In an LCR series circuit,if R = X_L = 2X_C,fi | vedclass

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

Question

(B) Given: $R = X_L$ and $R = 2X_C$,which implies $X_L = R$ and $X_C = \frac{R}{2}$.The impedance $Z$ of an $LCR$ series circuit is given by $Z = \sqr

Vedclass · Accepted Answer

$\frac{\sqrt{5}R}{2}, 	an^{-1}\left(\frac{1}{2}ight)$

Vedclass · Answer

(B) Given: $R = X_L$ and $R = 2X_C$,which implies $X_L = R$ and $X_C = \frac{R}{2}$.The impedance $Z$ of an $LCR$ series circuit is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.Substituting the values: $Z = \sqrt{R^2 + (R - \frac{R}{2})^2} = \sqrt{R^2 + (\frac{R}{2})^2} = \sqrt{R^2 + \frac{R^2}{4}} = \sqrt{\frac{5R^2}{4}} = \frac{\sqrt{5}}{2}R$.The phase difference $\phi$ is given by $	an \phi = \frac{X_L - X_C}{R}$.Substituting the values: $	an \phi = \frac{R - \frac{R}{2}}{R} = \frac{R/2}{R} = \frac{1}{2}$.Therefore,$\phi = 	an^{-1}(\frac{1}{2})$.Thus,the impedance is $\frac{\sqrt{5}}{2}R$ and the phase difference is $	an^{-1}(\frac{1}{2})$.

In an $LCR$ series circuit,if $R = X_L = 2X_C$,find the impedance and the phase difference between the current and the voltage.

Similar Questions

$A$ series $LR$ circuit is connected to an ac source of frequency $\omega$ and the inductive reactance is equal to $2R$. $A$ capacitance of capacitive reactance equal to $R$ is added in series with $L$ and $R$. The ratio of the new power factor to the old one is

When $L$, $C$, and $R$ are connected in series with an $AC$ source of frequency $f$, the current leads the voltage by a phase angle of $45^{\circ}$. What is the value of $C$?

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

In an $LCR$ series circuit $R=10 \,\Omega, X_L=8 \,\Omega$ and $X_C=6 \,\Omega$. The total impedance of the circuit is ........ $\Omega$.

$A$ resistor of resistance $30 \Omega$, an inductor of reactance $10 \Omega$, and a capacitor of reactance $10 \Omega$ are connected in series to an $AC$ voltage source $V = 300 \sqrt{2} \sin(\omega t)$. The current in the circuit is . . . . . . (in $\text{ A}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module