Question

Average power dissipation in a pure inductive ac circuit is?

Answer 1

Answer:

The average power dissipation in an AC circuit can be different from the DC circuit because of its alternating behavior. The alternating current uses sinusoidal voltage and current, which change its direction and magnitude every moment. How could one say that the power in an AC circuit has a fixed value?

Suppose a general load connected with an AC source, as presented in the diagram. Suppose it has the voltage

v=

V

m

sinωt…(1a)

v=Vmsin⁡ωt…(1a)

Which cause it to flow the following sinusoidal current

i=

I

m

sinωt+θ…(1b)

i=Imsin⁡ωt+θ…(1b)

The power flow of the source will be

p=vi=(

V

m

sinωt)(

I

m

sinωt+θ)

p=vi=(Vmsin⁡ωt)(Imsin⁡ωt+θ)

p=

V

m

I

m

sinωt.sinωt+θ…(1c)

p=VmImsin⁡ωt.sin⁡ωt+θ…(1c)

Using the following trigonometric identity

sinA.sinB=

1

2

cos(A−B)−

1

2

cos(A+B)

sin⁡A.sin⁡B=12cos⁡(A−B)−12cos⁡(A+B)

so

p=

V

m

I

m

2

[cos(ωt−ωt−θ)−cos(ωt+ωt+θ)]

p=VmIm2[cos⁡(ωt−ωt−θ)−cos⁡(ωt+ωt+θ)]

As we know from the following trigonometric identity that

cosθ=cos−θ

cos⁡θ=cos−θ

Now

p=

V

m

I

m

2

cosθ−

V

m

I

m

2

cos(2ωt+θ)

p=VmIm2cos⁡θ−VmIm2cos⁡(2ωt+θ)

Now observe the above equation, the power is made up of two cosine terms. The second term frequency is twice (

2ωt)

2ωt)

as that of current or voltage (both have same frequency of  

ωt

ωt

). This term has a component of time in it, which cause the cosine wave to oscillate and its average value over a cycle is zero. That’s why no energy is transfer due to the second term to the generic load attached to the ac source. Energy only float back and forth between load and source.

Where the first term has no element of time, so it gives a fixed value for a specific value of theta. This is called average power dissipated in the load. This average power is mostly called real or active power. This power depends upon the angle  

θ

θ

, which is the phase difference between current and voltage. So the equation for the real power is

P=

V

m

I

m

2

cosθ=

V

m

2

√

I

m

2

√

cosθ

P=VmIm2cos⁡θ=Vm2Im2cos⁡θ

Where the relationship between RMS and peak value is  

V

RMs

=

V

m

2

√

VRMs=Vm2

, putting it into the above equation will give us

P=VIcosθ…(1d)

P=VIcos⁡θ…(1d)

In Pure resistive AC circuit, the circuit current and voltage are in-phase, or the phase difference angle is zero i.e.  

θ=

0

∘

θ=0∘

. So, the average power delivered to the pure resistive load will

P=VIcos

0

∘

P=VIcos⁡0∘

P=VI(1)cos

0

∘

=1

P=VI(1)cos⁡0∘=1

P=VI…(2a)

P=VI…(2a)

The equation (2a) means that the power delivered to the load is maximum in a pure resistive circuit.

Where in pure inductive AC load, the load current lags by  

90

∘

90∘

, i.e.  

θ

∘

=−

90

∘

θ∘=−90∘

. So, the power delivered to the inductive load will be

P=VIcos(−

90

∘

)

P=VIcos⁡(−90∘)

P=VI(0)cos−

90

∘

=0

P=VI(0)cos−90∘=0

P=0…(2b)

P=0…(2b)

The equation (2b) means that no power is dissipated in a pure inductive load, the power only float back and forth between inductive load and source.

In pure capacitive AC circuit, the current leads source voltage by  

90

∘

90∘

i.e.  

θ=

90

∘

θ=90∘

. In this case, the power dissipated in the load is

P=VIcos

90

∘

P=VIcos⁡90∘

P=VI(0)cos90=0

P=VI(0)cos⁡90=0

P=0…(2c)

P=0…(2c)

According to the equation (2c), the power dissipated in a pure capacitive circuit is zero. Power only float back and forth between source and the capacitive load

Explanation: