In LCR series circuit, an alternating e.m.f. ‘ e ’ and current ‘ i ’ is given by equation e = 100 sin ( 100 t ) and i = 100 sin [ 100 t + (π/3) ] mA The average power dissipated in the circuit will be _ _ _ _ _ _ _ _ _ ( MHT - CET 2014 )

a ) 100 W              b ) 10 W                      c ) 5 W                        d ) 2.5 W

Solution :

Given :

e = 100 sin ( 100 t )

i = 100 sin [ 100 t + ( π / 3 ) ] mA

   = 100 sin [ 100 t + ( π / 3 ) ]  X 10 ^-3 A

Comparing these equation with

e = e_o sin ( w t )

 and 

i = i_o sin ( w t + φ )

we get,

 e_o = 100,         i_o = 100 X 10 ^-3,           w = 100           and      φ = π / 3

The average power dissipated in the circuit can be given as

P _avg = e _rms  X I _rms cos φ

We know that

e _rms  =  e_o / √( 2 )

i _rms  =  i_o / √( 2 )

Putting these value in above equation,

P _avg = [ e_o / √( 2 ) ] X [ i_o / √ ( 2 ) ] cos φ

P _avg = { [ e_o  X i_o ]/ ( 2 ) } cos φ

P _avg = { [ 100  X 100 X 10 ^-3  ] / ( 2 ) } cos ( π / 3 )

P _avg = 2.5 W

The average power dissipated in the circuit will be 2.5 W ( d )