Question

A voltage V = V0 sin ωt is applied to a series LCR circuit. Derive the expression for the average power dissipated over a cycle. Under what condition (i) no power is dissipated even though the current flows through the circuit, (ii) maximum power is dissipated in the circuit?

Answer 1

Voltage V=V0sinωt is applied to an series LCR circuit. Current is I=I0sin(ωt+ϕ) I0=V0Zϕ=tan-1XC-XLR Instantaneous power supplied by the source is P=VI =(V0sinωt)×(I0sin(ωt+ϕ) =V0I02cosϕ-cos(2ωt+ϕ) The average power over a cycle is average of the two terms on the R.H.S of the above equation. The second term is time dependent; so, its average is zero. P=V0I02cosϕ =V0I022cosϕ =VIcosϕP=I2Zcosϕ is called the power factor. Case I For pure inductive circuit or pure capacitive circuit, the phase difference between current and voltage is . Therefore, no power is dissipated. This current is sometimes referred to as wattless current. Case II For power dissipated at resonance in an LCR circuit, So, maximum power is dissipated.

Answer 2

Explanation:

Power factor of the circuit is given by:

cosϕ=

Z

R

=

R

2

+(ωL−

ωC

1

)

2

R

Current flowing in the circuit is given by:

I=

Z

V

I=

Z

V

o

sin(ωt)

Instantaneous real power dissipated in the circuit is:

P=I

2

R

P=

Z

2

V

o

2

sin

2

(ωt)

R

Average power dissipated in a cycle is given by:

<P>=

∫

0

2π/ω

dt

∫

0

2π/ω

Pdt

=

2Z

2

×2π/ω

V

o

2

R

∫

0

2π/ω

(1−cos(2ωt))dt

<P>=V

rms

I

rms

cos(ϕ)

(i)

No power is dissipated when P = 0

This implies cosϕ=0

ϕ=π/2

That is the circuit is purely inductive or capacitive.

(ii)

Maximum power is dissipated when P is maximum.

This implies cosϕ=1

ϕ=0

Circuit is purely resistive.