JEE Main Alternating Current PYQ

21. (JEE Main 2021 (Online) 27th July Evening Shift)

A 100 $Ω$ resistance, a 0.1 $μ$ F capacitor and an inductor are connected in series across a 250 V supply at variable frequency. Calculate the value of inductance of inductor at which resonance will occur. Given that the resonant frequency is 60 Hz.

A. 0.70 H

B. 70.3 mH

C. 7.03 $\times$ 10^{$-$ 5} H

D. 70.3 H

Correct Option is (D)

C = 0.1 $μ$ F = 10^{$-$ 7} F

Resonant frequency = 60 Hz.

$ω_{0} = \frac{1}{\sqrt{L C}}$

$2 π f_{0} = \frac{1}{\sqrt{L C}} \Rightarrow L = \frac{1}{4 π^{2} f_{0}^{2} C}$

by putting values $L ≃ 70.3$ Hz.

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

A 0.07 H inductor and a 12 $Ω$ resistor are connected in series to a 220V, 50 Hz ac source. The approximate current in the circuit and the phase angle between current and source voltage are respectively. [Take $π$ as $\frac{22}{7}$ ]

A. 8.8 A and $\tan^{- 1} (\frac{11}{6})$

B. 88 A and $\tan^{- 1} (\frac{11}{6})$

C. 0.88 A and $\tan^{- 1} (\frac{11}{6})$

D. 8.8 A and $\tan^{- 1} (\frac{6}{11})$

Correct Option is (A)

$ϕ = \tan^{- 1} (\frac{X_{L}}{R})$

$X_{L} = ω L$

$X_{L} = 2 \times \frac{22}{7} \times 50 \times 0.07 = 22 Ω$

$ϕ = \tan^{- 1} (\frac{22}{12})$

$R = 12 Ω$

$ϕ = \tan^{- 1} (\frac{11}{6})$

$Z = \sqrt{X_{L}^{2} + R^{2}} = 25.059$

$I = \frac{V}{Z} = \frac{220}{25.059} = 8.77 A$

23. (JEE Main 2021 (Online) 22th July Evening Shift)

Match List - I with List - II

	List - I		List - II
(a)	$ω L > \frac{1}{ω C}$	(i)	Current is in phase with emf
(b)	$ω L = \frac{1}{ω C}$	(ii)	Current lags behind the applied emf
(c)	$ω L < \frac{1}{ω C}$	(iii)	Maximum current occurs
(d)	Resonant frequency	(iv)	Current leads the emf

Choose the correct answer from the options given below

A. a(ii), b(i), c(iv), d(iii)

B. a(ii), b(i), c(iii), d(iv)

C. a(iii), b(i), c(iv), d(ii)

D. a(iv), b(iii), c(ii), d(i)

Correct Option is (A)

$ω L = \frac{1}{ω C}, X_{L} = X_{C}$

So current in phase with EMF

At resonance, current have maximum value.

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

In a series LCR resonance circuit, if we change the resistance only, from a lower to higher value :

A. The bandwidth of resonance circuit will increase.

B. The resonance frequency will increase.

C. The quality factor will increase.

D. The quality factor and the resonance frequency will remain constant.

Correct Option is (A)

$ω = \frac{1}{\sqrt{L C}}$

$\Rightarrow$ 2 $π$ f = $\frac{1}{\sqrt{L C}}$

$\Rightarrow$ f = $\frac{1}{2 π \sqrt{L C}}$

f does not depends on resistance(R).

Quality factor, $Q = \frac{ω L}{R}$

$\Rightarrow$ $Q \propto \frac{1}{R}$

So if R increase then Q will decrease.

Also, $Q = \frac{ω L}{R} = \frac{ω}{Δ β}$

where $Δ β$ = bandwidth

$Δ β = \frac{R}{L}$

So if R increase then $Δ β$ will increase too.

25. (JEE Main 2021 (Online) 26th February Evening Shift)

Find the peak current and resonant frequency of the following circuit (as shown in figure).

JEE Main 2021 (Online) 26th February Evening Shift Physics - Alternating Current Question 70 English

A. 2 A and 100 Hz

B. 2 A and 50 Hz

C. 0.2 A and 100 Hz

D. 0.2 A and 50 Hz

Correct Option is (D)

We know, z = $\sqrt{{(x_{L} - x_{C})}^{2} + R^{2}}$

$x_{L} = ω_{L} = 100 \times 100 \times 10^{- 3} = 10 Ω$

$x_{C} = \frac{1}{ω_{C}} = \frac{1}{100 \times 100 \times 10^{- 6}} = 10 Ω$

$∴$ $z = \sqrt{{(10 - 100)}^{2} + R^{2}} = \sqrt{90^{2} + 120^{2}}$

$= 30 \times 5 = 150 Ω$

$i_{p e a k} = \frac{Δ v}{z} = \frac{30}{150} = \frac{1}{5} a m p = 0.2$ amp

For resonant frequency,

$ω L = \frac{1}{ω C} \Rightarrow ω^{2} = \frac{1}{L C} \Rightarrow ω = \frac{1}{\sqrt{L C}}$

& $f = \frac{1}{2 π \sqrt{L C}} = \frac{1}{2 π \sqrt{100 \times 10^{- 3} \times 100 \times 10^{- 6}}}$

$= \frac{100 \sqrt{10}}{2 π} = \frac{100 π}{2 π} = 50$ Hz

as $\sqrt{10} \approx π$