Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • A voltage  applied to a series LCR circuit drives a current in the circuit given <img alt="i = i_{m}\sin \le

A voltage v = v_{m}\sin \omega t applied to a series LCR circuit drives a current in the circuit given i = i_{m}\sin \left ( \omega t+\phi \right ). Deduce the expression for the instantaneous power supplied by the source. Hence, obtain the expression for the average power.

Define the terms 'power factor' and wattless current', giving the examples where power factor is maximum and the circuit where there is wattless current.

 
 
 
 
 

Answers (1)

Given,

i = i_{m}\sin \left ( \omega t+\phi \right )

v = v_{m}\sin \omega t

So instantaneous power (Pi) will be:

Pi= VI

Pi= i_{m}\sin \left ( \omega t+\phi \right )v_{m}\sin \omega t

Pi= V_{m}i_{m}\sin \omega t \left [ \sin \omega t \cos \phi + \sin \phi \cos \omega t \right ]

\because \left [ \sin (\omega t +\phi)= \sin \omega t +\cos \phi+\sin \phi\cos \omega t \right ]

Now, Pi = V_{m}i_{m}\sin ^{2}\omega t \cos \phi + \frac{2V_{m}i_{m}}{2}\sin \omega t \sin \theta \cos \omega t

 \left [\because 2\sin \omega t \cos \omega t = \sin 2 \omega t \right ]

Pi = V_{m}i_{m}\sin ^{2}\omega t \cos \phi + \frac{V_{m}i_{m}}{2}\sin \phi \sin2\omega t

then, for average power (P_{av});

P_{av}= \int_{0}^{T}\frac{Pidt}{T}

P_{av}= \frac{1}{T}\left [ V_{m}i_{m}\int_{0}^{T}\left ( \frac{1-\cos \phi}{2} \right ) \cos \phi dt + \int_{0}^{T}V_{m}i_{m}\frac{\sin 2 \cot +\sin \phi dt}{2}\right ]

P_{av}= \frac{1}{T}\left [ V_{m}i_{m}\cos \phi \left ( \frac{1-\cos \phi}{2} \right ) \int_{0}^{T}dt+\frac{V_{m}i_{m}\sin \phi}{2} \int_{0}^{T}\sin 2 \cot.dt\right ]

P_{av}= \frac{1}{T}\left [ V_{m}i_{m}\cos \phi \left ( \frac{1-\cos \phi}{2} \right ) (T)+\frac{V_{m}i_{m}\sin \phi}{2}\right ]

                                                                                    \left [ \because \omega = 0, \sin 0^{\circ}=0, \cos 0^{\circ}=1 \right ]

then,

P_{av}= \frac{1}{T}\frac{\left [ V_{m}i_{m}\cos \phi \times T \right ]}{2}

P_{av}=\frac{\left [ V_{m}i_{m}\cos \phi \right ]}{2}

P_{av}=\frac{V_{m}}{\sqrt{2}}\times \frac{i_{m}}{\sqrt{2}}\times \cos \phi

Power factor, 

P_{av}=\frac{V_{m}}{\sqrt{2}}\times \frac{i_{m}}{\sqrt{2}}\times \cos \phi

So, here \cos \phi factor in (P_{av}) is power factor which decides power consumed by an AC circuit.

e.g., A resistance  circuit has a maximum power factor

as, \cos \phi= \frac{R}{Z}

where, R = resistance

            Z = impedance

\cos \phi=1

which implies,

R =Z

Wattles's current: A current which has a phase difference of \frac{\pi}{2} with voltage is called wattless current as there is no dissipation of power by this current.

Generally, the current is purely inductive or purely capacitive circuit is wattless.

Posted by

Safeer PP

View full answer

Similar Questions

Latest Question