}$. The average emf induced in the coil in $mV$ is from the image given below

A

$15$

B

$10$

C

$5$

D

$20$

Question 41: Palette 41/94

DPT MCQ

A coil having $500$ square loops each of side $10\text{ cm}$ is placed normal to a magnetic field which increases at a rate of $1\text{ T/s}$ from the image given below. The induced emf in volt is

A

$5$

B

$0.1$

C

$1$

D

$0.5$

Question 42: Palette 42/94

DPT MCQ

The magnetic field of $2 \times 10^{-2}\text{ Tesla}$ acts at right angle to a coil of area $100\text{ cm}^2$ with $50$ turns. The average emf induced in the coil is $0.1\text{ V}$ when it is removed from the field in time $t$. The value of $t$ is from the image given below

A

$1\text{ s}$

B

$0.1\text{ s}$

C

$20\text{ s}$

D

$0.01\text{ s}$

Question 43: Palette 43/94

DPT MCQ

A circular coil of $500$ turns of a wire has an enclosed area of $0.1\text{ m}^2$ per turn. It is kept perpendicular to a magnetic field of induction $0.2\text{ T}$ and rotated by $180^\circ$ about a diameter perpendicular to the field in $0.1\text{ sec}$ from the image given below. How much charge will pass when the coil is connected to a galvanometer with a combined resistance of $50\text{ ohms}$

A

$0.2\text{ C}$

B

$0.4\text{ C}$

C

$2\text{ C}$

D

$4\text{ C}$

Question 44: Palette 44/94

DPT MCQ

Flux $\phi$ (in weber) in a closed circuit of resistance $10\text{ ohm}$ varies with time $t$ (in sec) according to the equation $\phi = 6t^2 - 5t + 1$ from the image given below. What is the magnitude of the induced current at $t = 0.25\text{ s}$

A

$1.2\text{ A}$

B

$0.8\text{ A}$

C

$0.6\text{ A}$

D

$0.2\text{ A}$

Question 45: Palette 45/94

DPT MCQ

The variation of induced emf $E$ with time $t$ in a coil if a short bar magnet is moved along its axis with a constant velocity is best represented as from the image given below

A

B

C

D

Question 46: Palette 46/94

DPT MCQ

A square loop of side $a$ and resistance $R$ is placed in a transverse uniform magnetic field $B$. If it suddenly changes into circular form in time $t$ from the image given below, then magnitude of induced charge will be

A

$\frac{Ba^2}{R}(4\pi - 1)$

B

$\frac{Ba^2}{R}\left(1 - \frac{1}{4\pi}\right)$

C

$\frac{Ba^2}{R}\left(\frac{1}{4\pi} - 1\right)$

D

$\frac{Ba^2}{R}\left(\frac{4}{\pi} - 1\right)$

Question 47: Palette 47/94

DPT MCQ

A circular coil and a bar magnet placed near by are made to move in the same direction from the image given below. The coil covers a distance of $1\text{ m}$ in $0.5\text{ sec}$ and the magnet a distance of $2\text{ m}$ in $1\text{ sec}$. The induced emf produced in the coil

A

$1\text{ V}$

B

$0.5\text{ V}$

C

Zero

D

Cannot be determined from the given information

Question 48: Palette 48/94

DPT MCQ

A short-circuited coil is placed in a time-varying magnetic field. Electrical power is dissipated due to the current induced in the coil. If the number of turns were to be quadrupled and the wire radius halved, the electrical power dissipated would from the image given below

A

Halved

B

The same

C

Doubled

D

Quadrupled

Question 49: Palette 49/94

DPT MCQ

A conducting circular loop is placed in a uniform magnetic field $B = 40\text{ mT}$ with its plane perpendicular to the field. If the radius of the loop starts shrinking at a constant rate $0.2\text{ mm/s}$, then the induced emf in the loop at an instant when its radius is $1.0\text{ cm}$ is

A

$0.1\pi\text{ }\mu\text{V}$

B

$0.2\pi\text{ }\mu\text{V}$

C

$1.0\pi\text{ }\mu\text{V}$

D

$0.16\pi\text{ }\mu\text{V}$

Question 50: Palette 50/94

DPT MCQ

A solenoid has $2000$ turns wound over a length of $0.314\text{ m}$. Around its central section a coil of $100$ turns and area of cross-section $1 \times 10^{-3}\text{ m}^2$ is wound. If an initial current of $2\text{ A}$ in the solenoid is reversed in $0.25\text{ sec}$, the emf induced in the coil is equal to

