Solve this: 58 A series L-C-R circuit is connected to an AC source having voltage - Physics - Alternating Current

Question

Solve this:
  58   A   s e r i e s   L - C - R   c i r
c u i t   i s   c o n n e c t e d   t o   a n
  A C   s o u r c e   h a v i n g   v o l t a g
e               V = V m  
sin   ω t   D e r i v e   t h e   e x p r
e s s i o n   f o r   t h e   i n s tan tan e o u s
  c u r r e n t   I   a n d      
            i t s   p h a s e
  r e l a t i o n s h i p   t o   t h e   a p p
l i e d   v o l t a g e   O b t a i n   t h e  
c o n d i t i o n   f o r          
    r e s o n a n c e   t o   o c c u r  
D e f i n e   p o w e r   f a c t o r   S t a t e
  t h e   c o n d i t i o n s   u n d e r  
            w h i c h   i t
  i s         ( i )   m a x i m u m
  a n d         ( i i )   m i n i m u
m

Vijay Shankar Yadav · Accepted Answer

Dear Student,
â€‹The actual derivation involves a second order
differential equation involving complex variables Here I am giving
you a brief result of that Z is the impedance of the ckt
Im=VmZ=VmR2+ωL-1ωC2
And the phase angle,
φ=tan-1XL-XCR=tan-1ωL-1ωCR
Current is given by,
I(t)=Imsin(ωt-φ)
Instantaneous current,
I(0)=-Imsin(φ)
For resinance condition Z should be minimum hence it will be when 1wC=Lww=1LC
And Power factor
cosφ= RZ; where Z=R2+(XL-XC)2where R is the resitance, XL=effective resistance offered by indutorand XC=effective resistance offered by capacitor
 At resonance XL = XCSo cosφ= RZ=RR=1Hence power factor at resonance of LCR circuit is 1

Maximum power factor ( means maximum power consuming ) when it is
one
in above case
Power factor has R and Z , and R and Z can not give any negative
value
hence minimum value of power factor is 0 when circuit is purely
capacitive
Regards