**Electric Power**

In general
power is defined as the change in energy (expanding and absorbing) i.e.
transfer of energy with respect to time. The SI unit of power is watt that is
also expressed in joules per second.

p = dw/dt J/s
(Joule per Second ) ……………………..(1)

In equation
(1)

- p represents the power

- dw represents the change in energy

- dt represents the change in time

Similarly,
Electric Power is defined as the transfer of electrical energy with respect to
time. The SI unit of electric power is watt.

**Electric Power Formula **

Multiply
dq/dq in equation (1)

p(t) = (dw/dq).(dq/dt)

p(t) = v(t).i(t) Watts

P= V.I Watts
……………… (2)

Equation
(2) represents the formula for electric power in consumed by the load in circuit
that is connected with a source having voltage V and drawn a current I.

Equation (2)
can be combined with ohm’s law ( V = I.R) to derive the power dissipated in
case of **resistive load** or in case of power dissipated in DC circuit.

Put the
value of V in equation (2)

P =
(I.R).I = I^{2}.R = V^{2}/R
…………(3)

Equation
(3) represents the power dissipated in resistive load in case of power
dissipated in DC circuit.

So, the electric
power is also defined as product of voltage applied to the load terminal and
current drawn by it.

In the
above figure there is a load element X potential difference across this element
is V and current passing through this element is I. So, power across this
element is

P
= V.I Watts …………………..(2)

The polarity of applied voltage and direction of current in the circuit are important for determining whether the power is delivered or absorbed by the circuit or element.

- If P is positive in equation (3) the element absorb power or we can say power is dissipated in the circuit element.

- If P is negative in equation (3) then the circuit element delivers the power, or we can say it is generated the power in circuit.

**How can we
assure that the Power is absorbed or delivered?**

If the direction
of current in such a way that it enters the positive terminal of voltage across
through the element then the sign of power is positive and the power is
absorbed by the element.

If the
current enters negative terminal of voltage across through the element, then
the sign of p is negative and the is delivered by the element.

Let us
understand by the circuit, In the given figure direction of current in such a
way that it enters the positive terminal of Z and negative terminal of Y. So,
from both of above statement we can say the element Z is load and it absorb
power and element Y is source and it delivers power.

**DC Circuit Power Analysis**

In Dc
circuit energy storage elements **inductor** and capacitor does not store energy,
they behave like short circuit and open circuit simultaneously. In Dc circuit
only resistive load are taken into account. So, in dc circuit equation (2)
& (3) are valid.

P
= V.I Watts …………………..(2)

P =
(I.R).I = I^{2}.R = V^{2}/R
…………(3)

**AC Circuit Power Analysis**

Unlike DC, AC current changes its direction periodically and changes its magnitude continuously with time. The waveform of AC voltage and current are sinusoidal in nature and represented as in the given figure.

In AC
magnitude of current and voltage continuously changes with time, therefore we
have to calculate instantaneous power.

**Instantaneous Power in AC Circuit**

The power
absorbed by the element in ac circuit at any instant of time is called
instantaneous power. Let us understand by a circuit.

Let us say voltage
v(t) is applied to an element Z and current drawn by the element Z is i(t) as
shown in given circuit.

v(t) = V_{m} sin(wt + Î¸_{v})

i(t) = I_{m}
sin(wt ± Î¸_{i}) ………… (4)

p(t) = v(t).i(t)

p(t) = I_{m}.V_{m}.
sin(wt + Î¸_{v}). sin(wt ± Î¸_{i}).

p(t) = I_{m}.V_{m}
/2 . (cos(wt+ Î¸_{v})-(wt + Î¸_{i}))-cos((wt+ Î¸_{v})-(wt
+ Î¸_{i}))-)))

p(t) = I_{m}.V_{m}
/2 . (cos(2wt ± Î¸_{v} - Î¸_{i}) +
cos(Î¸_{v} - Î¸_{i}))

Let us
say Î¸
= Î¸_{v} - Î¸_{i
}

p(t) = I_{m}.V_{m}
/2 . cos(2wt ± Î¸) + I_{m}.V_{m} /2
cos(Î¸)) ………………. (5)

(i)
(ii)

In equation (5) Î¸ is the phase angle difference between voltage v(t) and
current i(t)

- If Î¸ is positive then it means current is lagging with respect to
voltage (
**inductive load**).

- If Î¸ is negative then it means current is leading with respect to voltage (capacitive load).

- If Î¸ is zero then it means current is phase with voltage (
**resistive load**).

The instantaneous power is the sum of two terms. The first term is time
dependent and the second term is time independent.

The instantaneous power changes with time, therefore it is difficult to
measure. There is more convenient way to measure power in ac circuit i.e.
average power.

**Average Power in AC Circuit**

The concept of average power is used in AC circuit because the magnitude
and direction of the voltage and current in an AC circuit are changed
continuously. Unlike in DC circuits they oscillate sinusoidaly. As a result,
instantaneous power also varies with time as shown in equation (5). By
calculating the average power over a complete cycle, we can determine the
actual power dissipated in the circuit.

The average power provides a measure net power dissipated or delivered
over a complete cycle, which is useful for determining the actual power
consumption in AC circuits.

