In an LCR series circuit,an inductor 30 , {mH | vedclass

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

Question

(A) The phase angle $\phi$ in an $LCR$ circuit is given by $	an \phi = \frac{X_C - X_L}{R}$.Since the current leads the voltage,the circuit is capaci

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(A) The phase angle $\phi$ in an $LCR$ circuit is given by $	an \phi = \frac{X_C - X_L}{R}$.Since the current leads the voltage,the circuit is capacitive,and the phase angle is $\phi = -45^{\circ}$.Thus,$	an(-45^{\circ}) = \frac{X_C - X_L}{R} \implies -1 = \frac{X_C - X_L}{R}$.This gives $X_L - X_C = R$.Given $L = 30 \, {mH} = 0.03 \, {H}$,$R = 1 \, \Omega$,and $\omega = 300 \, {rad/s}$.$X_L = \omega L = 300 	imes 0.03 = 9 \, \Omega$.Substituting the values: $9 - X_C = 1 \implies X_C = 8 \, \Omega$.Since $X_C = \frac{1}{\omega C}$,we have $\frac{1}{300 	imes C} = 8$.$C = \frac{1}{300 	imes 8} = \frac{1}{2400} = \frac{1}{2.4} 	imes 10^{-3} \, {F}$.Wait,re-evaluating the phase condition: If current leads voltage,$\phi$ is negative,so $	an \phi = \frac{X_C - X_L}{R} = 	an(-45^{\circ}) = -1$.$X_L - X_C = R \implies 9 - X_C = 1 \implies X_C = 8$. This leads to $C = \frac{1}{2400} \approx 0.416 	imes 10^{-3} \, {F}$.If the question implies the magnitude of the phase difference is $45^{\circ}$ and current leads,then $X_C > X_L$ is incorrect; it should be $X_L - X_C = R$ if current lags. For current to lead,$X_C - X_L = R$ is not possible if $X_C $X_C - 9 = 1 \implies X_C = 10 \, \Omega$.$\frac{1}{300 	imes C} = 10 \implies C = \frac{1}{3000} = \frac{1}{3} 	imes 10^{-3} \, {F}$.Thus,$x = 3$.

Similar Questions

$A$ resistor of $20 \ \Omega$ and a capacitor are connected in series with an $AC$ current source of $50 \ Hz$. What should be the capacitance to produce a phase difference of $30^\circ$ between the voltage and current?

$A$ coil of resistance $450 \Omega$ and self-inductance $1.5 \text{ H}$ is connected to an $A$.$C$. source of frequency $\frac{150}{\pi} \text{ Hz}$. The phase difference between voltage and current is

$A$ lamp is connected in series with a capacitor and an $AC$ source. What happens if the capacitance of the capacitor is reduced?

In an $LCR$ series circuit,the source voltage is $120 \ V$,the voltage across the inductor is $50 \ V$,and the voltage across the resistor is $40 \ V$. Determine the voltage across the capacitor.

$A$ charged capacitor discharges through a resistance $R$ with time constant $\tau$. The two are now placed in series across an $AC$ source of angular frequency $\omega = \frac{1}{\tau}$. The impedance of the circuit will be-

Vedclass Test Series

Exam Paper Generator

Online Exam Module