Home
Class 12
PHYSICS
Obtain the relation for the current I=I(...

Obtain the relation for the current `I=I_(o)sin(omegat-pi//2)` for a pure inductor across which an alternating `emfe=e_(o)sinomegat` is applied.

Text Solution

AI Generated Solution

To derive the relationship for the current \( I = I_0 \sin(\omega t - \frac{\pi}{2}) \) for a pure inductor when an alternating EMF \( e = E_0 \sin(\omega t) \) is applied, we can follow these steps: ### Step 1: Understand the Inductor's Behavior A pure inductor has a property where the induced EMF (back EMF) is proportional to the rate of change of current through it. This is given by the formula: \[ E = -L \frac{di}{dt} \] where \( L \) is the inductance and \( \frac{di}{dt} \) is the rate of change of current. ...
Promotional Banner

Similar Questions

Explore conceptually related problems

Obtain the relation I=I_(0) sin (omegat+ pi//2) and X_L= 1//omegaC for capacitor across which an a.c. emf of e= e_(0) sin ot is applied as shown in the figure. Draw a phasor diagram showing emf e_t current I and their phase difference phi .

If a current I given by l_(o) sin(omegat-pi//2) flows in an AC circuit across which an AC potential of E_(0) . sin (omegat) has been applied, then the power consumption P in the circuit will be

r.m.s. value of current i=3+4sin (omegat+pi//3) is

Explain the term 'inductive reactance'.Show graphically the variation of inductive reactance with frequency of the applied alternating voltage. An a.c. voltage E=E_(0)sin omegat is applied across a pure inductor L .Show mathematically that the current flowing through it lags behind the applied voltage by a phase angle of pi//2 .

Explain the term 'capacitive reactance'.Show graphically the variation of capacitive reactance with frequency of the applied alternating voltage. An a.c. voltage E=E_(0) sin omegat is applied a across a pure capacitor C .Show mathematically that the current flowing through it leads the applied voltage by a phase angle of pi//2

If the alternating current I =I_(1)cosomegat+I_(2) sinomegat , then the rms current is given by

Find the field at the origin O due to the current i=2A .

An alternating potential V = underset(o)(V) sin omega t is applied across a circuit. As a result the current, I = underset(o)(I)sin(omegat - pi/2) flows in it. The power consecutive in the circuit per cycle is

The average value of alternating current I=I_(0) sin omegat in time interval [0, pi/omega] is

The average power dissipated in a pure inductor L carrying an alternating current of rms value I is .