1. ⇒ (JEE Main 2023 (Online) 1st February Evening Shift)

A square shaped coil of area $70 {cm}^{2}$ having 600 turns rotates in a magnetic field of $0.4 {wbm}^{- 2}$ , about an axis which is parallel to one of the side of the coil and perpendicular to the direction of field. If the coil completes 500 revolution in a minute, the instantaneous emf when the plane of the coil is inclined at $60^{\circ}$ with the field, will be ____________ V. (Take $π = \frac{22}{7}$ )

Correct Answer is (44)

Area $(A) = 70 {cm}^{2} = 70 \times 10^{- 4} m^{2}$

$B = 0.4 T$

$f = \frac{500 revolution}{60 minute} = \frac{500}{60} \frac{rev.}{\sec .}$

Induced emf in rotating coil is given by

Volt $\begin{aligned} e = N ω B A \sin θ \\ = 600 \times 2 \times \frac{22}{7} \times \frac{500}{60} \times 0.4 \times 70 \times 10^{- 4} \sin 30^{\circ} \\ = 600 \times 2 \times \frac{22}{7} \times \frac{500}{6} \times 0.4 \times 70 \times 10^{- 4} \times \frac{1}{2} \\ = 44 Volt \end{aligned}$

2. ⇒ (JEE Main 2023 (Online) 30th January Evening Shift)

In an ac generator, a rectangular coil of 100 turns each having area $14 \times 10^{- 2} m^{2}$ is rotated at $360 rev / \min$ about an axis perpendicular to a uniform magnetic field of magnitude $3.0 T$ . The maximum value of the emf produced will be ________ $V$ .

$($ Take $π = \frac{22}{7})$

Correct Answer is (1584)

$ϕ = B . A$

$ϕ = BNA \cos ω t$

So, $E m f = \frac{- d ϕ}{d t} = N B A ω \sin ω t$

So maximum value of emf is

$E_{max} = N B A ω$

$= 100 \times 3 \times 14 \times 10^{- 2} \times \frac{360 \times 2 π}{60} = 1584$

3. ⇒ ( JEE Main 2022 (Online) 27th July Morning Shift)

A direct current of $4 A$ and an alternating current of peak value $4 A$ flow through resistance of $3 Ω$ and $2 Ω$ respectively. The ratio of heat produced in the two resistances in same interval of time will be

A. 3 : 2

B. 3 : 1

C. 3 : 4

D. 4 : 3

Correct Option is (B)

Ratio = $\frac{i_{1}^{2} R_{1}}{{(\frac{i_{2}}{\sqrt{2}})}^{2} R_{2}} = \frac{4^{2} \times 3}{{(\frac{4}{\sqrt{2}})}^{2} \times 2}$

$\Rightarrow$ Ratio = 3 : 1

4. ⇒ (JEE Main 2022 (Online) 27th June Morning Shift)

The current flowing through an ac circuit is given by

I = 5 sin(120 $π$ t)A

How long will the current take to reach the peak value starting from zero?

A. $\frac{1}{60}$ s

B. 60 s

C. $\frac{1}{120}$ s

D. $\frac{1}{240}$ s

Correct Option is (D)

$ω = 120 π$

$\Rightarrow T = \frac{1}{60} \sec$

The current will take its peak value in $\frac{T}{4}$ time

So $t = \frac{T}{4}$

$= \frac{1}{240} s$

5. ⇒ (JEE Main 2022 (Online) 24th June Morning Shift)

A resistance of 40 $Ω$ is connected to a source of alternating current rated 220 V, 50 Hz. Find the time taken by the current to change from its maximum value to the rms value :

A. 2.5 ms

B. 1.25 ms

C. 2.5 s

D. 0.25 s

Correct Option is (A)

$I = I_{0} \cos (ω t)$ say

$\Rightarrow$ At maximum $ω t_{1} = 0$ or $t_{1} = 0$

Then at rms value $I = I_{0} / \sqrt{2}$

$\Rightarrow ω t_{2} = π / 4$

$\Rightarrow ω (t_{2} - t_{1}) = π / 4$

$Δ t = \frac{π}{4 ω} = \frac{π T}{4 \times 2 π}$

$= \frac{1}{400} s$ or 2.5 ms

6. ⇒ (JEE Main 2021 (Online) 18th March Morning Shift)

An AC source rated 220 V, 50 Hz is connected to a resistor. The time taken by the current to change from its maximum to the rms value is :

A. 2.5 ms

B. 25 ms

C. 2.5 s

D. 0.25 ms

Correct Option is (A)

$I = I_{0} \sin ω t$

$\Rightarrow$ $\frac{I_{M}}{\sqrt{2}} = I_{M} \sin ω t$

$\Rightarrow$ $ω t = \frac{π}{4}$

$\Rightarrow$ $t = \frac{π}{4 ω}$ $= \frac{π}{4 (2 π f)}$

$\Rightarrow$ t = $\frac{1}{8 \times 30} = \frac{1}{400}$ = 2.5 ms

7. ⇒ (JEE Main 2021 (Online) 17th March Morning Shift)

An AC current is given by I = I₁ sin $ω$ t + I₂ cos $ω$ t. A hot wire ammeter will give a reading :

A. $\frac{I_{1} + I_{2}}{\sqrt{2}}$

B. $\sqrt{\frac{I_{1}^{2} - I_{2}^{2}}{2}}$

C. $\sqrt{\frac{I_{1}^{2} + I_{2}^{2}}{2}}$

D. $\frac{I_{1} + I_{2}}{2 \sqrt{2}}$

Correct Option is (C)

