Phasor Algebra Assignment Help | Phasor Algebra Homework Help

Phasor Algebra

Complex quantities, normally employed in ac circuit analysis, can be added and subtracted like coplanar vector. Such coplanar vectors which represent sinusoidally time varying. Quantities are called phasors.

Let the x-axis and y-axis projections of a phasor A be xA  is yA respectively. Then, its magnitude A is
        A = √(x2A + y2A)                                … (1)

The angle between the directions of phasor A and the direction of the positive x-axis is
        θ = tan-1(yA/xA)                                    … (2)

j is used to indicate the counter clockwise rotation of a phasor through 90o without changing its magnitude. It is assigned a value of √-1. When a given phasor A, the direction of which is along the x-axis, is multiplied by the operator j, a new phasor j A is obtained which will be 90o counter clock-wise from A i.e., along y-axis. If the operator j is applied now to the phasor jA, a new j2A is obtained which will be along –x-axis hand having same magnitude as of A. Thus J2 A = -A.
        J3A = -jA.
        J4A = (√-1)4 A=A.
Hence it is seen that the successive applications of the operator j to the phasor A produces successive 90o steps of rotation of the phasor in the counter clockwise direction without in any way affecting the magnitude of the phasor.
Thus in Cartesian form, a phasor A can be written as
        A = a + jb,
Where a is the x-component and b the y-component of phasor A.

Polar Form of Representation

In polar form, the phasor can be written as
        A = A cos θ + jA sin θ = A (cos θ + j sin θ).
Here A cos θ is the x-component of the phasor and A sin θ the y-component. If the angle is measured in the clockwise direction from the x-axis, the phasor A can be written as
        A = A (cos θ – j sin θ).

Thus we see that cos θ + j sin θ is an operator which when applied with a phasor A, along the x-direction, rotates it through an angle θ from its initial position. Similarly the operator cos θ – j sin θ rotates the original phasor through an angle – θ.

Exponential Form of Representation

In exponential form the phasor A can be written as
        A = Ae+ jθ = A (cos θ – j sin θ) = A+ θ

Addition of phasors. The sum of two phasors is a phasor which is defined in magnitude and phase position by he diagonal of parallelogram formed by the two phasors.
        A + B = C = B + A.

Subtraction of phasors. Phasors to be subtracted is rotated through 180o and the addition is done by completing the parallelogram

Multiplication of phasors. The product of two phasors is a third phasor having magnitude equal to the algebraic sum of the individual phasor angles of these phasors with respect to the same reference axis.

        A = AejθA = A|θA and B = BejθB = B|θB
. :        AB = ABej(θA + θB)  = AB|θA + θB

Division of Phasors. When one phasor A is divided by another phasor B, it result in a phasor having magnitude A/B and the phase position with respect to the reference axis is the algebraic difference between the individual phase angles of these phasors with respect to the same reference axis.
Thus            A /B = C = AejθA /BejθB = A/B ej(θA – θB) = A/B|θA - θB

Raising of Power. A phasor may be raised to a given power n (an integer), by multiplying the phasor by itself n times.
        (A)n = An|nθA 

Phasor Algebra

For more help in Phasor Algebra click the button below to submit your homework assignment.