The voltage drop across 4 series connected impedances are given:

$V1 = 60 \sin (wt + π/6) \\ V2= 15 \sin (wt - 5 π/6) \\ V3 = 100 \cos (wt + π/4) \\ V= V4m \sin (wt + ф/4)$

Calculate the value of V4m and $ф4$ if the voltage applied across the series circuit is $V= 140 \sin (wt + 3 π/5)$.

Writing the voltages in phase form

V1 = 60/ \/2 < $\bar{30}o$ = 42.43 < 30o

V2 = 75/ \/2 < $\bar{-150}o$ = 33.03< -150o

V3= 100/ \/2 < $\bar{-135}o$ = 70.71 < 135o

V= 140/ \/2 < $\bar{108}o$ = 98.99 < 108o

For series connected impedances,

V= V1 + V2 + V3 + V4

V4= V – V1 – V2 – V3

= 98.99 < 108o – 42.43 <30o – 53.03 < -150o -70.71 < 135o

= 57.13259.96o

V4 $= 57.13 \sqrt{2} \sin (wt + 59.96o) \\ = 80.79 \sin (wt + 59.96o)$

∴ V4M= 80.79

Ф= 59.96o