Ans ; no change in the reactance value
For better understanding let give an example. Suppose we have animpedance consisting of 50 ohms resistance and 100 ohms inductivereactance (Z = 50 +j100). If we assign a value of 100 ohm to prime center, we
normalize the above impedance by dividing each component of the impedance by 100 (
Normalization must be used, in order to facilitate the plotting of larger impedances. Each impedanceto be plotted is divided by a convenient number that will place thenew normalized impedance near the center of the chart where increased accuracy inplotting is obtained). The
normalized impedance is then 50/100 + j(100/100) = 0.5 + j1.0. This impedance is plotted on the Smith Chart at theintersection of 0.5 resistance circle and the +1.0 reactance circle. Instead of assigning 100 ohms to prime center, we assign a value of50 ohms. With this assignment, the 50 + j100 ohm is plotted at theintersection of the 50/50 = 1.0 resistance circle, and the 100/50 = 2.0 positive the sameimpedance value, 50 + j100 ohms. This example shows that the sameimpedance may be plotted at different points on the chart, dependingupon the value assigned to prime center. But two plotted points cannotrepresent the same impedance in the same time. Prime center is a point of special significance. It is customary whensolving problems to assign the Zo value of the line to this point on the chart, 50 ohms for a 50 ohms line