# AC theory (2 motors in parallel)

An induction motor with a stator resistance of 6 Ω and 8Ω inductive reactance is connected in parallel with a synchronous motor of 8 Ω resistance and 15Ω capacitive reactance. The system is connected to a 230 v mains supply. Draw clearly labelled phase diagrams to show the horizontal (in phase) and vertical (out of phase) current components with the phase angle for each of the sub units. Use the phase diagrams (vectors) and trigonometry to solve the following.
a. The current taken from the 230 v mains supply

b. The phase angle

c. The total impedance (i.e. the impedance as seen by the electrical supply)

d. The power factor

For the phase diagram please see attached figure.

$I = sqrt{[(U/R1+U/R2)^2 + (U/XL-U/X c)^2]} =$

$= 230*sqrt{(1/6+1/8)^2 +(1/8 -1/15)^2} =$

$=230*sqrt(0.085+0.0034)=68.385 A$

$tan(phi) = (1/XL-1/X C)/(1/R1+1/R2) = (1/8-1/15)/(1/6+1/8) =0.0583/0.2916=0.2$

$phi =11.31 degree$

$I = U/Z$

$Z = U/I = 230/68.385 =3.363 ohm$

another way to compute Z is

$1/Z = sqrt{[(1/R1 +1/R2)^2 + (X C parallel XL)^2]} =$

$= sqrt{[(R1+R2/R1*R2)^2 + (X C-XL)/(X C*XL))^2]} =$

$= sqrt{[(6+8)/(6*8))^2 +(15-8/(15*8))^2]} =$

$= sqrt{(0.085 +0.0034)}=0.297$

$Z =1/0.297 =3.363 ohm$

Power factor $= P/Pa$

$Pa = U*I = 230*68.385 =15728.55 W$

$P = U^2/R eq = 230*230/3.428 =15431.74$

$R eq = R1*R2/(R1+R2) = 6*8/(6+8) =3.428$

Power factor $= 15431.74/15728.55 =0.9811 =98.11%$