MHT CET 2024 11th May Morning Shift | Alternating Current Question 18 | Physics

An inductance of $\frac{300}{\pi} \mathrm{mH}$, a capacitance of $\frac{1}{\pi} \mathrm{mF}$ and a resistance of $20 \Omega$ are connected in series with an a.c. source of $240 \mathrm{~V}, 50 \mathrm{~Hz}$. The phase angle of the circuit is

$\tan ^{-1}(0)$

$\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)$

$\tan ^{-1}(1)$

$\tan ^{-1}(\sqrt{3})$

MHT CET 2024 11th May Morning Shift

MCQ (Single Correct Answer)

-0

With an alternating voltage source frequency ' f ', inductor ' $L$ ', capacitor ' $C$ ' and resistance ' $R$ ' are connected in series. The voltage leads the current by $45^{\circ}$. The value of ' $L$ ' is $\left(\tan 45^{\circ}=1\right)$

$\left(\frac{1+2 \pi \mathrm{fCR}}{4 \pi^2 \mathrm{f}^2 \mathrm{C}}\right)$

$\left(\frac{1-2 \pi f C R}{4 \pi^2 \mathrm{f}^2 \mathrm{C}}\right)$

$\left(\frac{4 \pi^2 \mathrm{f}^2 \mathrm{C}}{1+2 \pi \mathrm{fCR}}\right)$

$\left(\frac{4 \pi^2 \mathrm{f}^2 \mathrm{C}}{1-2 \pi \mathrm{fCR}}\right)$

MHT CET 2024 11th May Morning Shift

MCQ (Single Correct Answer)

-0

With gradual increase in frequency of an a.c. supply, the impedance of an LCR series circuit

increases.

decreases.

remains constant

first decreases, becomes minimum and then increases.

MHT CET 2024 10th May Evening Shift

MCQ (Single Correct Answer)

-0

An a.c. voltage source $\mathrm{V}=\mathrm{V}_0 \sin \omega \mathrm{t}$ is connected across resistance ' $R$ ' and capacitance ' $C$ ' in series. It is given that $R=\frac{1}{\omega c}$ and the peak current is $\mathrm{I}_0$. If the angular frequency of the voltage source is changed to $\left(\frac{\omega}{\sqrt{3}}\right)$, then the new peak current in the circuit is

$\frac{\mathrm{I}_0}{2}$

$\frac{\mathrm{I}_0}{\sqrt{2}}$

$ \sqrt{2} \mathrm{I}_0$

$\sqrt{3} \mathrm{I}_0$

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