$I_{R M S} = \sqrt{\frac{\int I^{2} d t}{\int d t}}$

$I_{R M S}^{2} = \int_{0}^{T} \frac{{(I_{1} \sin ω t + I_{2} \cos ω t)}^{2} d t}{T}$

$= \frac{1}{T} \int_{0}^{T} (I_{1}^{2} \sin^{2} ω t + I_{2}^{2} \cos^{2} ω t + 2 I_{1} I_{2} \sin ω t \cos ω t) d t$

$= \frac{I_{1}^{2}}{2} + \frac{I_{2}^{2}}{2} + 0$

$I_{R M S} = \sqrt{\frac{I_{1}^{2} + I_{2}^{2}}{2}}$

8. ⇒ (JEE Main 2021 (Online) 26th February Morning Shift)

An alternating current is given by the equation i = i₁ sin $ω$ t + i₂ cos $ω$ t. The rms current will be :

A. $\frac{1}{\sqrt{2}} {(i_{1}^{2} + i_{2}^{2})}^{\frac{1}{2}}$

B. $\frac{1}{\sqrt{2}} (i_{1} + i_{2})$

C. $\frac{1}{\sqrt{2}} (i_{1} + i_{2})^{2}$

D. $\frac{1}{2} {(i_{1}^{2} + i_{2}^{2})}^{\frac{1}{2}}$

Correct Option is (A)

$I_{0} = \sqrt{I_{1}^{2} + I_{2}^{2} + 2 I_{1} I_{2} \cos θ}$

$I_{0} = \sqrt{I_{1}^{2} + I_{2}^{2} + 2 I_{1} I_{2} \cos 90^{\circ}}$

$I_{0} = \sqrt{I_{1}^{2} + I_{2}^{2} + 2 I_{1} I_{2} (0)} = \sqrt{I_{1}^{2} + I_{2}^{2}}$

We know that,

$I_{r m s} = \frac{I_{0}}{\sqrt{2}}$

So, $I_{r m s} = \frac{\sqrt{I_{1}^{2} + I_{2}^{2}}}{\sqrt{2}}$

9. ⇒ (JEE Main 2021 (Online) 27th August Morning Shift)

The alternating current is given by $i = {\sqrt{42} \sin (\frac{2 π}{T} t) + 10} A$

The r.m.s. value of of this current is ................. A.

Correct Answer is (11)

$f_{r m s}^{2} = f_{1 r m s}^{2} + f_{2 r m s}^{2}$

$= {(\frac{\sqrt{42}}{\sqrt{2}})}^{2} + 10^{2}$

$= 121 \Rightarrow f_{r m s}$ = 11 A

10. ⇒ (JEE Main 2018 (Offline))

In an a.c. circuit, the instantaneous e.m.f. and current are given by
e = 100 sin 30 t
i = 20 sin $(30 t - \frac{π}{4})$
In one cycle of a.c., the average power consumed by the circuit and the wattless current are, respectively

A. 50, 0

B. 50, 10

C. $\frac{1000}{\sqrt{2}}, 10$

D. $\frac{50}{\sqrt{2}}$

Correct Option is (C)

Wattless current,

here $ϕ$ is the angle between i and e.

Average power,

P_av = V_rms I_rms cos $ϕ$

= $\frac{100}{\sqrt{2}} \times \frac{20}{\sqrt{2}}$ cos $\frac{π}{4}$

= $\frac{1000}{\sqrt{2}}$ watt.

11. ⇒ (AIEEE 2007)

In an $a . c .$ circuit the voltage applied is $E = E_{0} \sin ω t .$ The resulting current in the circuit is $I = I_{0} \sin (ω t - \frac{π}{2}) .$ The power consumption in the circuit is given by

A. $P = \sqrt{2} E_{0} I_{0}$

B. $P = \frac{E_{0} I_{0}}{\sqrt{2}}$

C. $P = z e r o$

D. $P = \frac{E_{0} I_{0}}{2}$

Correct Option is (C)

KEY CONCEPT : We know that power consumed in a.c. circuit is given by,

$P = E_{r m s} I_{r m s} \cos ϕ$

Here, $E = E_{0} \sin ω t$

$I = I_{0} \sin (ω t - \frac{π}{2})$

which implies that the phase difference, $ϕ = \frac{π}{2}$

$∴$ $P = E_{r m s} . I_{r m s} . \cos \frac{π}{2} = 0$

$($ as $\cos \frac{π}{2} = 0)$

12. ⇒ (AIEEE 2006)

In an $A C$ generator, a coil with $N$ turns, all of the same area $A$ and total resistance $R,$ rotates with frequency $ω$ in a magnetic field $B .$ The maximum value of $e m f$ generated in the coil is

A. $N . A . B . R .$ $ω$

B. $N . A . B$

C. $N . A . B . R .$

D. $N . A . B .$ $ω$

Correct Option is (D)

$e = - \frac{d ϕ}{d t} = - \frac{d (N \vec{B} . \vec{A})}{d t}$

$= - N \frac{d}{d t} (B A \cos ω t) = N B A ω \sin ω t$

$\Rightarrow e_{max} = N B A ω$

13. ⇒ (AIEEE 2004)

Alternating current can not be measured by $D . C .$ ammeter because

A. Average value of current for complete cycle is zero

B. $A . C .$ Changes direction

C. $A . C .$ can not pass through $D . C .$ Ammeter

D. $D . C .$ Ammeter will get damaged.

Correct Option is (A)

$D . C .$ ammeter measure average current in $A C$ current, average current is zero for complete cycle. Hence reading will be zero.

⇒Alternating Current

Topic 01 : AC Current, Voltage And Power