A series LCR circuit with resistance R=500 O | vedclass

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

Question

(D) When the capacitance is removed,the circuit becomes an $LR$ circuit. The phase difference is given by $	an \phi = \frac{X_L}{R}$.Given $\phi = 60

Vedclass · Accepted Answer

$500 \ \Omega$

Vedclass · Answer

(D) When the capacitance is removed,the circuit becomes an $LR$ circuit. The phase difference is given by $	an \phi = \frac{X_L}{R}$.Given $\phi = 60^{\circ}$,we have $	an 60^{\circ} = \frac{X_L}{R} = \sqrt{3}$,so $X_L = R\sqrt{3}$.When the inductance is removed,the circuit becomes an $RC$ circuit. The phase difference is given by $	an \phi = \frac{X_C}{R}$.Given $\phi = 60^{\circ}$,we have $	an 60^{\circ} = \frac{X_C}{R} = \sqrt{3}$,so $X_C = R\sqrt{3}$.Since $X_L = X_C$,the circuit is in resonance.At resonance,the impedance $Z = R$.Given $R = 500 \ \Omega$,the impedance $Z = 500 \ \Omega$.

Similar Questions

In an $L-C-R$ series $AC$ circuit,$L = 9 \ H$,$R = 10 \ \Omega$,and $C = 100 \ \mu F$. The $Q$-factor of the circuit is . . . . . . .

Which of the following combinations should be selected for better tuning of an $L-C-R$ circuit used for communication?

In an $LCR$ circuit,the inductance is changed from $L$ to $9 L$. For the same resonant frequency,the capacitance should be changed from $C$ to:

The frequency of $AC$ at which a $16 \mu F$ capacitor and a $\frac{10}{\pi^2} \ mH$ inductor will have the same reactance is: (in $kHz$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module