JEE Mains · Physics · STD 12 -Alternating current
- JEE Mains 2024
- Hard
- MCQ
Given below are two statements : Statement \(I\) : In an \(LCR\) series circuit, current is maximum at resonance. Statement \(II\) : Current in a purely resistive circuit can never be less than that in a series LCR circuit when connected to same voltage source. In the light of the above statements, choose the correct from the options given below :
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) Both Statement \(I\) and Statement \(II\) are true
Step-by-step Solution
Detailed explanation
Statement-\(I\)
\(\mathrm{I}_{\mathrm{m}}=\frac{\mathrm{V}_{\mathrm{m}}}{\sqrt{\mathrm{R}^2+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^2}}\) at resonance \(\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}\)
Thus, \(I_m=\frac{V_m}{R}\)
\(\because\) Impendence is minimum therefore \(I\) is maximum at resonance.
Statement-\(II\)
\(\mathrm{I}=\left(\frac{\mathrm{V}}{\mathrm{R}}\right)\) in purely resistive circuit.
More questions from Physics
- Four point charges \(-q, +q, +q\) and \(-q\) are placed on \(y\) axis at \(y = -2d\), \(y = -d, y = +d\) and \(y = +2d\), respectively. The magnitude of the electric field \(E\) at a point on the \(x -\) axis at \(x = D\), with \(D > > d\), will vary asJEE Mains 2019 Hard
- A wire of resistance \(20 \Omega\) is divided into \(10\) equal parts. A combination of two parts are connected in parallel and so on. Now resulting pairs of parallel combination are connected in series. The equivalent resistance of final combination is _______\(\Omega\).JEE Mains 2024 Hard
- Time periods of oscillation of the same simple pendulum measured using four different measuring clocks were recorded as \(4.62 \mathrm{~s}, 4.632 \mathrm{~s}, 4.6 \mathrm{~s}\) and \(4.64 \mathrm{~s}\). The arithmetic mean of these reading in correct significant figure is _______.JEE Mains 2024 Hard
- An electron rotates in a circle around a nucleus having positive charge \(\mathrm{Ze}\). Correct relation between total energy ( \(E\) ) of electron to its potential energy \((U)\) is _______.JEE Mains 2024 Hard
- In Young's double slit experiment, if the source of light changes from orange to blue thenJEE Mains 2021 Medium
- A line charge of length \(\frac{\text { ' } a \text { ' }}{2}\) is kept at the center of an edge \(B C\) of a cube \(A B C D E F G H\) having edge length ' \(a\) ' as shown in the figure. If the density of line charge is \(\lambda\) C per unit length, then the total electric flux through all the faces of the cube will be _______. (Take, \(\epsilon_0\) as the free space permittivity)
JEE Mains 2025 Hard
More PYQs from JEE Mains
- The value closest to the thermal velocity of a Helium atom at room temperature \((300\,K)\)in \(ms^{-1}\) is \([k_B\, = 1 .4\times10^{-23}\,J/K;\, m_{He}\, = 7\times10^{-27}\,kg]\)JEE Mains 2018 Medium
- The gravitational field, due to the 'left over part' of a uniform sphere (from which a part as shown, has been 'removed out'), at a very far off point , \(P\), located as shown, would be (nearly)
JEE Mains 2013 Hard - A ball of mass \(m\) is thrown vertically upward. Another ball of mass \(2\,m\) is thrown an angle \(\theta\) with the vertical. Both the balls stay in air for the same period of time. The ratio of the heights attained by the two balls respectively is \(\frac{1}{x}\). The value of \(x\) is \(.....\)JEE Mains 2022 Medium
- As shown in the figure, a combination of a thin plano concave lens and a thin plano convex lens is used to image an object placed at infinity. The radius of curvature of both the lenses is \(30\,cm\) and refraction index of the material for both the lenses is \(1.75\). Both the lenses are placed at distance of \(40\,cm\) from each other. Due to the combination, the image of the object is formed at distance \(x =..........cm\), from concave lens.
JEE Mains 2023 Hard - As shown in figure, when a spherical cavity (centred at \(\mathrm{O})\) of radius \(1\) is cut out of a uniform sphere of radius \(\mathrm{R} \text { (centred at } \mathrm{C}),\) the centre of mass of remaining (shaded) part of sphere is at \(G\), i.e, on the surface of the cavity. \(\mathrm{R}\) can be detemined by the equation
JEE Mains 2020 Hard