ELLOREG, Axlle Rhoess D.
BSEE 2A
1. In a series LCR AC circuit, the current is maximum when the impedance is equal to:
2. A 100 volt A.C. source of frequency 500 hertz is connected to a L-C-R circuit with
L = 8.1 millihenry, C = 12.5 microfarad and R = 10 ohm, all connected in series. The
potential difference across the resistance will be:
3. A resistor R, inductor L and a capacitor C are connected in series to an oscillator of
frequency v. If the resonant frequency is vr, then the current lags behind the voltage,
when:
4. .) 220 V, 50 Hz, AC source is connected to an inductance of 0.2 H and a resistance of
20 Ω in series. What is the current in the circuit?
XL=ωL= (2πf) L=20πΩ
√𝑋𝐿2 + 𝑅 2
= 65.94Ω
𝑉
220𝑉
Ic = 𝑋 = 65.94 = 3.33𝐴
5. Three loads, each of resistance 30, are connected in star to a 415 V, 3-phase supply.
Determine
(a) The system phase voltage, (b) the phase current and (c) the line current.
A ‘415 V, 3-phase supply’ means that 415 V is the line voltage, VL
(a) For a star connection, VL =√3Vp Hence phase voltage, Vp = VL/√3
= 415 /√3
= 239.6 V or 240 V
(b) Phase current, Ip = Vp/Rp
= 240/30
=8A
6. Three identical coils, each of resistance 10ohm and inductance 42mH are connected (a) in
star and (b) in delta to a 415V, 50 Hz, 3-phase supply. Determine the total power
dissipated in each case.
Solution:
(a) Star connection
Inductive reactance,
XL =2πf L =2π (50) (42×10−3) =13.19
Phase impedance,
Zp =√ (R2 +XL2)
=√ (102 +13.192) =16.55
Line voltage, VL =415 V
And phase voltage,
VP =VL/√3=415/√3=240 V.
Phase current,
Ip =Vp/Zp =240/16.55=14.50 A. Line current,
IL =Ip =14.50 A.
Power factor= cos φ=Rp/Zp =10/16.55 =0.6042 lagging.
Power dissipated,
P =√3 VLIL cos φ =√3 (415) (14.50) (0.6042) = 6.3kW (Alternatively,
P =3I2R =3(14.50)2(10) =6.3kW)
(b) Delta connection
VL = Vp = 415 V,
Zp = 16.55_, cos φ = 0.6042 lagging (from above). Phase current,
Ip =Vp/Zp =415/16.55=25.08A. Line current,
IL =√3Ip =√3 (25.08) =43.44A.
Power dissipated,
P =√3 VLIL cos φ
=√3 (415) (43.44) (0.6042) = 18.87kW
(Alternatively,
P =3I2R
=3(25.08)2(10) =18.87 kW)
7. A 415V, 3-phase a.c. motor has a power output of 12.75kW and operates at a power
factor of 0.77 lagging and with an efficiency of 85 per cent. If the motor is deltaconnected, determine (a) the power input, (b) the line current and (c) the phase current.
(a) Efficiency=power output/power input.
(85/100)=12.750 power input from which, Power input = 12. 750 × 10085
= 15 000W or 15Kw
(b) Power, P=√3 VLIL cos φ, hence
(c) Line current,
IL = P/ √3 (415) (0.77)
= 15 000/ √3 (415) (0.77)
= 27.10A
(d) For a delta connection, IL =√3 Ip,
Phase current, Ip = IL/√3
= 27.10 /√3
= 15.65A
8. In a 3-phase, 3-wire system with star-connected load the impedance of each phase is(3 +j4)Ω. If
the line voltage is 230 V, calculate the line current.
Zph = √Rph2+Xph2= √32+42= 5 ohms
IL = Iph= EphZph= 2305√3
= 26.56 A
9. A balanced delta connected load having an impedance of 20 - j15 ohm is connected to a delta
connected, positive-sequence generator having Vab= 330∠0 V. Calculate the phase currents of the
load.
