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# Addition of Voltages With Phase Displacement

Author: E.E. Kimberly

In Fig. 4-2 let Ea and Eb be two electromotive forces generated in conductors a and b with a time phase displacement angle θ2 similar to that in Fig. 4-1.

 Fig. 4-2. Addition of Voltages in Conductors in Series

If in Fig. 4-1 the farther end of conductor a be connected to the nearer end of conductor b by a conductor which lies outside the magnetic field, the electromotive forces Ea and Eb will act in series to produce a difference of potential at their free ends which at every instant will be the sum of the instantaneous values of Ea and Eb. The resultant sum will also vary sinusoidally, as shown by the curve marked e in Fig. 4-2 (b), In Fig. 4-2 (a) the vector E is the vector sum of Ea and Eb and is seen to be the diagonal of a parallelogram which has Ea and Eb as adjacent sides. The instantaneous value e of E is always equal to the algebraic sum of ea and eb at that instant. The sinusoid e of Fig. 4-2 (b) is thus a plot of E sin θE.

In the application of vectors in computations, it is customary to use effective values which are maximum values times 0.707, as shown by equation (3-1). Vectors are assumed to rotate counter-clockwise.

Example 4-1. - A conductor with an effective generated emf Ea of 0.10 volt is connected in series with another conductor with an effective generated emf Eb of 0.20 volt. The angle θ1 by which Ea leads Eb in time phase is 45°. What is the resultant emf E that would be shown by a voltmeter?

 Fig. 4-3. Vector Sum of Two Voltages

Solution. - The resultant emf E is the vector sum of Ea and Eb and it can be found by constructing a parallelogram of vectors, as shown in Fig. 4-3. This problem is one in simple trigonometry, as it is merely required to find the length of a diagonal of a parallelogram in which two sides and the included angle are known. The effective values may be used in the vector diagram. The computations follow:

In solving Example 4-1, one of the vectors Eb is laid out along the horizontal axis of a rectangular-coordinate system. Also Ea is resolved into its horizontal component Ea cos θ1 and its vertical component Ea sin θ1. Then, in the equation for E, the horizontal components Eb and Ea cos θ1 are added to make the base of the triangle of which E is the hypotenuse. The value of E is then found to be the square root of the sum of the squares of the base and the perpendicular side of a right triangle.

If, instead of only two electromotive forces Ea and Eb, there had been a sheaf of emf vectors resulting from a large number of conductors moving in the circular path, the sum of all the electromotive forces in series could most easily be found as follows: Resolve each emf into its horizontal and vertical components; add algebraically all horizontal components and all vertical components; and combine the algebraic sums by taking the square root of the sum of their squares. If it is desired to find both the magnitude of the resultant emf and its phase displacement from a particular one of the given electromotive forces, it is convenient to locate the sheaf of vectors on a coordinate system the z-axis of which will be along that particular emf vector.

Example 4-2. - Five conductors connected in series have the following electromotive forces: Ea = 20 volts, Eb = 40 volts lagging, Ea by 45°, Ec = 50 volts lagging Ea by 80°, Ed = 45 volts leading Ea by 55°, and Ee = 50 volts leading Ea by 75°. Find the magnitude of the resultant voltage and its phase displacement from Ea.

 Fig. 4-4. Vector Addition of Several Voltages

Solution. - Fig. 4-4 shows graphically the arrangement of the vectors given. Their components are as follows:

Hence

and

The resultant emf is 96 volts and it leads Ea by 4°34'.

Last Update: 2010-10-06