A

$6 \times 10^{-4}\text{ V}$

B

$12.8\text{ mV}$

C

$6 \times 10^{-2}\text{ V}$

D

$12.8\text{ V}$

Question 51: Palette 51/94

DPT MCQ

A square coil ABCD lying in x-y plane with its centre at origin. A long straight wire passing through origin carries a current $i = 2t$ in negative z-direction from the image gven below

A

Clockwise

B

Anticlockwise

C

Alternating

D

Zero

Question 52: Palette 52/94

DPT MCQ

In the following figure, the magnet is moved towards the coil with a speed $v$ and induced emf is $e$ from the image gven below. If magnet and coil recede away from one another each moving with speed $v$, the induced emf in the coil will be

A

$e$

B

$2e$

C

$e/2$

D

$4e$

Question 53: Palette 53/94

DPT MCQ

Two coils P and Q are lying a little distance apart coaxially. If an anticlockwise current $i$ is suddenly set up in the coil P then the direction of current from the image given below

A

Clockwise

B

Towards north

C

Towards south

D

Anticlockwise

Question 54: Palette 54/94

DPT MCQ

A rectangular loop is drawn from left to right across a uniform magnetic field perpendicular into the plane of the loop from the image given below

A

The direction of current in position 1 is clockwise

B

The direction of current in position 2 is clockwise

C

The direction of current in position 3 is anti-clockwise

D

The direction of current in position 4 is clockwise

Question 55: Palette 55/94

DPT MCQ

A small loop lies outside a circuit. The key of the circuit is closed and opened alternately. The closed loop will show from the image given below

A

Clockwise pulse followed by another clockwise pulse

B

Anticlockwise pulse followed by another anticlockwise pulse

C

Anticlockwise pulse followed by a clockwise pulse

D

Clockwise pulse followed by an anticlockwise pulse

Question 56: Palette 56/94

DPT MCQ

Consider the arrangement shown in figure in which the north pole of a magnet is moved away from a thick conducting loop containing capacitor. Then excess positive charge will arrive on from the image gven below

A

Plate a

B

Plate b

C

Both plates simultaneously

D

None of the above

Question 57: Palette 57/94

DPT MCQ

A square loop of side $1\text{ m}$ is placed in a perpendicular magnetic field. Half of the area of the loop inside the magnetic field. A battery of emf $10\text{ V}$ and negligible internal resistance is connected in the loop. The magnetic field changes with time according to relation $B = 0.01 - 2t\text{ Tesla}$ from the image given below. The resultant emf in the loop will be

A

$1\text{ V}$

B

$11\text{ V}$

C

$10\text{ V}$

D

$9\text{ V}$

Question 58: Palette 58/94

DPT MCQ

A short magnet is allowed to fall along the axis of a horizontal metallic ring from the image given below. Starting from rest, the distance fallen by the magnet in one second may be

A

$4\text{ m}$

B

$5\text{ m}$

C

$6\text{ m}$

D

$7\text{ m}$

Question 59: Palette 59/94

DPT MCQ

A conducting wire frame is placed in a magnetic field which is directed into the paper from the image gven below. The magnetic field is increasing at a constant rate. The directions of induced current in wires AB and CD are

A

From A to B and C to D

B

From B to A and C to D

C

From A to B and D to C

D

From B to A and D to C

Question 60: Palette 60/94

DPT MCQ

Self-inductance of a coil is defined as the magnetic flux linkage per unit current, given by the formula $L = \frac{N\phi}{I}$, where the magnetic field at the center of the circular coil is $B = \frac{\mu_{0}NI}{2R}$ and total flux linkage is $N\phi = \frac{\mu_{0}N^{2}I\pi R}{2}$.

From the image given below, find the self-inductance L of a circular coil of N turns and radius R carrying current I.

A

$L = \frac{\mu_{0}N^{2}\pi R}{2}$

B

$L = \frac{\mu_{0}N\pi R}{2}$

C

$L = \frac{\mu_{0}N^{2}R}{2\pi}$

D

$L = \frac{\mu_{0}N^{2}\pi R}{4}$

Question 61: Palette 61/94

DPT MCQ

The self-inductance of a solenoid is defined as the magnetic flux linkage per unit current, given by the formula $L = \frac{N\phi}{I}$, where the magnetic field inside the solenoid is $B = \mu_{0}ni = \mu_{0}\left(\frac{N}{l}\right)I$ and the total flux linkage is given by $N\phi = \frac{\mu_{0}N^{2}A}{l}I$.

