(A) The phase angle $\phi$ in an $LCR$ series circuit is given by $\tan \phi = \frac{X_C - X_L}{R}$.Substituting the given values: $\tan \phi = \frac{

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Leads the voltage by $\tan^{-1}(3/4)$

(A) The phase angle $\phi$ in an $LCR$ series circuit is given by $\tan \phi = \frac{X_C - X_L}{R}$.Substituting the given values: $\tan \phi = \frac{8 - 5}{4} = \frac{3}{4}$.Therefore,$\phi = \tan^{-1}(3/4)$.Since $X_C > X_L$,the circuit is capacitive in nature.In a capacitive circuit,the current leads the voltage by the phase angle $\phi$.Thus,the current leads the voltage by $\tan^{-1}(3/4)$.

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

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In a given series $LCR$ circuit,$R = 4\, \Omega, X_L = 5\, \Omega$ and $X_C = 8\, \Omega$. The current :-

A
Leads the voltage by $\tan^{-1}(3/4)$
B
Leads the voltage by $\tan^{-1}(5/8)$
C
Lags the voltage by $\tan^{-1}(3/4)$
D
Lags the voltage by $\tan^{-1}(5/8)$

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Alternating Current

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RL, RC and LC AC Circuits

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The following series $L-C-R$ circuit,when driven by an emf source of angular frequency $70 \text{ krad/s}$,behaves effectively as:

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In a series $LCR$ circuit,$R=300 \Omega$,$L=0.9 \text{ H}$,$C=2 \mu\text{F}$,and $\omega=1000 \text{ rad/s}$. The impedance of the circuit is: (in $Omega$)

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The impedance of a circuit,when a resistance $R$ and an inductor of inductance $L$ are connected in series in an $AC$ circuit of frequency $f$ is

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In a given series $LCR$ circuit,$R = 4\, \Omega, X_L = 5\, \Omega$ and $X_C = 8\, \Omega$. The current :-

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An electric bulb rated $100 \,W - 300 \,V$ is connected to an $AC$ supply of $500 \,V$ and $\frac{150}{\pi} \,Hz$. The required inductance to save the electric bulb is .......... $H$.

$A$ $12\,\Omega$ resistor and a $0.21\,H$ inductor are connected in series to an $ac$ source operating at $20\,V$,$50\,Hz$. The phase angle between the current and the source voltage is.....$^o$

The following series $L-C-R$ circuit,when driven by an emf source of angular frequency $70 \text{ krad/s}$,behaves effectively as:

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