(D) The inductive reactance $X_L$ is given by $X_L = L\omega = L(2\pi f)$.Substituting the given values: $X_L = \frac{1}{\pi} \times 2\pi \times 200 =

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$\tan^{-1}\left(\frac{4}{3}\right)$

(D) The inductive reactance $X_L$ is given by $X_L = L\omega = L(2\pi f)$.Substituting the given values: $X_L = \frac{1}{\pi} \times 2\pi \times 200 = 400 \text{ } \Omega$.The phase angle $\phi$ in an $LR$ series circuit is given by $\tan \phi = \frac{X_L}{R}$.Substituting the values: $\tan \phi = \frac{400}{300} = \frac{4}{3}$.Therefore,$\phi = \tan^{-1}\left(\frac{4}{3}\right)$.

Class 12 Physics Alternating Current RL, RC and LC AC Circuits

Medium

$A$ coil having an inductance of $\frac{1}{\pi} \text{ H}$ is connected in series with a resistance of $300 \text{ } \Omega$. If $20 \text{ V}$ from a $200 \text{ Hz}$ source are impressed across the combination,the value of the phase angle between the voltage and the current is

MHT CET 2023 All MHT CET

A
$\tan^{-1}\left(\frac{5}{4}\right)$
B
$\tan^{-1}\left(\frac{4}{5}\right)$
C
$\tan^{-1}\left(\frac{3}{4}\right)$
D
$\tan^{-1}\left(\frac{4}{3}\right)$

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RL, RC and LC AC Circuits

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$A$ resistor of $200 \; \Omega$ and a capacitor of $15.0 \; \mu F$ are connected in series to a $220 \; V, 50 \; Hz$ $ac$ source.
$(a)$ Calculate the current in the circuit.
$(b)$ Calculate the voltage $(rms)$ across the resistor and the capacitor. Is the algebraic sum of these voltages more than the source voltage? If yes,resolve the paradox.

Medium

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In a series $LCR$ circuit,the $rms$ voltage across the resistor and the capacitor are $30 \ V$ and $90 \ V$ respectively. If the applied voltage is $V = 50 \sqrt{2} \sin \omega t$,then the peak voltage across the inductor is

EasyWBJEE 2024

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$A$ resistor and an inductor are connected in series to an $AC$ source of voltage $V = 150 \sin (100 \pi t + \pi) \text{ V}$. If the current in the circuit is $I = 5 \sin (100 \pi t + \frac{2 \pi}{3}) \text{ A}$,then the average power dissipated and the resistance of the resistor are respectively:

MediumAP EAMCET 2018

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In an $L-R$ circuit,the inductive reactance is equal to the resistance $R$ in the circuit. An emf $E = E_0 \cos \omega t$ is applied to the circuit. The power consumed in the circuit is

MediumMHT CET 2023

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For an $LCR$ circuit driven at frequency $\omega $,the equation reads $L\frac{di}{dt} + Ri + \frac{q}{C} = V_i = V_m \sin \omega t$.
$(a)$ Multiply the equation by $i$ and simplify where possible.
$(b)$ Interpret each term physically.
$(c)$ Cast the equation in the form of a conservation of energy statement.
$(d)$ Integrate the equation over one cycle to find that the phase difference between $V$ and $i$ must be acute.

Medium

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$A$ coil having an inductance of $\frac{1}{\pi} \text{ H}$ is connected in series with a resistance of $300 \text{ } \Omega$. If $20 \text{ V}$ from a $200 \text{ Hz}$ source are impressed across the combination,the value of the phase angle between the voltage and the current is

Similar Questions

In a series $LCR$ circuit,the $rms$ voltage across the resistor and the capacitor are $30 \ V$ and $90 \ V$ respectively. If the applied voltage is $V = 50 \sqrt{2} \sin \omega t$,then the peak voltage across the inductor is

$A$ resistor and an inductor are connected in series to an $AC$ source of voltage $V = 150 \sin (100 \pi t + \pi) \text{ V}$. If the current in the circuit is $I = 5 \sin (100 \pi t + \frac{2 \pi}{3}) \text{ A}$,then the average power dissipated and the resistance of the resistor are respectively:

In an $L-R$ circuit,the inductive reactance is equal to the resistance $R$ in the circuit. An emf $E = E_0 \cos \omega t$ is applied to the circuit. The power consumed in the circuit is

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