From the image given below, find the expression for the self-inductance L of a long solenoid of length l, cross-sectional area A, and total turns N carrying current I.

A

$L = \frac{\mu_{0}N^{2}A}{l}$

B

$L = \frac{\mu_{0}NA}{l}$

C

$L = \frac{\mu_{0}N^{2}A}{2l}$

D

$L = \frac{\mu_{0}N^{2}l}{A}$

Question 62: Palette 62/94

DPT MCQ

The magnitude of the induced emf in a coil or solenoid is given by Faraday's law in terms of self-inductance $L$ and the rate of change of current, represented by the formula $\vert{}E\vert{} = L \frac{dI}{dt}$, where the self-inductance of the solenoid is $L = \frac{\mu_{0}N^{2}A}{l}$ and cross-sectional area is $A = \pi R^{2}$.

From the image given below, find the emf induced in a solenoid of 240 turns, having a length of 12 cm and a radius of 2 cm, when the current changes at the rate of $0.8 \text{ As}^{-1}$.

A

$6 \times 10^{-4} \text{ V}$

B

$3 \times 10^{-4} \text{ V}$

C

$1.2 \times 10^{-3} \text{ V}$

D

$6 \times 10^{-3} \text{ V}$

Question 63: Palette 63/94

DPT MCQ

The magnitude of the induced emf in a coil is given by Faraday's law in terms of self-inductance $L$ and the rate of change of current, represented by the formula $\vert{}E\vert{} = L \frac{dI}{dt}$, where $\frac{dI}{dt} = \frac{I_{2} - I_{1}}{\Delta t}$.

From the image given below, find the value of induced emf in a coil of self inductance $5 \text{ H}$ when the current changes from $1 \text{ A}$ to $2 \text{ A}$ in $5 \text{ sec}$.

A

$10 \text{ V}$

B

$0.1 \text{ V}$

C

$1.0 \text{ V}$

D

$100 \text{ V}$

Question 64: Palette 64/94

DPT MCQ

The magnitude of the induced emf in a coil is given by Faraday's law in terms of self-inductance $L$ and the rate of change of current, represented by the formula $E = -L \frac{dI}{dt}$, where $\frac{dI}{dt} = \frac{I_{2} - I_{1}}{\Delta t}$ and the current changes from $I$ to $-I$.

From the image given below, find the induced emf in a coil of self inductance $2 \text{ H}$ carrying a $2 \text{ A}$ current when the direction of current is reversed in $1 \text{ sec}$.

A

$-8 \text{ V}$

B

$8 \text{ V}$

C

$-4 \text{ V}$

D

Zero

Question 65: Palette 65/94

DPT MCQ

The induced emf in a coil is given by Faraday's law in terms of self-inductance $L$ and the rate of change of current, represented by the formula $E = -L \frac{dI}{dt}$, where the emf becomes zero when the time derivative of the current $\frac{dI}{dt} = 0$.

From the image given below, for a coil having self-inductance $L = 2 \text{ mH}$, current flow through it is $I = t^{2}e^{-t}$, find the time at which the emf becomes zero.

A

$2 \text{ sec.}$

B

$1 \text{ sec.}$

C

$4 \text{ sec.}$

D

$3 \text{ sec.}$

Question 66: Palette 66/94

DPT MCQ

The induced emf in a coil is related to the current by Faraday's law of electromagnetic induction, given by the formula $E = -L \frac{dI}{dt}$, which means the induced emf is proportional to the negative slope of the current versus time graph.

From the image given below, if current through the coil varies according to the given graph, then find the corresponding induced emf versus time graph.

A

B

C

D

Question 67: Palette 67/94

DPT MCQ

The self-inductance of a solenoid in terms of turn density $n$, cross-sectional area $A$, and length $l$ is given by the formula $L = \mu_{0} n^{2} A l$, where $L$ is directly proportional to the length $l$ when turn density and area are constant.

A solenoid has the self inductance $2 \text{ H}$. If length of the solenoid is doubled having turn density and area constant then new self inductance is :-

A

$4 \text{ H}$

B

$1 \text{ H}$

C

$8 \text{ H}$

D

$0.5 \text{ H}$

Question 68: Palette 68/94

DPT MCQ

The self-inductance of a solenoid is given by the formula $L = \mu_{0} n^{2} A l$, where $n$ is the number of turns per unit length, $A$ is the cross-sectional area, and $l$ is the length of the solenoid. Since area is proportional to the square of linear dimensions ($A \propto x^{2}$) and length is proportional to linear dimensions ($l \propto x$), the self-inductance is proportional to the cube of the linear dimensions ($L \propto x^{3}$).