Therefore, the average power corresponding to the p(t) can be calculated
by integrating the instantaneous power shown in equation (5) over one period of
sinusoidal signal.

P_{av} = 1/T _{0}Êƒ^{T}
p(t).dt

P_{av} = 1/T _{0}Êƒ^{T} (I_{m}.V_{m}
/2 . cos(2wt ± Î¸) + I_{m}.V_{m} /2 cos(Î¸))

The average
of the sinusoid over a complete cycle is zero. Hence the integral of first part
is zero. The Second term is constant so the average power.

P_{av} = I_{m}.V_{m} /2 cosÎ¸

P_{av} = (I_{m}/√2).(V_{m}/√2).cosÎ¸

P_{av} = V_{rms}.I_{rms} cosÎ¸ ………………………………. (6)

In equation
(6) cosÎ¸ is **power factor** which is defined as the cosine of phase angle
difference between voltage and current.

Î˜ is zero
in case of pure resistive load so, average power consumed by pure resistive
load

P_{av} = V_{rms}.I_{rms} cos0 =
V_{rms}.I_{rms } ( cos0 = 1 )

Î˜ is +90^{0
}and -90^{0} in case of pure inductive and capacitive load
respectively. So, average power absorbed by pure inductive and capacitive load

P_{av} = V_{rms}.I_{rms} cos90^{0} =
0
(cos90^{0 } =
0)

** **

**Types of electric Power in AC Circuit**

In AC circuit
analysis we deal with three types of power these are

1. Complex
power or Apparent power

2. Active
Power or Real Power or True Power

3. Reactive
Power

** **

** **

**Apparent Power**

In an AC
(alternating current) circuit, apparent power is a measure of the total power
supplied to the circuit or element, considering both the real power and the
reactive power. It is expressed as the product of V_{rms }and I_{rms} and its unit volt-amperes
(VA).

** **

**Apparent Power Formula**

** **

Consider a
network

In above
circuit V∟Î¸_{v} is applied voltage to the element a
and I∟Î¸_{i }is the current drawn by element X.

S = V_{rms}.I^{*}_{rms} =
V_{rms}∟Î¸_{v}.I_{rms}∟Î¸_{i } (I^{*} means Conjugate of I)

=
V_{rms}.I_{rms}∟Î¸_{v }- Î¸_{i} ……………………………… (7)

Converting
the equation (7) from polar form to rectangular form gives

S = V_{rms}.I_{rms}
(cos(Î¸_{v}- Î¸_{i})
+ j.sin(Î¸_{v}- Î¸_{i}))

S =
V_{rms}.I_{rms} cos(Î¸_{v}- Î¸_{i}) +
j. V_{rms}.I_{rms}.sin(Î¸_{v}- Î¸_{i})

Let us
say Î¸
= Î¸_{v}- Î¸_{i}

S =
V_{rms}.I_{rms} cosÎ¸
+ j. V_{rms}.I_{rms}.sinÎ¸ …………………. (8)

(i)
(ii)

The
apparent power consists of two parts, in which part (i) is real and part (ii)
is imaginary. The real part of S is called real power or active power and it is
similar to the average power that we are calculated previously it is denoted by
P, and imaginary part of the S is called reactive power and is denoted by Q.

S =
P + j Q
………………………… (9)

Equation
(9) shows that the apparent power (S) in an AC circuit is the vector sum of the
real power (P) and the reactive power (Q).

The
practical significance of apparent power is as a rating unit of generators and
transformers that is in a KVA or MVA.

**Reactive Power**

The reactive
power is the portion of the complex power that is exchange between reactive
load (inductor and capacitor) and source. Let us understand by the given
circuit.

In the
given circuit voltage v(t) = Vm.sinwt is applied to an inductor. In this
circuit inductor stores the energy when positive half of v(t) is applied to
inductor means the flow of energy is from source to load. When negative half
v(t) is applied to inductor, inductor releases the stored energy during
positive half. This oscillation of energy between inductor and source is going
on until the voltage is applied. This energy (power) is called reactive power
and it is measured in VARs (Volt Ampere Reactive).

If the
inductor is ideal, means it is purely inductive load then 100% of complex power
oscillates between load and source and there is no active power absorbed by
inductor. (We discussed this Inductor).

Reactive
power has no effect on net power absorbed by circuit or element. But it has its
own significance.

As we know
that the different electrical machines like electric motor, transformer etc establish
a magnetic domain in the form magnetic flux for performing their operation. To
established this magnetic domain, electric machines use a small fraction of
complex power, this fraction of complex power is known reactive power. It is
denoted by Q.

Q = V_{rms}.I_{rms}.sinÎ¸ ……………………… (10)

Î˜ is the
phase angle difference between applied voltage and current drawn

**Active power**

Active
power or real power is the fraction of complex power which is used to fulfill
the load demand of consumers. It is the actual power which is used or
dissipated by the load in circuit. It is also called true power or real power and
is measured in watts. It is similar with the average power dissipated in DC
circuit.

P =
V_{rms}.I_{rms}.cosÎ¸
……………………… (11)

In equation
(11) cosÎ¸ is the **power factor** of the
circuit.

The
relation between these three types of power is studied with the help of Power Triangle.

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