(a) Power factor is leading because current leads the voltage.
(b) p.f. = cos 30° = 0.86 (lead)
(c) The circuit is capacitive.
(i) Circuit current can be found by dividing voltage drop V1 by Z1
I = (10 + j0)/(3 + j4)
= (10∠ 0°)/(5 ∠ 53.1°) = 2 ∠ - 53.1º
= 2(cos53.1º - jsin53.1º)
= 2 (0.6 −j 0.8) = 1.2 −j 1.6
Z2 = 2 + j 3.46; V2 = IZ2
= (1.2 −j 1.6) (2 + j3.46) = (7.936 + j 0.952) volt
V3 = (1.2 − j 1.6) (1− j 7.46)
= (− 10.74 −j 10.55) volt
(ii) V = V1 + V2 + V3
= (10 + j0) + (7.936 + j 0.952) + (− 10.74 −j 10.55)
= (7.2 −j 9.6) = 12 ∠ − 53.1°
Incidentally, it shows that current I and voltage V are in phase with each other