A solenoid is wound over a rectangular frame. If all the linear dimensions of the frame are increased by a factor of 3 and the number of turns per unit length remains the same, the self inductance increased by a factor of :-

A

3

B

9

C

27

D

63

Question 69: Palette 69/94

DPT MCQ

The total magnetic flux linkage $N\phi$ through a coil of self-inductance $L$ carrying a current $I$ is related by the formula $N\phi = LI$, where $N\phi$ represents the magnetic flux linkage in weber.

A coil of inductance $2 \text{ H}$ has a current of $5.8 \text{ A}$. The flux in weber through the coil is :-

A

$0.29$

B

$2.9$

C

$3.12$

D

$11.6$

Question 70: Palette 70/94

DPT MCQ

The potential difference across an RL circuit component involving an inductor with a rate of change of current $\frac{di}{dt}$ is analyzed using Kirchhoff's Voltage Law (KVL), where the voltage drop across the inductor is $L \frac{di}{dt}$ and the induced emf/polarity depends on whether the current is increasing or decreasing.

From the image given below, find the potential difference $V_{A} - V_{B}$ in the given circuit if the current is $I = 5 \text{ A}$, the emf is $E = 15 \text{ V}$, resistance is $R = 1 \text{ \Omega}$, inductance is $L = 5 \text{ mH}$, and the current is decreasing at the rate of $10^{3} \text{ A/s}$.

A

$15 \text{ V}$

B

$5 \text{ V}$

C

$20 \text{ V}$

D

$10 \text{ V}$

Question 71: Palette 71/94

DPT MCQ

The potential difference across an RL circuit component involving an inductor with a rate of change of current $\frac{di}{dt}$ is analyzed using Kirchhoff's Voltage Law (KVL), where the voltage drop across the inductor is $L \frac{di}{dt}$ and the induced emf opposes the change in current.

From the image given below, find $V_{A} - V_{B}$ in the given circuit if the current is $I = 5 \text{ A}$, the emf is $E = 15 \text{ V}$, resistance is $R = 1 \text{ \Omega}$, inductance is $L = 5 \text{ mH}$, and the current is increasing at the rate of $10^{3} \text{ A/s}$.

A

$25 \text{ V}$

B

$15 \text{ V}$

C

$20 \text{ V}$

D

$5 \text{ V}$

Question 72: Palette 72/94

DPT MCQ

The potential difference across an RL circuit component involving an inductor when the current is constant is analyzed using Kirchhoff's Voltage Law (KVL), where the rate of change of current $\frac{di}{dt} = 0$, meaning the induced emf across the inductor is zero.

From the image given below, find $V_{A} - V_{B}$ in the given circuit if the current is $I = 5 \text{ A}$, the emf is $E = 15 \text{ V}$, resistance is $R = 1 \text{ \Omega}$, inductance is $L = 5 \text{ mH}$, and the current is constant.

A

$20 \text{ V}$

B

$15 \text{ V}$

C

$10 \text{ V}$

D

$5 \text{ V}$

Question 73: Palette 73/94

DPT MCQ

The potential difference across an RL circuit component involving a time-dependent current is analyzed using Kirchhoff's Voltage Law (KVL), where the voltage drop across the inductor is $L \frac{di}{dt}$ and the rate of change of current is $\frac{di}{dt} = \frac{d}{dt}(t^{2} + 2) = 2t$.

From the image given below, find the potential difference $V_{A} - V_{B}$ in the given circuit at $t = 2 \text{ s}$ if the current is given by $I = (t^{2} + 2) \text{ A}$, inductance is $L = 0.5 \text{ H}$, and resistances are $4 \text{ \Omega}$ and $3 \text{ \Omega}$.

A

$44 \text{ V}$

B

$22 \text{ V}$

C

$11 \text{ V}$

D

$88 \text{ V}$

Question 74: Palette 74/94

DPT MCQ

To find the potential difference between two points $A$ and $B$, we apply Kirchhoff's Voltage Law (KVL) along a path from $A$ to $B$. The potential drop across a resistor is $-iR$ (in the direction of current), across a battery it depends on the terminal traversed ($-E$ or $+E$), and across an inductor it is $-L\frac{di}{dt}$ in the direction of increasing current. The formula used is $V_A - i R_1 - E_1 - i R_2 - L\frac{di}{dt} + E_2 = V_B$.

From the image given below, find $V_A - V_B$ if $\frac{di}{dt} = +3\ \text{A/s}$ and current in the circuit is $6\ \text{A}$.

