In an ac circuit,the current lags behind the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) In an $ac$ circuit,if the current lags behind the voltage by a phase angle $\phi$,the circuit must be an inductive circuit containing a resistor $

Vedclass · Accepted Answer

$R$ and $L$

Vedclass · Answer

(A) In an $ac$ circuit,if the current lags behind the voltage by a phase angle $\phi$,the circuit must be an inductive circuit containing a resistor $(R)$ and an inductor $(L)$.In an $RL$ series circuit,the voltage across the resistor $(V_R)$ is in phase with the current $(I)$,while the voltage across the inductor $(V_L)$ leads the current by $90^\circ$.The resultant voltage $V$ leads the current by a phase angle $\phi$,which is given by:$	an \phi = \frac{V_L}{V_R} = \frac{I_0 X_L}{I_0 R} = \frac{X_L}{R}$Since the current lags behind the voltage,the circuit must contain $R$ and $L$.

In an $ac$ circuit,the current lags behind the voltage by $\pi /3$. The components in the circuit are

Similar Questions

$A$ series $L, R$ circuit connected with an ac source $E = (25 \sin 1000 t) \ V$ has a power factor of $\frac{1}{\sqrt{2}}$. If the source of emf is changed to $E = (20 \sin 2000 t) \ V$,the new power factor of the circuit will be:

In an $LCR$ series circuit, an alternating voltage source of frequency $F$ is connected. The current leads the voltage by $45^{\circ}$. The value of $L$ is

An $AC$ voltage source of variable angular frequency $\omega$ and fixed amplitude $V_0$ is connected in series with a capacitance $C$ and an electric bulb of resistance $R$ (inductance zero). When $\omega$ is increased,

An ideal inductor of $\left(\frac{1}{\pi}\right) H$ is connected in series with a $300 \Omega$ resistor. If a $20 \ V, 200 \ Hz$ alternating source is connected across the combination,the phase difference between the voltage and current is

In an $L-R$ circuit,the inductive reactance is equal to $\sqrt{3}$ times the resistance $R$ of the circuit. An e.m.f. $E = E_0 \sin(\omega t)$ is applied to the circuit. The power consumed in the circuit is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module