1. A single-phase motor draws a current of 6.5A from 230V, 50Hz line. The power factor of the motor is 65%. Calculate (with support of the diagram:

i) The active power absorbed by the motor.
ii) The reactive power supplied by the line.

2. A 20μF paper capacitor is placed across the motor terminals for above motor.
Calculate:

iii) The reactive power generated by the capacitor.
iv) The active power absorbed by the motor.
v) The reactive power absorbed from the line.
vi) The new line current.


i) The active power absorbed by the motor.

Power factor is equal to: 65%
Power drown from the circuit is equal to I=6.5A,V=230V,f=50Hz
\[
Power factor=65\%=0.65
\]
As we know the power factor formula we can now count angle :
\[
\cos{\left(\phi{}\right)}=0.65
\]
\[
\phi{}={cos}^{-1}\left(0.65\right)
\]
\[
\phi{}=49.458^\circ{}
\]
In next step I’m going to apply angle to formula to count active power absorbed by the motor.
\[
P=VI\times{}cos(\phi{})
\]
\[
P=230\times 6.5\times\cos{\left(49.458^\circ{}\right)}
\]
\[
P=971.757W
\]
Answer:
Active power absorbed by motor is equal to 971.757W.


ii) The reactive power supplied by the line.

Reactive power, supplied by the line we can count by applying the fallowing formula:

\[
Q=VI\times{}\sin{\left(\phi{}\right)}
\]
\[
Q=230\times 6.5\times \sin{\left(49.458^\circ{}\right)}
\]
\[
Q=1136.094VAR
\]
Answer:
Reactive power supplied by the line is equal to 1136.094VAR

To check my calculations, I need to draw a graph with represents power triangle. I’m going to use basic trigonometric calculation just to have different point of view on this aspect.

Power triangle diagram:

Side (c) between vertexes (A) and (B) of the triangle represents active power = P
Side (a) between vertexes (B) and (C) of the triangle represents reactive power = Q
Side (b) between vertexes (A) and (C) of the triangle represents apparent power = S


\[
c=P=971.757
\]
\[
a=Q=1136.094
\]
\[
\phi{}=49.458^\circ{}
\]
\[
b=S=\ ?
\]
By applying Sine Law I can find angle of vertex (C) of the triangle.
\[
\frac{\sin{\left(C\right)}}{c}=\frac{\sin(\phi)}{a }
\]
\[
\frac{\sin(C)}{971.757}=\frac{\sin 49.458^{\circ }}{1136.094}
\]
\[
\sin{\left(C\right)}=0.650
\]
\[
C={sin}^{-1}\left(0.650\right)
\]
\[
C=40.541
\]
Answer:
Angle of vertex B of the triangle is equal to 90

Sum of triangle’s angles is equal to 180 so:
\[
49.458+40.541+B=180
\]
\[
B=90
\]
Answer:
Angle of vertex B of the triangle is equal to 90

By applying formula from Sine Law, I will be able to obtain the apparent power.
\[
\frac{b}{sin⁡(B)}=\frac{a}{sin⁡(\phi{})}
\]
\[
\frac{b}{1}=\frac{1136.094}{0.759}
\]
\[
b=S=1494.998
\]
By mathematic calculations above I have confirmed that apparent power is:
\[
S=VI
\]
\[
S=2306.5
\]
\[
S=1495VA
\]
Answer:
Apparent power is equal to 1495VA
Additionally, by drawing of the Phasor diagram, I can confirm my calculations.

Phasor diagram
\[
Vsin\left(\phi{}\right)=230\times{}\sin{\left(49.4 58^\circ{}\right)}=174.783
\]
\[
174.783\times{}6.5=1136.094VAR
\]
\[
Vcos\left(\phi{}\right)=230\times{}cos\left(49.458 ^\circ{}\right)=149.501
\]
\[
149.501\times{}6.5=971.757W
\]
\[
V\times{}I=230\times{}6.5
\]
\[
VI=1495VA
\]


iii) The reactive power generated by the capacitor.

To count reactive power generated by the capacitor I need to apply specific formula.

Where:
\[
Q_c=\omega{}CV^2
\]
Angular frequency is equal to:
\[
\omega{}=2\pi{}f
\]
\[
\omega{}=2\pi{}\times{}50Hz
\]
\[
\omega{}=314.159rad/s
\]
Voltage V=230Volts
Capacitance C=20 microFarads
\[
Q_c=314.159\times{}20\times{}{10}^{-6}\times{}{230}^2
\]
\[
Q_c=332.380VAR
\]
Answer:
Reactive power generated by the capacitor is equal to 332.380VAR


iv) The active power absorbed by the motor.

The active power absorbed by the motor won’t change. To prove my thesis I will plot power triangle diagram.

Power triangle with capasitor

Answer:
Active power absorbed by motor is equal to 971.757W.


v) The reactive power absorbed from the line.

Formulea:
\[
Q_m=Q-Q_c
\]
\[
Q_m=1136.094-332.380
\]
\[
Q_m=803.714VAR
\]
Answer:
The reactive power absorbed from the line is equal to 803.714VAR

vi) The new line current.

To count new current line, with capacitor in the circuit. I need to find new apparent power.
\[
S^{'}=\sqrt{P+Q_m}
\]
\[
S^{'}=\sqrt{{\left(971.757\right)}^2+{(803.714)}^2 }
\]
\[
S^{'}=1261.058VA
\]
\[
230\times{}I^{'}=1261.058
\]
\[
I^{'}=\frac{1261.058}{230}
\]
\[
I^{'}=5.482A
\]
Answer:
New line current is equal to 5.482 Amper.


Additional calculations:

New power factor in the circuit with capacitor.
\[
cos⁡(\varphi{})=\frac{{Q_m}^2-{S^{'}}^2+P^2}{2\times{}Q_m\times{}P}
\]
\[
\varphi{}=90^\circ{}
\]
\[
\cos{\left(\beta{}\right)}=\frac{{Q_m}^2+{S^{'}}^2-P^2}{2\times{}Q_m\times{}S^{'}}
\]
\[
\beta{}=50.406^\circ{}
\]
\[
{-\phi{}}^{'}=(50.406^\circ{}+90^\circ{})-180^\circ{}
\]
\[
{\phi{}}^{'}=39.594^\circ{}
\]
\[
\cos{\left({\phi{}}^{'}\right)}=New\ power\ factor
\]
\[
\cos{\left(39.594^\circ{}\right)}=0.77
\]
Answer:
New power factor is equal to 0.77