A

$50\ \text{V}$

B

$65\ \text{V}$

C

$80\ \text{V}$

D

$95\ \text{V}$

Question 75: Palette 75/94

DPT MCQ

To find the potential difference between two points $A$ and $B$, we apply Kirchhoff's Voltage Law (KVL) along a path from $A$ to $B$. The potential drop across a resistor is $-iR$, across a battery it is $-E$ or $+E$ depending on the terminal traversed, and across an inductor it is $-L\frac{di}{dt}$ in the direction of increasing current. The potential change across a capacitor with charge $q$ or given by $C$ is accounted for by its potential difference $\frac{q}{C}$ or standard capacitor potential drops. The formula used is $V_A - i R_1 + E_1 - \frac{q}{C} - L\frac{di}{dt} - E_2 - i R_2 - E_3 = V_B$.

From the image given below, find $V_A - V_B$ if $\frac{di}{dt} = +2\ \text{A/s}$ and current in the circuit is $10\ \text{A}$.

A

$-30\ \text{V}$

B

$-50\ \text{V}$

C

$-70\ \text{V}$

D

$-90\ \text{V}$

Question 76: Palette 76/94

DPT MCQ

An inductor acts as an open circuit (infinite resistance) just after the key is closed, meaning no current flows through the branch containing the inductor. After a long time (steady state), an inductor acts as a short circuit (zero resistance / plain wire) under steady direct current conditions.
Ohm's Law: $I = \frac{V}{R_{eq}}$

from the image given below, calculate current which is given by battery for the following circuit: (a) Just after closing the key. (b) after some time of closing the key.

A

(a) $2\text{ A}$, (b) $2.5\text{ A}$

B

(a) $1.5\text{ A}$, (b) $3\text{ A}$

C

(a) $2.5\text{ A}$, (b) $2\text{ A}$

D

(a) $3\text{ A}$, (b) $1.5\text{ A}$

Question 77: Palette 77/94

DPT MCQ

When a switch is closed in a circuit containing an inductor, a back e.m.f. is induced in the inductor according to Lenz's law, which opposes the growth of current in that branch. Therefore, current grows more slowly in the branch with the inductor than in the purely resistive branch. At steady state, the inductive effect becomes negligible, and identical components with equal resistances will experience the same current and show equal brightness.

from the image given below, figure shows an inductor L, a resistor R connected in parallel to a battery through a switch. The resistance of resistor R is same as that of the coil that makes L. Two identical bulb are put in each arm of the circuit. (a) Which of two bulbs lights up earlier when S is closed? (b) Will the bulbs be equally bright after some time?

A

(a) Bulb P, (b) Yes

B

(a) Bulb Q, (b) Yes

C

(a) Bulb Q, (b) No

D

(a) Bulb P, (b) No

Question 78: Palette 78/94

DPT MCQ

Dimensional analysis of $RC$ circuits, $RL$ circuits, and $LC$ circuits. The dimensions of time constant in an $RC$ circuit is $\tau = RC$, so $\frac{1}{RC}$ has dimensions of frequency ($[T^{-1}]$). For an $RL$ circuit, the time constant is $\tau = \frac{L}{R}$, so $\frac{R}{L}$ has dimensions of frequency. For an $LC$ circuit, the angular frequency of oscillation is $\omega = \frac{1}{\sqrt{LC}}$, so $\frac{1}{\sqrt{LC}}$ has dimensions of frequency. Therefore, the combination $\frac{C}{L}$ has dimensions of $\frac{[T^2]}{[\text{Dimensionless/other]}}$, which is not equivalent to frequency.

L, C and R respectively indicate inductance, capacitance and resistance. Select the combination, which does not have dimensions of frequency:

A

$1/RC$

B

$R/L$

C

$1/\sqrt{LC}$

D

$C/L$

Question 79: Palette 79/94

DPT MCQ

Growth of current in an $L-R$ circuit as a function of time is given by the formula: $I = I_{0} (1 - e^{-t/\tau})$ where $I_{0} = \frac{V}{R}$ is the steady-state current, $\tau = \frac{L}{R}$ is the time constant of the circuit, and $t$ is the time elapsed.

A coil of $10\text{ H}$ inductance and $5\ \Omega$ resistance is connected to $5\text{ volt}$ battery in series. The current in ampere in circuit $2\text{ seconds}$ after switched is on:

A

$e^{-1}$

B

$1 - e^{-1}$

C

$1 - e$

D

$e$

Question 80: Palette 80/94

DPT MCQ

The time constant of an $L-R$ circuit is defined as the time during which the current grows to $63.2\%$ of its maximum steady-state value. The formula for the time constant is $\tau = \frac{L}{R}$ where $L$ is the inductance and $R$ is the resistance.

