Question

9. In Fig.2

v1= √2(16)sin Wt V,

V2 =√2(24) sin(Wt +90°) V, and

v3=√2(15) sin(Wt -90°) V.

Determine the source voltage, e

Answer 1

• Gives

$V_1=\sqrt2(16)sin(wt)$

$V_2=\sqrt24sin(wt+90°)$

$V_3=\sqrt2(15)sin(wt-90°)$

$V=V_{RMS}sin(wt+\phi)$

$$V=16sin(wt+0)+24sin(wt+90)+15sin(wt-90)$$

$$V_1=16/(\theta=0°);V_2=24/(\theta=90°);V_3=15/(\theta=-90°)$$

Now resultant voltage

$V=V_1+V_2+V_3$

$$V=(V_1=16/(\theta=0°);V_2=24/(\theta=90°);V_3=15/(\theta=-90°))$$

$$V=16(cos0°+sin0°)+24(cos90°+sin90°)+15(sin(-90°)+cos(-90°))$$

$V=16+0-15+j(0+24+0)$

V=1+j(24)

Converting poler form

V=5+j24

Resultant velocity

$V=5\sqrt{2}sin(wt+24°)$