JEE Main Alternating Current PYQ

6. (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

$E_{max} = N B A ω$

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

7. ( 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

8. (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 $t = \frac{T}{4}$

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

9. (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

10. (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