An L-R circuit consists of an inductance of $8\text{ mH}$ and a resistance of $4\ \Omega$. The time constant of the circuit is:

A

$2\text{ ms}$

B

$12\text{ ms}$

C

$32\text{ ms}$

D

$500\text{ s}$

Question 81: Palette 81/94

DPT MCQ

The time constant of an $L-R$ circuit is the time $t = \tau = L/R$ at which the current reaches about $63.2\%$ of its maximum steady-state value. Substituting $t = \tau$ into the growth of current equation $I = I_{0} (1 - e^{-t/\tau})$ gives $I = I_{0} (1 - e^{-1}) = I_{0} (1 - 0.368) = 0.632 I_{0}$.

In an $L-R$ circuit, time constant is that time in which current grows from zero to the value where $I_{0}$ is the steady state current:

A

$0.63 I_{0}$

B

$0.50 I_{0}$

C

$0.37 I_{0}$

D

$I_{0}$

Question 82: Palette 82/94

DPT MCQ

The growth of current in an $L-R$ circuit is given by the differential equation $E - L \frac{dI}{dt} - IR = 0$. Initially, at $t = 0$, the current $I = 0$, so the initial rate of change of current is given by $\frac{dI}{dt}\Big\vert{}_{t=0} = \frac{E}{L}$.

An inductor of $20\text{ H}$ and a resistance of $10\ \Omega$, are connected to a battery of $5\text{ volt}$ in series, then initial rate of change of current is:

A

$0.5\text{ amp/s}$

B

$2.0\text{ amp/s}$

C

$2.5\text{ amp/s}$

D

$0.25\text{ amp/s}$

Question 83: Palette 83/94

DPT MCQ

The growth of current in an $L-R$ circuit as a function of time is given by the differential equation $L \frac{dI}{dt} + IR = E$, which solves to $I = \frac{E}{R} (1 - e^{-t/\tau})$. Differentiating this with respect to time gives the rate of change of current at any time $t$: $\frac{dI}{dt} = \frac{E}{L} e^{-t/\tau}$ where $\tau = \frac{L}{R}$ is the time constant.

A coil of $L = 5 \times 10^{-3}\text{ H}$ and $R = 18\ \Omega$ is abruptly supplied a potential of $5\text{ volts}$. What will be the rate of change of current in $0.001\text{ second}$? ($e^{-3.6} = 0.0273$)

A

$27.3\text{ amp/sec.}$

B

$27.8\text{ amp/sec.}$

C

$2.73\text{ amp/sec.}$

D

$2.78\text{ amp/sec.}$

Question 84: Palette 84/94

DPT MCQ

Growth of current in an $L-R$ circuit as a function of time is given by the formula $I = I_{0} (1 - e^{-t/\tau})$ where $I_{0} = \frac{V}{R}$ is the steady-state current and $\tau = \frac{L}{R}$ is the time constant of the circuit. Rearranging this gives $t = -\tau \ln\left(1 - \frac{I}{I_{0}}\right)$.

A coil of inductance $8.4\text{ mH}$ and resistance $6\ \Omega$ is connected to a $12\text{ V}$ battery in series. The current in the coil is $1.0\text{ A}$ at approximately the time:

A

$500\text{ s}$

B

$20\text{ s}$

C

$35\text{ ms}$

D

$1\text{ ms}$

Question 85: Palette 85/94

DPT MCQ

Dimensional analysis of electrical components: Inductance $L$, capacitance $C$, voltage $V$, resistance $R$, and current $I$. We can express these quantities in terms of fundamental dimensions. Since $V = IR$, $CV = Q$ (charge), and the time constant of an $LC$ circuit involves $\sqrt{LC}$ or $L/R$, simplifying the expression $\frac{L}{CVR}$ helps match it with physical quantities like charge, current, or their inverses.

