(C) The phase difference $\phi$ in an $RL$ series circuit is given by the formula: $\tan \phi = \frac{X_L}{R} = \frac{\omega L}{R}$.Given:Resistance $

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

$\tan^{-1} \left( \frac{\pi}{4} \right)$

(C) The phase difference $\phi$ in an $RL$ series circuit is given by the formula: $\tan \phi = \frac{X_L}{R} = \frac{\omega L}{R}$.Given:Resistance $R = 1 \Omega$Inductance $L = 2.5 \text{ mH} = 2.5 \times 10^{-3} \text{ H}$Frequency $\nu = 50 \text{ Hz}$Angular frequency $\omega = 2 \pi \nu = 2 \pi \times 50 = 100 \pi \text{ rad/s}$.Substituting the values:$\tan \phi = \frac{100 \pi \times 2.5 \times 10^{-3}}{1}$$\tan \phi = 100 \times 2.5 \times 10^{-3} \times \pi$$\tan \phi = 0.25 \pi = \frac{\pi}{4}$.Therefore,$\phi = \tan^{-1} \left( \frac{\pi}{4} \right)$.

Class 12 Physics Alternating Current RL, RC and LC AC Circuits

Easy

$A$ circuit is made of $1 \Omega$ resistance and $2.5 \text{ mH}$ inductance in series. If an alternating source of $200 \text{ V}$ and $50 \text{ Hz}$ is connected to the circuit,then the phase difference between current and voltage is . . . . . . .

GUJCET 2012 All GUJCET

A
$\tan^{-1} \pi$
B
$\tan^{-1} \left( \frac{\pi}{2} \right)$
C
$\tan^{-1} \left( \frac{\pi}{4} \right)$
D
$\tan^{-1} \left( \frac{\pi}{3} \right)$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Physics Questions

Chapter

Alternating Current

Subtopic

RL, RC and LC AC Circuits

Previous Year

GUJCET 2012 Paper All GUJCET years →

An inductor of inductance $L$ and resistor $R$ are joined together in series and connected to a source of frequency $\omega$. The power dissipated in the circuit is

EasyKCET 2019

View Solution

An inductor and a resistor are connected in series to an $AC$ source of voltage $V = 144 \sin \left(100 \pi t + \frac{\pi}{2}\right) \text{ V}$. If the current in the circuit is $I = 6 \sin \left(100 \pi t + \frac{\pi}{6}\right) \text{ A}$,then the resistance of the resistor is: (in $\Omega$)

DifficultAP EAMCET 2024

View Solution

In a series $L-C-R$ circuit,the current in the circuit is $11 \, A$ when the applied voltage is $220 \, V$. The voltage across the capacitor is $200 \, V$. If the value of the resistor is $20 \, \Omega$,then the voltage across the unknown inductor is.......$V$

Medium

View Solution

$A$ series combination of resistor $R$ and capacitor $C$ is connected to an $A$.$C$. source of angular frequency $\omega$. Keeping the voltage same,if the frequency is changed to $\frac{\omega}{3}$,the current becomes half of the original current. Then the ratio of capacitive reactance and resistance at the former frequency is

DifficultKCET 2013

View Solution

$A$ 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

EasyTS EAMCET 2023

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

$A$ circuit is made of $1 \Omega$ resistance and $2.5 \text{ mH}$ inductance in series. If an alternating source of $200 \text{ V}$ and $50 \text{ Hz}$ is connected to the circuit,then the phase difference between current and voltage is . . . . . . .

Similar Questions

An inductor of inductance $L$ and resistor $R$ are joined together in series and connected to a source of frequency $\omega$. The power dissipated in the circuit is

An inductor and a resistor are connected in series to an $AC$ source of voltage $V = 144 \sin \left(100 \pi t + \frac{\pi}{2}\right) \text{ V}$. If the current in the circuit is $I = 6 \sin \left(100 \pi t + \frac{\pi}{6}\right) \text{ A}$,then the resistance of the resistor is: (in $\Omega$)

In a series $L-C-R$ circuit,the current in the circuit is $11 \, A$ when the applied voltage is $220 \, V$. The voltage across the capacitor is $200 \, V$. If the value of the resistor is $20 \, \Omega$,then the voltage across the unknown inductor is.......$V$

$A$ 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