With a gradual increase in the frequency of a | vedclass

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

Question

(D) The impedance $Z$ of an $LCR$ series circuit is given by the formula: $Z = \sqrt{R^2 + (X_L - X_C)^2}$.We know that inductive reactance $X_L = L\o

Vedclass · Accepted Answer

first decreases,becomes minimum,and then increases.

Vedclass · Answer

(D) The impedance $Z$ of an $LCR$ series circuit is given by the formula: $Z = \sqrt{R^2 + (X_L - X_C)^2}$.We know that inductive reactance $X_L = L\omega = 2\pi fL$ and capacitive reactance $X_C = \frac{1}{C\omega} = \frac{1}{2\pi fC}$.As the frequency $f$ increases,$X_L$ increases linearly while $X_C$ decreases.At low frequencies,$X_C$ dominates,so $Z$ decreases as $f$ increases.At the resonant frequency $f_0 = \frac{1}{2\pi\sqrt{LC}}$,$X_L = X_C$,making the impedance $Z = R$,which is the minimum value.As the frequency increases beyond $f_0$,$X_L$ dominates,causing the impedance $Z$ to increase.Therefore,the impedance first decreases,becomes minimum at resonance,and then increases.

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

Similar Questions

For the series $LCR$ circuit shown in the figure,what is the resonance frequency and the amplitude of the current at the resonating frequency?

The resonant frequency of an $L-C$ circuit is

In an $LCR$ series resonance circuit, choose the wrong statement.

Obtain the resonant frequency $\omega_{r}$ of a series $LCR$ circuit with $L=2.0 \;H, C=32\; \mu F$ and $R=10\; \Omega$. What is the $Q$-value of this circuit?

For an $R-L-C$ circuit,driven with voltage of amplitude $V_m$ and frequency $\omega_0 = \frac{1}{\sqrt{LC}}$,the current exhibits resonance. The quality factor $Q$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module