The dimensions of combination $\frac{L}{CVR}$ are same as dimensions of:

A

Charge

B

Current

C

Charge$^{-1}$

D

Current$^{-1}$

Question 86: Palette 86/94

DPT MCQ

When a switch is closed in a circuit containing an inductor, the inductor offers infinite impedance (acts as an open circuit) at $t = 0$ (immediately after closing the key). Therefore, no current flows through the branch containing the inductor ($i_{3} = 0$). The total current $i_{1}$ from the battery splits into $i_{2}$ through $R_{2}$, calculated using Ohm's Law with the equivalent resistance of the active branches.

from the image given below, in the circuit shown in adjoining figure $E = 10\text{ V}$, $R_{1} = 1\ \Omega$, $R_{2} = 2\ \Omega$, $R_{3} = 3\ \Omega$ and $L = 2\text{ H}$. Calculate the value of current $i_{1}$, $i_{2}$ and $i_{3}$ immediately after key S is closed:

A

$3.3\text{ amp}$, $3.3\text{ amp}$, $0\text{ amp}$

B

$3.3\text{ amp}$, $3.3\text{ amp}$, $3.3\text{ amp}$

C

$3.3\text{ amp}$, $0\text{ amp}$, $0\text{ amp}$

D

$3.3\text{ amp}$, $3.3\text{ amp}$, $1.1\text{ amp}$

Question 87: Palette 87/94

DPT MCQ

The time constant of an RL circuit is given by $\tau = \frac{L}{R_{eq}}$, where $L$ is the inductance and $R_{eq}$ is the equivalent resistance connected across the inductor. In the steady state of an DC circuit, an inductor acts as a short circuit (zero resistance / conducting wire) or an open circuit depending on the connection, or more precisely, the current through the inductor becomes constant so that $\frac{di}{dt} = 0$, meaning the induced emf is zero and it behaves like a plain conducting wire.

For the circuit shown from the image given below, which of the following statement(s) is (are) correct?
(A) Its time constants is 2 second.
(B) In steady state, current through inductance will be 1A.
(C) In steady state, current through $4\Omega$ resistance will be $2/3\text{A}$.
(D) In steady state, current through $8\Omega$ resistance will be zero.

A

Its time constants is 2 second.

B

In steady state, current through inductance will be 1A.

C

In steady state, current through $4\Omega$ resistance will be $2/3\text{A}$.

D

In steady state, current through $8\Omega$ resistance will be zero.

Question 88: Palette 88/94

DPT MCQ

At $t = 0$, an inductor offers infinite resistance and acts as an open circuit because current through it cannot change abruptly. At $t = \infty$, the inductor acts as a short circuit (a simple conducting wire) as steady state is reached. Kirchhoff's Voltage Law (KVL) and Kirchhoff's Current Law (KCL) are applied to analyze the circuit branches.

From the image given below, in the given circuit, the switch is closed at $t = 0$. Find the currents $i_1$, $i_2$, $i_3$ and $\frac{di_3}{dt}$ at $t = 0$ and at $t = \infty$, given that initially all currents are zero.

A

At $t=0$: $i_1 = i_2 = \frac{\varepsilon}{2R}$, $i_3 = 0$, $\frac{di_3}{dt} = \frac{\varepsilon}{L}$; At $t=\infty$: $i_1 = i_3 = \frac{\varepsilon}{2R}$, $i_2 = 0$, $\frac{di_3}{dt} = 0$

B

At $t=0$: $i_1 = i_2 = \frac{\varepsilon}{R}$, $i_3 = 0$, $\frac{di_3}{dt} = \frac{\varepsilon}{2L}$; At $t=\infty$: $i_1 = i_3 = \frac{\varepsilon}{R}$, $i_2 = \frac{\varepsilon}{R}$, $\frac{di_3}{dt} = 0$

C

At $t=0$: $i_1 = i_2 = 0$, $i_3 = \frac{\varepsilon}{2R}$, $\frac{di_3}{dt} = 0$; At $t=\infty$: $i_1 = i_3 = 0$, $i_2 = \frac{\varepsilon}{2R}$, $\frac{di_3}{dt} = \frac{\varepsilon}{L}$

D

At $t=0$: $i_1 = i_2 = \frac{\varepsilon}{2R}$, $i_3 = \frac{\varepsilon}{R}$, $\frac{di_3}{dt} = 0$; At $t=\infty$: $i_1 = i_3 = \frac{\varepsilon}{R}$, $i_2 = 0$, $\frac{di_3}{dt} = \frac{\varepsilon}{2L}$

Question 89: Palette 89/94

DPT MCQ

When a switch connected to an inductor is toggled, the current through the inductor cannot change instantaneously due to self-inductance. We apply Kirchhoff's Voltage Law (KVL) around the new loop containing the inductor, the resistors, and the capacitor after the state change to determine the rate of change of current $\frac{di}{dt}$.

From the image given below, in the circuit shown, $S_1$ remains closed for a long time and $S_2$ remains open. Now $S_2$ is closed and $S_1$ is opened. Find out the $\frac{di}{dt}$ just after that moment.

