Question

A series LCR circuit is connected to an ac

source having voltage V = Vm sinwt. Derive the

expression for the instantaneous current I and

its phase relationship to the applied voltage.

Obtain the condition for resonance to occur.

Dene ‘power factor’. State the conditions under

which it is (i) maximum and (ii) minimum.

Answer 1

SOLUTION

v = vm sin ωt

Let the current in the circuit be led the applied voltage by an angleΦ.

i=imsin(ωt+ϕ)

The Kirchhoff’s voltage law gives

L(didt+Ri+qC=v).

It is given that v = vm sin ωt (applied voltage)

Ld2qdt2+Rdqdt+qC=vmsinωt ......(1)

On solving the equation, we obtain

q=qmsin(ωt+θ)

dpdt=qmωcos(ωt+θ)

(d2)qdt2 =-qmω2sin(ωt+θ)

On substituting these values in equation (1), we obtain

qmω[Rcos(ωt+θ)+(Xc-XL)sin(ωt+θ)]=vmsinωt

Xc=1ωC Xc=ωL

Z=R2+(Xc-XL)2

qmωZ[RZcos(ωt+θ)+(Xc-XL)Zsin(ωt+θ)]=vmsinωt ...........(2)

Let

cosϕ=R2 and

Xc-XLZ=sinϕ

This gives

tanϕ=Xc-XLR

On substituting this in equation (2), we obtain

qmωZcos(ωt+θ-ϕ)=vmsinωt

On comparing the two sides, we obtain

Vm=qmωZ=i+mZ

im=qmω

and

(θ-ϕ)=-π2

I=dpdt=qmωcos(ωt+θ)

=imcos(ωt+θ)

Or

i=imsin(ωt+θ)

Where,

im=vmZ=vmR2+(Xc-XL)2

And

ϕ=tan-1(Xc-XLR)

The condition for resonance to occur

im=vmR2+(XC-XL)2

For resonance to occur, the value of im has to be the maximum.

The value of im will be the maximum when

Xc=XL

1ωC=ωL

ω2=1LC

ω=1LC

2πf=1LC

f=102πLC

Power factor = cos Φ

Where,

cosϕ=RZ=RR2+(Xc-XL)2

(i) Conditions for maximum power factor (i.e., cos Φ = 1)

XC = XL

Or

R = 0

(ii) Conditions for minimum power factor

When the circuit is purely inductive

When the circuit is purely capacitive.

❤Sweetheart❤