In the circuit shown,C = frac{sqrt{3}}{2} tim | vedclass

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

Question

(C) Given: $\omega = 100 \, rad/s$,$L = \frac{\sqrt{3}}{10} \, H$,$R_1 = 10 \, \Omega$,$C = \frac{\sqrt{3}}{2} 	imes 10^{-3} \, F$,$R_2 = 20 \, \Omeg

Vedclass · Accepted Answer

$90^\circ$

Vedclass · Answer

(C) Given: $\omega = 100 \, rad/s$,$L = \frac{\sqrt{3}}{10} \, H$,$R_1 = 10 \, \Omega$,$C = \frac{\sqrt{3}}{2} 	imes 10^{-3} \, F$,$R_2 = 20 \, \Omega$.For the $L-R_1$ branch:Inductive reactance $X_L = \omega L = 100 	imes \frac{\sqrt{3}}{10} = 10\sqrt{3} \, \Omega$.The phase angle $\phi_1$ of current $I_1$ with respect to voltage $V$ is given by $	an \phi_1 = -\frac{X_L}{R_1} = -\frac{10\sqrt{3}}{10} = -\sqrt{3}$.Thus,$\phi_1 = -60^\circ$.For the $C-R_2$ branch:Capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{100 	imes \frac{\sqrt{3}}{2} 	imes 10^{-3}} = \frac{1}{\frac{\sqrt{3}}{20}} = \frac{20}{\sqrt{3}} \, \Omega$.The phase angle $\phi_2$ of current $I_2$ with respect to voltage $V$ is given by $	an \phi_2 = \frac{X_C}{R_2} = \frac{20/\sqrt{3}}{20} = \frac{1}{\sqrt{3}}$.Thus,$\phi_2 = 30^\circ$.The phase difference between $I_1$ and $I_2$ is $|\phi_2 - \phi_1| = |30^\circ - (-60^\circ)| = 90^\circ$.

Similar Questions

In the circuit shown in the figure,if both the bulbs $B_1$ and $B_2$ are identical,then:

An inductance of $\left(\frac{200}{\pi}\right) \text{mH}$,a capacitance of $\left(\frac{10^{-3}}{\pi}\right) \text{F}$,and a resistance of $10 \, \Omega$ are connected in series with an $AC$ source of $220 \, \text{V}, 50 \, \text{Hz}$. The phase angle of the circuit is:

In an $LCR$ series circuit, the potential difference across the terminals of the inductor, capacitor and resistor is $60 \text{ V}$, $30 \text{ V}$ and $40 \text{ V}$ respectively. Then the supply voltage will be equal to (in $\text{ V}$)

In the adjoining figure,the impedance of the circuit will be $... \Omega$.

The $L-C-R$ circuit shown below is driven by an ideal $AC$ voltage source. At frequency $f=\frac{1}{2 \pi \sqrt{L C}}$,choose the correct statement.

Vedclass Test Series

Exam Paper Generator

Online Exam Module