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In this article we will discuss about:- 1. Introduction to Phasor Algebra of AC Circuits 2. Addition and Subtraction of Phasors 3. Multiplication and Division 4. Powers and Roots 5. Power Determination 6. Phasor Algebra Applied to Single Phase Parallel Circuits.

**Contents: **

- Introduction to Phasor Algebra of AC Circuits
- Addition and Subtraction of Phasors
- Multiplication and Division of Phasors
- Powers and Roots of Phasors
- Power Determination of AC Circuits
- Phasor Algebra Applied to Single Phase Parallel Circuits

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**1. Introduction to Phasor Algebra of AC Circuits: **

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We have seen that ac circuits cannot be solved by use of simple algebra since geometrical relations are to be taken into consideration. Though phasor diagram method is quite satisfactory for fairly simple circuits but becomes complicated when a circuit or network is made up of several branches. AC circuits can be conveniently solved by use of complex algebra. The results obtained by this method are of the same order of accuracy as those obtained by trigonometrical methods although calculations are usually much simpler.

The above system enables equations representing alternating voltages and currents and their phase relationships to be expressed in simple algebraic form. It is based upon the idea that a phasor can be resolved into two components along two axis at right angles to each other.

**There are four forms or methods of representing the vector or phasor quantities viz.:**

(i) Symbolic form,

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(ii) Trigonometrically form,

(iii) Polar form,

(iv) Exponential form.

All of these methods of expressing vector or phasor quantities enable those operations which are carried out graphically in a phasor diagram, to be performed analytically.

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**(i) Symbolic Form: **

A phasor, can be resolved into two components along two axis at right angles to each other. It is preferable and convenient to resolve it along X-axis and Y-axis.

In symbolic form, a phasor is resolved into two components at right angles to each other and the component lying along Y axis is designated by + j or – j according to its direction.

In Fig. 5.1 phasors OA_{1}, OA_{2}, OA_{3} and OA_{4} are shown lying in first, second, third and fourth quadrants respectively. The phasor OA_{1} lying in first quadrant may be resolved into two components a_{1} along X-axis (sometimes called the axis of real’s) and b_{1} along Y-axis (sometimes called the axis of imaginaries or j-axis). The component along Y-axis should be designated by j and should be written as jb_{1}. Thus we represent the phasor OA_{1}.

OA_{1} = a_{1 }+ j b_{1}

The phasor, when expressed in above form, is said to be expressed in complex form. In electrical engineering, the horizontal component (i.e., a_{1}) is called the in-phase or active component while the vertical component is called the quadrature or reactive component.

**The length of line OA _{1} represents, to adopt scale, the absolute value or modulus (magnitude) of the phasor and is equal to **

**√**

**a**

^{2}_{1}+ b^{2}_{1}, and its angel with the X-axis,**θ**

_{1}**, called the argument of the phasor, is given by the expression:**

**Similarly phasors OA _{2}, OA_{3} and OA_{4} can be resolved into two components along axis of real and axis of imaginaries and can be expressed as:**

**Significance of Operator j:**

The letter j used in the above expressions is an operator, the application of which to phasor a, or phasor component a, causes its rotation through 90° in the positive (counter-clockwise) direction. It is assigned the numerical value √-1. The double operation of j on phasor a rotates it in counterclockwise direction through 180° and hence reverses its sense because j j = j^{2} = (√-1)^{2} = – 1. The triple operation of j on phasor a rotates it in counter-clockwise direction through 270° and so on.

For example, if phasor V lying along X-axis is operated with j, the phasor V will rotate through an angle of 90° in counter-clockwise direction and lie along Y-axis. If this phasor V is operated twice with operator j it will rotate through 180° in counter-clockwise direction from X-axis and will lie along X-axis in negative direction, as illustrated in Fig. 5.2.

Application of the operator j thrice to this phasor V will rotate it through 270° in counter-clockwise direction from X- axis and now the phasor V will lie along Y-axis in negative direction. On operating the phasor V with j four times, it will rotate through 360° in counter-clockwise direction from the reference axis and so will come to the original position.

