JEE Main Alternating Current PYQ

1. (JEE Main 2024 (Online) 31st January Evening Shift )

An AC voltage $V = 20 \sin 200 π t$ is applied to a series LCR circuit which drives a current $I = 10 \sin (200 π t + \frac{π}{3})$ . The average power dissipated is:

A.21.6 W

B.200 W

C.173.2 W

D.50 W

Correct option is (d)

$\begin{aligned} < P >= IV \cos ϕ \\ = \frac{20}{\sqrt{2}} \times \frac{10}{\sqrt{2}} \times \cos 60^{\circ} \\ = 50 W \end{aligned}$

2.(JEE Main 2024 (Online) 30th January Evening Shift )

An alternating voltage $V (t) = 220 \sin 100 π t$ volt is applied to a purely resistive load of $50 Ω$ . The time taken for the current to rise from half of the peak value to the peak value is:

A.7.2 ms

B.3.3 ms

C.5 ms

D.2.2 ms

Correct option is (b)

Rising half to peak

$\begin{aligned} t = T / 6 \\ t = \frac{2 π}{6 ω} = \frac{π}{3 ω} = \frac{π}{300 π} = \frac{1}{300} = 3.33 ms \end{aligned}$

3.(JEE Main 2024 (Online) 29th January Evening Shift )

In an a.c. circuit, voltage and current are given by:

$V = 100 \sin (100 t) V$ and $I = 100 \sin (100 t + \frac{π}{3}) mA$ respectively.

The average power dissipated in one cycle is:

A.5 W

B.25 W

C.2.5 W

D.10 W

Correct option is (c)

$\begin{aligned} P_{avg} = V_{rms} I_{r m s} \cos (Δ ϕ) \\ = \frac{100}{\sqrt{2}} \times \frac{100 \times 10^{- 3}}{\sqrt{2}} \times \cos (\frac{π}{3}) \\ = \frac{10^{4}}{2} \times \frac{1}{2} \times 10^{- 3} \\ = \frac{10}{4} = 2.5 W \end{aligned}$

4.(JEE Main 2024 (Online) 4th April Morning Shift )

A alternating current at any instant is given by $i = [6 + \sqrt{56} \sin (100 π t + π / 3)]$ A. The $r m s$ value of the current is ______ A.

Correct answer is 8

The given alternating current (AC) can be represented as $i = 6 + \sqrt{56} \sin (100 π t + π / 3)$ A, where $6$ is the DC component and $\sqrt{56} \sin (100 π t + π / 3)$ is the AC component of the current. The RMS (Root Mean Square) value of an alternating current is a measure of the equivalent direct current (DC) that will produce the same power in a resistor. The RMS value is mostly relevant for the AC component of the current, as the DC component's effective value is just its magnitude itself.

The RMS value of the total current is not straightforward because the presence of the DC component affects how we calculate the RMS value. However, when calculating RMS values for a signal consisting of a superposition of AC and DC components, one notable property is that the RMS value of the combined signal is the square root of the sum of the squares of the RMS values of the separate AC and DC components.

First, let's acknowledge the components separately:

The DC component is: $6$ A

The AC component is: $\sqrt{56} \sin (100 π t + π / 3)$ A

For the DC component, the RMS value is simply its magnitude:

$I_{R M S, D C} = 6$ A

For the AC component, the RMS value is calculated using the formula for the RMS value of a sinusoidal function, which is $I_{R M S} = \frac{I_{m a x}}{\sqrt{2}}$ , where $I_{m a x}$ is the peak value of the current. In this case, $I_{m a x} = \sqrt{56}$ .

Therefore, the RMS value of the AC component is:

$I_{R M S, A C} = \frac{\sqrt{56}}{\sqrt{2}} = \frac{\sqrt{56}}{\sqrt{2}} = \sqrt{\frac{56}{2}} = \sqrt{28}$ A.

Finally, to find the total RMS value of the current, combine the DC and AC components as follows:

$I_{R M S} = \sqrt{{(I_{R M S, D C})}^{2} + {(I_{R M S, A C})}^{2}}$

Substituting the values:

$I_{R M S} = \sqrt{{(6)}^{2} + {(\sqrt{28})}^{2}}$

$= \sqrt{36 + 28}$

$= \sqrt{64}$

$= 8$ A.

Therefore, the RMS value of the current is $8$ A.

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