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Home Electromotive Forces in Conductors in Series Addition of Voltages with Phase Displacement  
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Addition of Voltages With Phase DisplacementAuthor: E.E. Kimberly In Fig. 42 let E_{a} and E_{b} be two electromotive forces generated in conductors a and b with a time phase displacement angle θ_{2} similar to that in Fig. 41.
If in Fig. 41 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 E_{a} and E_{b} 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 E_{a} and E_{b}. The resultant sum will also vary sinusoidally, as shown by the curve marked e in Fig. 42 (b), In Fig. 42 (a) the vector E is the vector sum of E_{a} and E_{b} and is seen to be the diagonal of a parallelogram which has E_{a} and E_{b} as adjacent sides. The instantaneous value e of E is always equal to the algebraic sum of e_{a} and e_{b} at that instant. The sinusoid e of Fig. 42 (b) is thus a plot of E sin θ_{E}. 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 (31). Vectors are assumed to rotate counterclockwise. Example 41.  A conductor with an effective generated emf E_{a} of 0.10 volt is connected in series with another conductor with an effective generated emf E_{b} of 0.20 volt. The angle θ_{1} by which E_{a} leads E_{b} in time phase is 45°. What is the resultant emf E that would be shown by a voltmeter?
Solution.  The resultant emf E is the vector sum of E_{a} and E_{b} and it can be found by constructing a parallelogram of vectors, as shown in Fig. 43. 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 41, one of the vectors E_{b} is laid out along the horizontal axis of a rectangularcoordinate system. Also E_{a} is resolved into its horizontal component E_{a} cos θ_{1} and its vertical component E_{a} sin θ_{1}. Then, in the equation for E, the horizontal components E_{b} and E_{a} cos θ_{1} 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 E_{a} and E_{b}, 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 zaxis of which will be along that particular emf vector. Example 42.  Five conductors connected in series have the following electromotive forces: E_{a} = 20 volts, E_{b} = 40 volts lagging, E_{a} by 45°, E_{c} = 50 volts lagging E_{a} by 80°, E_{d} = 45 volts leading E_{a} by 55°, and E_{e} = 50 volts leading E_{a} by 75°. Find the magnitude of the resultant voltage and its phase displacement from E_{a}.
Solution.  Fig. 44 shows graphically the arrangement of the vectors given. Their components are as follows:
Hence andThe resultant emf is 96 volts and it leads E_{a} by 4°34'.


Home Electromotive Forces in Conductors in Series Addition of Voltages with Phase Displacement 