**Hence it is seen that j is an operator which turns phasor a through an angle of 90° in counter-clockwise direction, each time that it is applied and has a numerical value of ****√****-1 so that:**

**Conjugate Complex Numbers: **

Two complex numbers are said to be conjugate when they are of same modulus and have arguments of equal magnitude and opposite sign. For example, the numbers (a + j b) and (a – j b) are conjugate.

The sum of two conjugate numbers gives in-phase or active component and their difference gives quadrature or reactive component. The product of two conjugate numbers gives in-phase or active component only (no quadrature component) equal to the sum of the square of their magnitudes.

i.e. (a + j b) + (a – j b) = 2a, active component

(a + j b) – (a – j b) = j 2b, reactive component

(a + j b) x (a – j b) = (a^{2} + b^{2}), active component

**(ii) Trigonometrically form of Phasor Representation: **

As may be seen from Fig. 5.3, X-component of V = V cos θ and Y-component of V= V sin θ.

**Hence phasor V can be expressed in the trigonometrically form as:**

V = V cos θ + j V sin θ = V (cos θ + j sin θ)

This is equivalent to the symbolic or rectangular form V = (a + j b) because a – V cos θ and b = V sin θ

In general V = V (cos θ ± j sin θ)

The above expression represents a phasor of numerical value V and having phase angle of ± θ with the reference axis, positive angle being measured in counter-clockwise direction from the reference axis and negative angle in clockwise direction from the reference axis.

**The phasors OA _{1}, OA_{2}, OA_{3} and OA_{4} (Fig 5.1) may be represented thus: **

OA_{1} = OA_{1} (cos θ_{1} + j sin θ_{1})

OA_{2} = OA_{2} (cos θ_{2} + j sin θ_{2})

OA_{3} = OA_{3} (cos θ_{3}, – j sin θ_{3})

OA_{4} = OA_{4} (cos θ_{4} – j sin θ_{4})

**(iii) Polar Form of Phasor Representation:**

The phasor V (Fig. 5.3) may be alternatively expressed as V ∠θ where V is the magnitude of the phasor and θ is the angle measured in counter-clockwise direction from reference axis, called the argument of the phasor.

**In general, the phasor V may be represented as: **

V = V ∠ ± θ

The above form of phasor representation is known as polar form.

The phasors OA_{1}, OA_{2}, OA_{3} and OA_{4} (Fig. 5.1) may be represented thus O A_{1} = OA_{1} ∠θ_{1}; O A_{2} = OA_{2} ∠θ_{2}; OA_{3}, = OA_{3} ∠- θ_{3}; OA_{4} = OA_{4 }∠- θ_{4}.

**(iv) Exponential Form of Phasor Representation: **

By Euler’s equation e ^{j }^{θ} = cos θ + j sin θ

**Hence phasor V may be expressed as:**

V = V e ^{j }^{θ} where e is the base of natural logarithm.

**In general, the phasor V may be represented as: **

V = Ve ^{± j }^{θ}

The above form is known as exponential form of representing a phasor. It represents a phasor of modulus (or magnitude) V and argument ± θ.

**Note: **

The phasor representation in either of four mathematical forms convey the same information i.e., magnitude (modulus) of the phasor and its phase angle (argument). So it is possible to convert one form to the other.

**Example 1:**

Give the voltage expressed in the symbolic form V = (5 + j 12) volts in trigonometrical, polar and exponential forms.

**Solution: **

**The modulus or magnitude of the given voltage:**

V = √5^{2} + 12^{2} = 13 V

**The argument or phase angle of the given voltage:**

θ = Tan^{-1} 12/5 = 67.38° or 0.374 π radians.

**So given voltage in: **

(i) Trigonometrically form is 13 (cos 67.38° + j sin 67.38°) volts Ans.

(ii) Polar form is 13 ∠67.38° volts Ans.

(iii) Exponential form is 13 e ^{+}j ^{0.374}^{π} volts (Ans.).

** 2. Addition and Subtraction of Phasors: **

The symbolic or rectangular form is most suitable form for addition or subtraction of phasors. If the phasors are given in polar or exponential form, they should be first converted to rectangular form and then addition or subtraction be carried out.

**Addition of Phasors:**

For the addition of phasors in the rectangular form in-phase (or active) components are added together and quadrature components are added together.

