An R-L-C circuit consists of a 150 Omega resi | vedclass

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

Question

(D) In an $R-L-C$ series circuit,the given values are:$R = 150 \Omega$$C = 20 \mu F = 20 	imes 10^{-6} F = 2 	imes 10^{-5} F$$L = 500 mH = 0.5 H$$\o

Vedclass · Accepted Answer

$	an^{-1}(0.5)$

Vedclass · Answer

(D) In an $R-L-C$ series circuit,the given values are:$R = 150 \Omega$$C = 20 \mu F = 20 	imes 10^{-6} F = 2 	imes 10^{-5} F$$L = 500 mH = 0.5 H$$\omega = 400 rad s^{-1}$First,calculate the inductive reactance $X_L$:$X_L = \omega L = 400 	imes 0.5 = 200 \Omega$Next,calculate the capacitive reactance $X_C$:$X_C = \frac{1}{\omega C} = \frac{1}{400 	imes 2 	imes 10^{-5}} = \frac{1}{800 	imes 10^{-5}} = \frac{10^5}{800} = \frac{1000}{8} = 125 \Omega$The phase angle $\phi$ is given by the formula:$	an \phi = \frac{X_L - X_C}{R}$Substituting the values:$	an \phi = \frac{200 - 125}{150} = \frac{75}{150} = 0.5$Therefore,the phase angle is $\phi = 	an^{-1}(0.5)$.

An $R-L-C$ circuit consists of a $150 \Omega$ resistor,$20 \mu F$ capacitor,and a $500 mH$ inductor connected in series with a $100 V$ $AC$ supply. The angular frequency of the supply voltage is $400 rad s^{-1}$. The phase angle between the current and the applied voltage is

Similar Questions

In a series $LCR$ circuit,$R$ represents the resistance of an electric bulb. If the frequency of the $A.C.$ supply is doubled,what should be the values of inductance $L$ and capacitance $C$ to maintain the same current?

With an increase in the frequency of an $a.c.$ supply,the impedance of an $LCR$ series circuit:

An inductor of reactance $100 \ \Omega$, a capacitor of reactance $50 \ \Omega$, and a resistor of resistance $50 \ \Omega$ are connected in series with an $AC$ source of $10 \ V, 50 \ Hz$. Average power dissipated by the circuit is . . . . . . $W$.

An inductance of $\frac{300}{\pi} \text{ mH}$,a capacitance of $\frac{1}{\pi} \text{ mF}$ and a resistance of $20 \ \Omega$ are connected in series with an a.c. source of $240 \text{ V}, 50 \text{ Hz}$. The phase angle of the circuit is

In an $R-L-C$ series circuit, the potential difference across each element is $20 \, V$. If the value of the resistance $R$ is doubled, what will be the potential difference across $R, L$, and $C$ respectively?

Vedclass Test Series

Exam Paper Generator

Online Exam Module