MHT CET 2024 10th May Morning Shift | Alternating Current Question 38 | Physics

An e.m.f. $E=E_0 \cos \omega t$ is applied to the $L-R$ circuit. The inductive reactance is equal to the resistance ' $R$ ' of the circuit. The power consumed in the circuit is

$\frac{E_0^2}{\sqrt{2} R}$

$\frac{E_0^2}{2 R}$

$\frac{E_0^2}{4 R}$

$\frac{E_0^2}{R}$

MHT CET 2024 10th May Morning Shift

MCQ (Single Correct Answer)

-0

A series resonant circuit consists of inductor ' $L$ ' of negligible resistance and a capacitor ' $C$ ' which produces resonant frequency ' $f$ '. If $L$ is changed to 3 L and ' C ' is changed to 6 C , the resonant frequency will become.

$\frac{\mathrm{f}}{6}$

$\frac{\mathrm{f}}{3}$

$\frac{\mathrm{f}}{2 \sqrt{2}}$

$\frac{\mathrm{f}}{3 \sqrt{2}}$

MHT CET 2024 10th May Morning Shift

MCQ (Single Correct Answer)

-0

The inductive reactance of a coil is ' $R$ ' $\Omega$. If inductance of a coil is tripled and frequency of a.c supply is also tripled, then new inductive reactance will be

$\frac{\mathrm{R}}{9}$

$\frac{\mathrm{R}}{3}$

$\mathrm{3 R}$

$\mathrm{9 R}$

MHT CET 2024 9th May Evening Shift

MCQ (Single Correct Answer)

-0

An e.m.f. $E=E_0 \cos \omega t$ is applied to circuit containing L and R in series. If $\mathrm{X}_{\mathrm{L}}=2 \mathrm{R}$, then the power dissipated in the circuit is

$\frac{\mathrm{E}_0{ }^2}{12 \mathrm{R}}$

$\frac{\mathrm{E}_0{ }^2}{10 \mathrm{R}}$

$\frac{\mathrm{E}_0{ }^2}{8 \mathrm{R}}$

$\frac{\mathrm{E}_0{ }^2}{6 \mathrm{R}}$

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