Let the two voltage phasors V_{1} and V_{2} represented as V_{1} = a_{1} + j b_{1}, and V_{2} = a_{2} – j b_{2}, be added.

**Then resultant voltage will be given as: **

Addition process by graphic representation is illustrated in Fig. 5.5.

**Subtraction of Phasors:**

**By the rule of phasor subtraction:**

V = V_{1} + V_{2} = (a_{1} + j b_{1}) + (a_{2} – j b_{2}) = (a_{1} + a_{2}) + j (b_{1} – b_{2})

Subtraction process by graphic representation is illustrated in Fig. 5.6.

** 3. Multiplication and Division of Phasors: **

**The (a + j b) method of representing a phasor is particularly well suited to the process of addition and subtraction, as illustrated above, but it also lends itself readily to multiplication and division as illustrated below: **

However, it is worth noting that it is easier to multiply and divide phasors when they are expressed in polar or exponential form.

**Consider two phasors given by:**

**Multiplication of Phasors: **

**(i) Rectangular Form:**

**(ii) Polar and Exponential Form:**

**For multiplication of phasors in polar forms, their magnitude or modulus is multiplied and arguments or phase angles are added as proved below:**

We thus see that, the product of two phasors is a phasor having modulus equal to the product of the modulus of the two phasors and the argument is equal to the algebraic sum of the arguments of the two phasors.

**Division of Phasors:**

**(i) Rectangular Form:**

**Multiplying both the numerator and denominator by conjugate of (a _{2} + j b_{2}) we have:**

**(ii) Polar and Exponential Form:**

We thus see that, when we divide one phasor by another phasor, the quotient is a phasor having modulus equal to the quotient of the modulus of the two phasors and an argument equal to the algebraic difference of the arguments of the two phasors.

**In polar form we have: **

** 4. Powers and Roots of Phasors: **

Powers and roots of phasors can be determined conveniently in polar form. If the phasor is not in polar form then it will be advisable to convert it first to polar form and then carry out algebraic operations.

**Powers of Phasors:**

Let us find the fourth power of the phasor 5 ∠20˚, for this purpose; the phasor has to be multiplied by itself four times.

**Roots of Phasors:**

Let us find the one-fourth power of the phasor 5 ∠20˚.

**Phasor Algebra Applied to Single Phase Series Circuits: **

A simple series circuit having non-inductive resistance R and inductive reactance X_{L} is shown in Fig. 5.7 (a). Let the current phasor be taken as reference phasor and be represented as I = I + j 0 or I ∠0°.

Since voltage drop across resistance R, V_{R} = IR in phase with current I, it is expressed as (IR + j 0) and the voltage drop across the inductive reactance, V_{L} = I X_{L} in quadrature with current and leading, therefore, it is expressed as (0 + j IX_{L}).

**Hence the voltage applied across the circuit can be expressed as:**

Thus impedance can be written as complex quantity though it is not a complex quantity. Since impedance causes resistance and reactance voltage drops to be at right angle to each other, it is actually a complex operator.

Now let the applied voltage be taken as reference phasor and expressed as V (1 + j 0) or V∠0° then current be determined by using impedance as phasor.

In Fig. 5.8 (a) circuit having resistance R and capacitive reactance X_{C} connected in series are shown. Taking current I as reference phasor i.e., along X-axis, as shown in phasor diagram [Fig. 5.8 (b)].

**Taking applied voltage as reference phasor, as shown in Fig. 5.8 (c), the current flowing through the circuit is given as: **

**Thus general expression for phasor impedance can, be written as Z = (R ± j X) in complex notation: **

Plus sign is used for inductive circuit as inductive reactance is considered positive and minus sign is used for capacitive circuit as capacitive reactance is considered negative.

** 5. Power Determination of AC Circuits: **

**If the voltage and current in a circuit are expressed in complex form, power can be determined as follows: **

Let the current be expressed as I = 1 – j I_{2} and voltage as V = V_{1} + j V_{2}.

From Fig. 5.16 it is evident that the real component of voltage, V_{1} and real component of current I_{1} are in phase. Since components of voltage and current may be treated as if they were acting alone, the power contributed by V_{1} and I_{1}, P_{1} = V_{1} I_{1}.

