JEE Main Alternating Current PYQ

1.(JEE Main 2024 (Online) 29th January Morning Shift )

A capacitor of capacitance $100 μ F$ is charged to a potential of $12 V$ and connected to a $6.4 mH$ inductor to produce oscillations. The maximum current in the circuit would be :

A.2.0 A

B.3.2 A

C.1.5 A

D.1.2 A

Correct option is (c)

By energy conservation

$\begin{aligned} \frac{1}{2} {CV}^{2} = \frac{1}{2} {LI}_{max}^{2} \\ I_{max} = \sqrt{\frac{C}{L}} V \\ = \sqrt{\frac{100 \times 10^{- 6}}{6.4 \times 10^{- 3}}} \times 12 \\ = \frac{12}{8} = \frac{3}{2} = 1.5 A \end{aligned}$

2. (JEE Main 2023 (Online) 25th January Morning Shift)

In an LC oscillator, if values of inductance and capacitance become twice and eight times, respectively, then the resonant frequency of oscillator becomes $x$ times its initial resonant frequency $ω_{0}$ . The value of $x$ is :

A. 1/4

B. 1/16

C. 4

D. 16

Correct Option is (A)

The resonance frequency of LC oscillations circuit is

$\begin{aligned} ω_{0} = \frac{1}{\sqrt{LC}} \\ L \to 2 L \\ C \to 8 C \\ ω = \frac{1}{\sqrt{2 L \times 8 C}} = \frac{1}{4 \sqrt{LC}} \\ ω = \frac{ω_{0}}{4} \end{aligned}$

So $x = \frac{1}{4}$

3. (JEE Main 2023 (Online) 8th April Morning Shift)

An oscillating LC circuit consists of a $75 mH$ inductor and a $1.2 μ F$ capacitor. If the maximum charge to the capacitor is $2.7 μ C$ . The maximum current in the circuit will be ___________ $mA$

Correct Answer is (9)

The maximum current in an LC circuit can be found using the following formula related to simple harmonic motion:

$I_{max} = ω Q_{max}$ ,

where:

$I_{max}$ is the maximum current,
$ω$ is the angular frequency, and
$Q_{max}$ is the maximum charge on the capacitor.

The angular frequency $ω$ for an LC circuit is given by:

$ω = \frac{1}{\sqrt{L C}}$ ,

where:

$L$ is the inductance, and
$C$ is the capacitance.

Given that $L = 75 mH = 75 \times 10^{- 3} H$ , $C = 1.2 μ F = 1.2 \times 10^{- 6} F$ ,

and $Q_{max} = 2.7 μ C = 2.7 \times 10^{- 6} C$ ,

we can substitute these values into the formulas to find $I_{max}$ :

$ω = \frac{1}{\sqrt{(75 \times 10^{- 3}) (1.2 \times 10^{- 6})}} = 3333.33 rad/s$ ,

$I_{max} = ω Q_{max} = 3333.33 \times 2.7 \times 10^{- 6} = 0.009 A$ .

Therefore, the maximum current in the circuit is $0.009 A$ , or equivalently, $9 mA$

4. (JEE Main 2022 (Online) 29th July Evening Shift)

A capacitor of capacitance 500 $μ$ F is charged completely using a dc supply of 100 V. It is now connected to an inductor of inductance 50 mH to form an LC circuit. The maximum current in LC circuit will be _______ A.

Correct Answer is (10)

JEE Main 2022 (Online) 29th July Evening Shift Physics - Alternating Current Question 24 English Explanation 1

At steady state charge stored on the capacitor,

$q_{max} = C V$

$= 500 \times 10^{- 6} \times 100$

$= 5 \times 10^{- 2} C$

JEE Main 2022 (Online) 29th July Evening Shift Physics - Alternating Current Question 24 English Explanation 2

Energy stored in the capacitor,

$U_{max} = \frac{q_{max}^{2}}{2 C}$

Now, when electrostatic energy of capacitor converted to magnetic field energy then all energy of capacitor is transferrd to the inductor.

$∴$ Maximum energy stored in the inductor

$U_{L max} = \frac{1}{2} L I_{max}^{2}$

$∴$ $\frac{1}{2} L I_{max}^{2} = \frac{q_{max}^{2}}{2 C}$

$\Rightarrow I_{max} = \frac{q_{max}}{\sqrt{L C}}$

$= \frac{5 \times 10^{- 2}}{\sqrt{50 \times 10^{- 3} \times 500 \times 10^{- 6}}}$

$= \frac{5 \times 10^{- 2}}{5 \times 10^{- 3}}$

$= 10 A$

5. ( 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