A

$\frac{di}{dt} = \frac{3\varepsilon}{L}$

B

$\frac{di}{dt} = \frac{5\varepsilon}{L}$

C

$\frac{di}{dt} = \frac{4\varepsilon}{L}$

D

$\frac{di}{dt} = \frac{2\varepsilon}{L}$

Question 90: Palette 90/94

DPT MCQ

The growth of current in an RL circuit is given by the relation $i = i_0 (1 - e^{-t/\lambda})$, where $\lambda$ represents the time constant. The initial slope of the growth curve is given by $\frac{di}{dt} = \frac{i_0}{\lambda}$, which shows that a steeper initial slope corresponds to a smaller time constant.

From the image given below, which of the two curves shown has less time constant?

A

Curve 1

B

Curve 2

C

Both have the same time constant

D

Cannot be determined

Question 91: Palette 91/94

DPT MCQ

The current in an RL circuit as a function of time is derived using Kirchhoff's Voltage Law (KVL) and Kirchhoff's Current Law (KCL) to set up a differential equation of the form $\frac{di}{dt} + P i = Q$, which is then solved by integrating both sides with the initial condition $i = 0$ at $t = 0$.

From the image given below, in the following circuit the switch is closed at $t = 0$. Initially there is no current in inductor. Find out current in the inductor coil as a function of time.

A

$i = \frac{\varepsilon}{3R} (1 - e^{-\frac{3Rt}{2L}})$

B

$i = \frac{\varepsilon}{2R} (1 - e^{-\frac{3Rt}{2L}})$

C

$i = \frac{2\varepsilon}{3R} (1 - e^{-\frac{3Rt}{2L}})$

D

$i = \frac{\varepsilon}{R} (1 - e^{-\frac{3Rt}{2L}})$

Question 92: Palette 92/94

DPT MCQ

The relationship between voltage, inductance, resistance, and current in an RL circuit is described by Kirchhoff's Voltage Law: $V - L \frac{di}{dt} - iR = 0$. By integrating this equation with respect to time over the interval from $0$ to $t$, we relate the total charge flown through the circuit to the current and time.

From the image given below, in the following circuit the switch is closed at $t = 0$. Initially there is no current in inductor. Find charge flown at time $t$ when current flowing in the circuit is $i$.

A

$Q = \frac{Vt + Li}{R}$

B

$Q = \frac{Vt - Li}{R}$

C

$Q = \frac{Vt - 2Li}{R}$

D

$Q = \frac{2Vt - Li}{R}$

Question 93: Palette 93/94

DPT MCQ

The total thermal energy dissipated (heat loss) in a resistor $R$ carrying a time-varying current $i(t)$ is given by integrating the instantaneous power $i^2 R$ with respect to time from $t = 0$ to $t = \infty$: $H = \int_{0}^{\infty} i^2 R dt$. In a decaying RL circuit, the current as a function of time is given by $i = i_0 e^{-\frac{Rt}{L}}$.

From the image given below, what is the total heat energy dissipated (decay energy loss) in the resistor from $t = 0$ to $t = \infty$ for an inductor of inductance $L$ carrying an initial current $i_0$?

A

$\frac{1}{4} L i_0^2$

B

$\frac{1}{2} L i_0^2$

C

$L i_0^2$

D

$2 L i_0^2$

Question 94: Palette 94/94

DPT MCQ

In a circuit containing resistors, a DC voltage source, and an inductor, closing the switch at $t = 0$ changes the circuit configuration. The transient growth or decay of current through any branch containing an inductor or resistor is governed by solving the mesh or node differential equations using Kirchhoff's laws, yielding an exponential function of the form $i(t) = i_f + (i_i - i_f) e^{-t/\tau}$, where $\tau = \frac{L}{R_{eq}}$.

From the image given below, a switch was open for a long time and now it is closed at $t = 0$. Find the current from resistor $R_1$ as a function of time.

A

$i_{R1}(t) = \frac{V_0}{R_1} + \left(\frac{V_0}{R_2} - \frac{V_0}{R_1}\right) e^{-\frac{(R_1 + R_2)t}{L}}$

B

$i_{R1}(t) = \frac{V_0}{R_1 + R_2} \left(1 - e^{-\frac{R_1 t}{L}}\right)$

C

$i_{R1}(t) = \frac{V_0}{R_1} e^{-\frac{R_2 t}{L}}$

D

$i_{R1}(t) = \frac{V_0}{R_1 + R_2} + \frac{V_0}{R_1} e^{-\frac{t}{\tau}}$