The coefficient of V_{2} is + j and coefficient I_{2} is – j, these two components of voltage and current are opposite in phase and therefore, their product gives negative power.

So power P = -V_{2} I_{2 }

Voltage component V_{1} and current component I_{2} are in quadrature so power contributed by V_{1} and I_{2} is zero; similarly voltage component V_{2 }and current component I_{1} are in quadrature and, therefore, contribute no power.

.** ^{.}**. Total power in the circuit, P = P

_{1}+ P

_{2}= V

_{1}I

_{1}– V

_{2}I

_{2}.

The total power is the algebraic sum of the product of the real quantities V_{1} and I_{1} and the product of imaginary quantities V_{2} and I_{2}. The sign is determined in the ordinary algebraic method. The operator j must not be included in the multiplication.

**Example 2:**

Determine the power consumed in a circuit carrying a current of (3 – j 4) A and connected across (80 + j 40) volt supply mains.

**Solution: **

Supply voltage, V = (80 + j 40) volts

Circuit current, I = (3 – j 4) amperes

Power consumed, P = V_{1} I_{1}+ V_{2 }1_{2} = 80 × 3 + 40 (- 4) = 80 watts Ans.

**Conjugate Method of Power Determination: **

In any circuit let the applied voltage and current flowing be V = V (cos α + j sin α) and I = I (cos β + j sin β) respectively, (α – β) is the phase difference between V and I so power factor of the circuit is cos (α – β).

**Multiplying V and I we have: **

VI = V (cos α + j sin α) × I (cos β + j sin β) = VI [cos (α + β) + j sin (α + β)].

The real part of the above expression is VI cos (α + β) and evidently it cannot be true power of the circuit, because power factor of the circuit is cos (α – β). Similarly VI sin (α + β) is not correct expression of the reactive volt-amperes. Hence to determine the power by j method, it is necessary to change the sign of the imaginary part of either V or I.

**Thus if we change the sign of imaginary part of V, leaving I as before, we have: **

V_{C} = V (cos α – j sin α)

Complex quantity so obtained is known as conjugate of the original complex quantity.

Now V_{C}I = V (cos α – j sin α) × I (cos β + j sin β)

= V I [cos (α – β) – j sin (α – β)].

The real part of the above expression V 1 cos (α – β) gives the true power and imaginary part V I sin (α – β) gives the reactive power. It should, however, be noted that the real power obtained by the conjugate method is the same regardless of whether V or I is reversed but sign of reactive volt-amperes will depend upon it.

When the conjugate of voltage is used, capacitive VARs have a positive sign and inductive VARs a negative sign. If the conjugate of the current rather than of the voltage is used the capacitive VARs have a negative sign and inductive VARs a positive sign.

The product of V. I_{C} is called the complex power where I_{C} is the conjugate of I.

** 6. Phasor Algebra Applied to Single Phase Parallel Circuits: **

When two or more circuits are connected in parallel, the current in each branch may be found in complex form and then total current in the circuit may be determined by addition of complex numbers.

In the circuit shown in Fig. 5.19.

**Equivalent Parallel Impedance: **

The ac parallel circuit may be solved in the same way as dc parallel circuit except that complex impedances rather than simple resistances are involved.

**With several impedances connected in parallel the equivalent impedance is expressed as: **

**Since admittance is reciprocal of impedance, therefore: **

**After finding equivalent impedance or admittance, current in the circuit may be determined from the relation: **

The negative sign is used for inductive susceptance and positive sign for capacitive susceptance.

**Symbolic Method for Solution of Single Phase Series- Parallel Circuits: **

Single phase ac series-parallel circuits may be solved in the same manner as dc series-parallel circuits are solved except that complex impedances instead of simple resistances are involved.

**The equivalent impedance of a parallel circuit is determined first from the relation: **

Then it is added (in complex form) to the impedance of the circuit connected in series with the combined parallel circuit and thus total impedance of the whole circuit is determined. ~~ ~~

**Example 3: **

**A 15 mH inductor is in series with a parallel combination of an 80 ****Ω**** resistor and 20 ****µ**** F capacitor. If the angular frequency of the applied voltage is ****ω**** = 1,000 radians/second, find the admittance of the network. **

**Solution:**