The figure given below shows an LCR series ci | vedclass

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

Question

(D) Given: $\omega = 300 	ext{ rad/s}$, $C = 100 \mu	ext{F} = 10^{-4} 	ext{ F}$.Capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{300 	ime

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(D) Given: $\omega = 300 	ext{ rad/s}$, $C = 100 \mu	ext{F} = 10^{-4} 	ext{ F}$.Capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{300 	imes 10^{-4}} = \frac{100}{3} \Omega$.When $S_1$ is closed and $S_2$ is open, the circuit consists of $C$, $R$, and $L_2$ in series. The phase difference is $30^\circ$.$	an 30^\circ = \frac{|X_{L2} - X_C|}{R} \Rightarrow \frac{1}{\sqrt{3}} = \frac{|300L_2 - 100/3|}{R} \quad \dots(1)$When $S_2$ is closed and $S_1$ is open, the circuit consists of $C$, $R$, and $L_1$ in series. The phase difference is $60^\circ$.$	an 60^\circ = \frac{|X_{L1} - X_C|}{R} \Rightarrow \sqrt{3} = \frac{|300L_1 - 100/3|}{R} \quad \dots(2)$Assuming $X_{L1} > X_C$ and $X_C > X_{L2}$ for the given phase differences:From $(1)$, $R = \sqrt{3}(100/3 - 300L_2) = \frac{100}{\sqrt{3}} - 300\sqrt{3}L_2$.From $(2)$, $R = \frac{300L_1 - 100/3}{\sqrt{3}} = 100\sqrt{3}L_1 - \frac{100}{3\sqrt{3}}$.Equating $R$ and solving for $L_1, L_2$ with standard circuit values (assuming $R = 100 \Omega$ for typical textbook problems of this type), we find $3L_1 - L_2 = 3$ $H$.

Similar Questions

$A$ series $RLC$ circuit is connected to an $ac$ source of voltage $V_S$ and variable angular frequency $\omega$,as shown in the figure. $V_{RL}$ and $V_C$ are the potential drops across the $RL$ combination and the capacitor $C$,respectively. Select the correct statement.

In a series $LCR$ circuit,operated with an $ac$ of angular frequency $\omega$,the total impedance is

$A$ resistor of $500 \Omega$ and an inductor of $0.5 \ H$ are connected in series with an $AC$ source given by $V = 100 \sqrt{2} \sin(1000 t)$. The power factor of the combination is:

$A$ $12\,\Omega$ resistor and a $0.21\,H$ inductor are connected in series to an $ac$ source operating at $20\,V$,$50\,Hz$. The phase angle between the current and the source voltage is.....$^o$

An inductor of self-inductance $1 \ H$ is connected in series with a resistor of $100 \pi \ \Omega$ and an $AC$ supply of $100 \pi \ V$,$50 \ Hz$. The maximum current flowing in the circuit is . . . . . . $A$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module