Alternating Current – JEE Main Previous Year Questions with Solutions

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Q. In a series LCR circuit R = $200 \Omega$ and the voltage and the frequency of the main supply is 220 V and 50 Hz respectively. On taking out the capacitance from the circuit the current lages behind the voltage by $30^{\circ}$. On taking out the inductor from the circuit the current leads the voltage by $30^{\circ}$. The power dissipated in the LCR circuit is :

(1) 242 W

(2) 305 W

(3) 210 W

(4) Zero W

[AIEEE – 2010]

Sol. (1)


Q. A fully charged capacitor C with intial charge $\mathrm{q}_{0}$ is connected to a coil of self inductance L at t = 0. The time at which the energy is stored equally between the electric and the magnetic fields is :-

(1) $2 \pi \sqrt{\mathrm{LC}}$

(2) $\sqrt{1 C}$

(3) $\pi \sqrt{\mathrm{LC}}$

(4) $\frac{\pi}{4} \sqrt{\mathrm{LC}}$

[AIEEE – 2011]

Sol. (4)


Q. In an LCR circuit as shown below both switches are open initially. Now switch S1 is closed, $\mathrm{S}_{2}$ kept open, (q is charge on the capacitor and $\tau=\mathrm{RC}$ is Capacitive time constant). Which of the following statement is correct?

(1) Work done by the battery is half of the energy dissipated in the resistor

(2) At t $=\tau, q=C \sqrt{/ 2}$

(3) At $\mathrm{t}=2 \pi, \mathrm{q}=\mathrm{CV}\left(1-\mathrm{e}^{-2}\right)$

(4) At $\mathrm{t}=\frac{\tau}{2}, \mathrm{q}=\mathrm{CV}\left(1-\mathrm{e}^{-1}\right)$

[JEE Main-2013]

Sol. (3)


Q. An LCR circuit is equivalent to a damped pendulum. In an LCR circuit the capacitor is charged to $\mathrm{Q}_{0}$ and then connected to the L and R as shown below. If a student plots graphs of the square of maximum charge on the capacitor with time (t) for two different values $\mathrm{L}_{1}$ and $\mathrm{L}_{2}\left(\mathrm{L}_{1}>\mathrm{L}_{2}\right)$ of L then which of the following represents this graph correctly ? (plots are schematic and not drawn to scale)

[JEE Main-2015]

Sol. (3)

As damping is happening its amplitude would vary as

The oscillations decay exponentially and will be proportional to $\mathrm{e}^{-\gamma t}$ where $\gamma$ depends inversely on L.

So as inductance increases decay becomes slower

 for


Q. An arc lamp requires a direct current of 10A at 80V to function. If it is connected to a 220V (rms), 50 Hz AC supply, the series inductor needed for it to work is close to :-

(1) 0.065 H

(2) 80 H

(3) 0.08 H

(4) 0.044 H

[JEE Main-2016]

Sol. (1)

As damping is happening its amplitude would vary as

The oscillations decay exponentially and will be proportional to $\mathrm{e}^{-\gamma t}$ where $\gamma$ depends inversely on L.

So as inductance increases decay becomes slower

 for


Q. For an RLC circuit driven with voltage of amplitude $\mathrm{v}_{\mathrm{m}}$ and frequency $\mathrm{w}_{0}=\frac{1}{\sqrt{\mathrm{LC}}}$ the current exhibits resonance. The quality factor, Q is given by :

(1) $\frac{\omega_{0} \mathrm{R}}{\mathrm{L}}$

( 2)$\frac{\mathrm{R}}{\left(\omega_{0} \mathrm{C}\right)}$

(3) $\frac{\mathrm{CR}}{\omega_{0}}$

(4) $\frac{\omega_{0} \mathrm{L}}{\mathrm{R}}$

[JEE Main-2018]

Sol. (4)

Quality factor $=\frac{\omega_{0} \mathrm{L}}{\mathrm{R}}$


Q. In an a. c. circuit, the instantaneous e.m.f. and current are given by e = 100 sin 30 t i = 20

sin $\left(30 t-\frac{\pi}{4}\right)$

In one cycle of a.c., the average power consumed by the circuit and the wattless current are, respectively.

(1) $\frac{1000}{\sqrt{2}}, 10$

(2) $\frac{50}{\sqrt{2}}, 0$

(3) 50, 0

(4) 50, 10

[JEE Main-2018]

Sol. (1)


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Comments
  • sam
    December 25, 2019 at 8:47 pm

    was really interesting