Question 15 Solution

Point A: Voltage divider with equal resistors ($R_1=R_2$). $$ V_A = \frac{V}{2} $$ Point B: Voltage divider with $C$ and $R$. $$ V_B = V \frac{R}{R – jX_C} $$ Potential Difference: $$ V_{AB} = V_B – V_A = V \left( \frac{R}{R – jX_C} – \frac{1}{2} \right) = \frac{V}{2} \left( \frac{R + jX_C}{R – jX_C} \right) $$ Magnitude: $$ |V_{AB}| = \frac{V}{2} \left| \frac{R + jX_C}{R – jX_C} \right| = \frac{V}{2} \cdot 1 = \frac{V}{2} $$

In the RC branch, the voltage across the resistor ($V_R$) and the voltage across the capacitor ($V_C$) are always perpendicular ($90^\circ$ phase shift). Their vector sum is the source voltage $V$.
Geometrically, this implies that the locus of point B (the junction of R and C) describes a semi-circle with the source voltage vector as the diameter.
Point A is the midpoint of the source voltage (since $R_1=R_2$).
The vector $V_{AB}$ connects the center of the diameter (A) to a point on the circumference (B).
The length of this vector is simply the radius of the circle: $$ |V_{AB}| = \text{Radius} = \frac{\text{Diameter}}{2} = \frac{V}{2} $$ This magnitude is constant regardless of the value of $R$.