BIUST
Faculty of Engineering
Electrical, Computer and Telecommunications Department
EEN221
Tutorial for week 4
Covers material on chapter 13
Question 1
A single-phase motor takes 8.3 A at a power factor of 0.866 lagging when connected to a 230
V, 50 Hz supply. Two similar capacitors are connected in parallel with each other to form a
capacitance bank. This capacitance bank is now connected in parallel with the motor to raise
the power factor to unity. Determine the capacitance of each capacitor.
Answer:
S = VI = 230 × 8.3 = 1907 VA
P = S cos φ = 1653 W
Q = 1907 2 − 16532 = 951 var
For each capacitor,
QC =
951
= 475 var
2
= VIC
IC =
475
= 2.067 A
230
∴ XC =
∴C =
V
230
=
= 111.26 Ω
I C 2.067
1
= 28.6 µ F
2πfX C
Question 2
A 25 kVA single-phase motor has a power factor of 0.8 lag. A 10 kVA capacitor is connected
for powerfactor correction. Calculate the input apparent power in kVA taken from the mains
and its power factor when the motor is
(a) on half load;
(b) on full load.
Sketch a phasor diagram for each case.
Answer:
(a) On half load, P =
25
× 0.8 = 10 kW
2
For motor, QM =
25
× 0.6 = 7.5 kvar
2
For capacitor,.QC = 10 kvar
S=
cos φ =
[102 + (10 − 7.52 ] = 10.3 kVA
10
= 0.97 leading
10.3
(b) On full load, P = 20 kW and QM = 15 kvar
S=
cos φ =
[202 + (15 − 10)2 ] = 20.6 kVA
20
= 0.97 lag
20.6
Question 3
A 230 V, single-phase supply feeds the following loads:
(a) incandescent lamps taking a current of 8 A at unity power factor;
(b) fluorescent lamps taking a current of 5 A at 0.8 leading power factor;
(c) a motor taking a current of 7 A at 0.75 lagging power factor.
Sketch the phasor diagram and determine the total current, active power and reactive power
taken from the supply and the overall power factor.
Answer:
IA = 8.0∠0° A
IB = 5.0∠36.9° A
IC = 7.0∠−41.4° A
∴Ip = 8 cos 0° + 5 cos 36.9° + 7 cos(−41.4°) = 17.25 A
Iq = 8 sin 0° + 5 sin 36.9° + 7 sin(−41.4°) = −1.63 A
I=
(17.252 + 1.632 ) = 17.35 A
P = 230 × 17.25 = 3967.5 W
Q = 230 × 1.63 = 375 var
cos Φ =
17.25
17.35
= 0.995 lag
Question 4
The load taken from an a.c. supply consists of: (a) a heating load of 15 kW; (b) a motor load
of 40 kVA at 0.6 power factor lagging; (c) a load of 20 kW at 0.8 power factor lagging.
Calculate the total load from the supply (in kW and kVA) and its power factor. What would
be the kvar rating of a capacitor to bring the power factor to unity and how would the
capacitor be connected?
Answer:
59 kW, 75.5 kVA, 0.78 lagging; 47 kvar in parallel
Question 5
A cable is required to supply a welding set taking a current of 225 A at 110 V alternating
current, the average power factor being 0.5 lagging. An available cable has a rating of 175 A
and it is decided to use this cable by installing a capacitor across the terminals of the welding
set. Find:
(a) the required capacitor current and reactive power to limit the cable current to 175 A;
(b) the overall power factor with the capacitor in circuit.
Answer:
Without the capacitor, Ip = 225 × 0.5 = 112.5 A
and Iq = 225 × 0.866 = 194.9 A
With the capacitor, Ip is still 112.5 A
but Iq =
(1752 − 112.52 = 134.1 A
∴Capacitor current = 194.9 − 134.1 = 60.8 A
Qc = VIC = 110 × 60.8 = 6.69 kvar
cos =
112.5
= 0.643 lag
175