Hello Benita dear, as per phasor diagram R will be along x axis and Xc - capacitive reactance will be along negative y axis.
So net impedance Z = (R^2 + Xc^2)
Recall Xc = 1/ 2 pi f C
f = 50 Hz, C = 15 uF = 15 * 10^-6 F
So 2 pi f C = 2 * 3.14 * 50 * 15 * 10^-6 = 15 * 3.14 * 10^-4
Now 1/(2pi* f *C) = 10^4 / (15 * 3.14) = 212.3 ohm
So R^2 + Xc^2 = 200^2 + (212.3 )^2
= 100^2 (2^2 + 2.123^2) = 100^2 (4 + 4.51) = 8.51 * 100^2 [using calculator it will be easier]
Z = ./(8.51 * 100^2) = 2.9 * 100 = 290 ohm
Therefore r m s value of current = 220/290 = 0.759 A
Now drop across R = 200 * 0.759 = 151.8 V
Drop across Xc = 212.3 * 0.759 = 161 V
As we add these voltage drops it would exceed 220 V. So straight addition is not allowed
As per phasor diagram these two voltages are perpendicular to one another
Hence we have to follow Pythagoras theorem and that would definitely be almost 220 V