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

An emf $E = 6 \cos(6000t) \ V$ is applied to an $L-R$ circuit of inductance $L = 4 \ mH$ and resistance $R = 7 \ \Omega$. The amplitude of the current in the circuit is . . . . . . (in $A$)

$A$ $20 V$ $AC$ is applied to a circuit consisting of a resistor and a coil with negligible resistance. If the voltage across the resistor is $12 V$,the voltage across the coil is (in $V$)

The power factor of an $R-L$ circuit is $\frac{1}{\sqrt{2}}$. If the frequency of $AC$ is doubled,the power factor will now be

When $80 \ V$ $d.c.$ is applied across a solenoid,a current of $0.8 \ A$ flows in it. When $80 \ V$ $a.c.$ is applied across the same solenoid,the current becomes $0.4 \ A$. If the frequency of $a.c.$ source is $50 \ Hz$,the impedance and inductance of the solenoid are nearly:

$A$ coil has a resistance of $30 \Omega$ and an inductive reactance of $20 \Omega$ at $50 \text{ Hz}$ frequency. If an $AC$ source of $200 \text{ V}$,$100 \text{ Hz}$ is connected across the coil,the current in the coil is

Vedclass Test Series

Exam Paper Generator

Online Exam Module