An alternating voltage of frequency omega is | vedclass

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

Question

(D) In a parallel $LC$ circuit,the total impedance $Z$ is given by $\frac{1}{Z} = \sqrt{(\frac{1}{X_L} - \frac{1}{X_C})^2}$.At resonance,$\omega^2 = \

Vedclass · Accepted Answer

voltage is maximum when $\omega^2 = \frac{1}{LC}$

Vedclass · Answer

(D) In a parallel $LC$ circuit,the total impedance $Z$ is given by $\frac{1}{Z} = \sqrt{(\frac{1}{X_L} - \frac{1}{X_C})^2}$.At resonance,$\omega^2 = \frac{1}{LC}$,which implies $X_L = X_C$.At this frequency,the net current drawn from the source is minimum because the impedance $Z$ becomes infinite.Since the circuit is connected in parallel,the voltage across the inductance $L$ and capacitance $C$ is the same as the source voltage.However,in a parallel $LC$ circuit at resonance,the circulating current between $L$ and $C$ is maximum,leading to maximum voltage across the components.

An alternating voltage of frequency $\omega$ is induced in an electric circuit consisting of an inductance $L$ and capacitance $C$,connected in parallel. Then across the inductance coil:

Similar Questions

In the given $LCR$ $AC$ circuit,the effective current flowing through the circuit will be:

The figure shows the variation of $R$,$X_L$,and $X_C$ with frequency $f$ in a series $L-C-R$ circuit. For what frequency point is the circuit capacitive?

Which one of the following curves represents the variation of impedance $(Z)$ with frequency $(f)$ in a series $LCR$ circuit?

Find the impedance of the circuit in $\Omega$.

When a capacitor is connected in series with an $LR$ circuit, the alternating current flowing in the circuit:

Vedclass Test Series

Exam Paper Generator

Online Exam Module