JEE Main Alternating Current PYQ

26. ( JEE Main 2021 (Online) 20th July Evening Shift)

A series LCR circuit of R = 5 $Ω$ , L = 20 mH and C = 0.5 $μ$ F is connected across an AC supply of 250 V, having variable frequency. The power dissipated at resonance condition is ______________ $\times$ 10² W.

Correct Answer is (125)

X_L = X_C (due to resonance)

Z = R so $i_{r m s} = \frac{V}{Z} = \frac{V}{R}$

$\frac{V^{2}}{R} = \frac{250 \times 250}{5} = 125 \times 10^{2} W$

27. (JEE Main 2021 (Online) 20th July Morning Shift)

In an LCR series circuit, an inductor 30 mH and a resistor 1 $Ω$ are connected to an AC source of angular frequency 300 rad/s. The value of capacitance for which, the current leads the voltage by 45 $^{\circ}$ is $\frac{1}{x} \times 10^{- 3}$ F. Then the value of x is ____________.

Correct Answer is (3)

Given,

Inductance, L = 30 mH

Resistance, R = 1 $Ω$

Angular frequency, $ω$ = 300 rad/s

We know that in L-C-R circuit, $\tan ϕ = \frac{X_{C} - X_{L}}{R}$

where, $ϕ$ = phase angle = 45 $^{\circ}$

X_C = capacitive reactance = $\frac{1}{ω C}$

X_L = inductive reactance = $ω$ L

$\Rightarrow$ $\tan 45^{\circ} = \frac{X_{C} - X_{L}}{R}$

$\Rightarrow X_{C} - X_{L} = R$ [ $∵$ tan 45 $^{\circ}$ = 1]

$\Rightarrow \frac{1}{ω C} - ω L = R \Rightarrow \frac{1}{ω C} - 300 \times 30 \times 10^{- 3} = 1$

$\Rightarrow \frac{1}{ω C} = 10 \Rightarrow ω C = \frac{1}{10}$

$\Rightarrow C = \frac{1}{10 ω} \Rightarrow C = \frac{1}{10 \times 300}$

$\Rightarrow C = \frac{1}{3} \times 10^{- 3} F$ .... (i)

According to question, the value of capacitance is $\frac{1}{x} \times 10^{- 3} F$ . So, on comparing it with Eq. (i), we can say x = 3.

28. (JEE Main 2021 (Online) 25th February Morning Shift)

A transmitting station releases waves of wavelength 960 m. A capacitor of 2.56 $μ$ F is used in the resonant circuit. The self inductance of coil necessary for resonance is __________ $\times$ 10^{$-$ 8} H.

Correct Answer is (10)

$λ$ = 960 m

C = 2.56 $μ$ F = 2.56 $\times$ 10^{$-$ 6} F

c = 3 $\times$ 10⁸ m/s

L = ?

Now at resonance, $ω_{0} = \frac{1}{\sqrt{L C}}$

[Resonant frequency]

$2 π f_{0} = \frac{1}{\sqrt{L C}}$

On substituting $f_{0} = \frac{c}{λ}$ , we have $2 π \frac{c}{λ} = \frac{1}{\sqrt{L C}}$

Squaring both sides : $4 π^{2} \frac{c^{2}}{λ^{2}} = \frac{1}{L C}$

$= \frac{4 \times 10 \times {(3 \times 10^{8})}^{2}}{{(960)}^{2}} = \frac{1}{L \times 2.56 \times 10^{- 6}}$

$\Rightarrow \frac{1}{L} = \frac{4 \times 10 \times 9 \times 10^{16} \times 2.56 \times 10^{- 6}}{960 \times 960}$

$\Rightarrow L = 10 \times 10^{- 8}$ H

29. ( JEE Main 2021 (Online) 24th February Morning Shift)

A resonance circuit having inductance and resistance 2 $\times$ 10^{$-$ 4} H and 6.28 $Ω$ respectively oscillates at 10 MHz frequency. The value of quality factor of this resonator is ___________. [ $π$ = 3.14]

Correct Answer is (2000)

Given, L = 2 $\times$ 10^{$-$ 4} H, R = 6.28 $Ω$ , f₀ = 10 MHz = 10 $\times$ 10⁶ Hz

$∴$ Quality factor $= ω_{0} \frac{L}{R} = 2 π f_{0} \frac{L}{R}$

$= 2 π \times 10 \times 10^{6} \times \frac{2 \times 10^{- 4}}{6.28}$

$= 2 \times 10^{3} = 2000$

30. (JEE Main 2020 (Online) 6th September Morning Slot)

An AC circuit has R= 100 $Ω$ , C = 2 $μ$ F and L = 80 mH, connected in series. The quality factor of the circuit is :

A. 20

B. 2

C. 0.5

D. 400

Correct Option is (B)

$Q = \frac{1}{R} \sqrt{\frac{L}{C}}$

= $\frac{1}{100} \sqrt{\frac{80 \times 10^{- 3}}{2 \times 10^{- 6}}}$

= $\frac{1}{100} \sqrt{40 \times 10^{3}}$

= $\frac{200}{100}$ = 2