ZΔ=20−j15=25∠−36.87°
ΩIab = VABZΔ= 330∠025∠−36.87= 13.2∠36.87°
AIbc = Iab∠-120 = 13.2∠-83.13°
AIca = Iab∠120 = 13.2∠156.87°A
10. The input power to a 3-phase a.c. motor is measured as 5kW. If the voltage and current to the
motor are 400V and 8.6A respectively, determine the power factor of the system?
Power
P=5000W,
VL = 400 V,
IL = 8.6A and
P =√3 VLIL cos φ
= cos φ = P √3 VLIL
= 5000 √3 (400) (8.6)
= 0.839
11. Two wattmeters are connected to measure the input power to a balanced 3-phase load by the twowattmeter method. If the instrument readings are 8kW and 4kW, determine (a) the total power
input and (b) the load power factor.
(a)Total input power,
P=P1 +P2 =8+4=12kW
(b) tan φ =√3(P1 − P2)/(P1 + P2)
=√3 (8 – 4) / (8 + 4)
=√3 (4/12)
=√3(1/3)
= 1/ √3
φ= tan−1 1 √3
=30◦
cos φ= cos 30◦ =0.866
12. Two wattmeters connected to a 3-phase motor indicate the total power input to be 12kW. The
power factor is 0.6. Determine the readings of each wattmeter.
P1 + P2 =12kW
(1)
tan φ =√3(P1 − P2)/(P1 + P2)
And power factor=0.6= cos φ.
Angle φ= cos−10.6=53.13◦,
tan 53.13◦ =1.3333.
1.3333 =√3(P1 − P2)/12
P1 − P2
= 12(1.3333) /√3
i.e. P1 −P2 =9.237kW (2)
Adding Equations (1) and (2) gives:
2P1 = 21.237
i.e P1 = 21.237/2
= 10.62kW Hence wattmeter 1 reads 10.62kW From Equation (1), wattmeter 2 reads
(12−10.62)=1.38kW
13. Three loads, each of resistance 30, are connected in star to a 415 V, 3-phase supply. Determine
(a) the system phase voltage, (b) the phase current and (c) the line current.
VL =√3Vp Hence phase voltage, Vp = VL/√3
= 415 /√3
= 239.6 V or 240 V
(b) Phase current, Ip = Vp/Rp
= 240/30
=8A
(c) For a star connection,
Ip = IL Hence the line current, IL = 8 A
14. Three identical coils, each of resistance 10ohm and inductance 42mH are connected (a) in star
and (b) in delta to a 415V, 50 Hz, 3-phase supply. Determine the total power dissipated in each
case.
15. (a) Star connection
Inductive reactance,
XL =2πf L =2π (50) (42×10−3) =13.19
Phase impedance,
Zp =√(R2 +XL2)
=√(102 +13.192) =16.55
Line voltage, VL =415 V
And phase voltage,
VP =VL/√3=415/√3=240 V.
Phase current,
Ip =Vp/Zp =240/16.55=14.50 A. Line current,
IL =Ip =14.50 A.
Power factor= cos φ=Rp/Zp =10/16.55 =0.6042 lagging.
Power dissipated,
P =√3 VLIL cos φ =√3 (415) (14.50)(0.6042) = 6.3kW (Alternatively,
P =3I2R =3(14.50)2(10)=6.3kW)
(b) Delta connection
VL = Vp = 415 V,
Zp = 16.55_, cos φ = 0.6042 lagging (from above). Phase current,
Ip =Vp/Zp =415/16.55=25.08A. Line current,
IL =√3Ip =√3(25.08)=43.44A.
Power dissipated,
P =√3 VLIL cos φ
=√3 (415)(43.44)(0.6042) = 18.87kW
(Alternatively,
P =3I2R
=3(25.08)2(10) =18.87 kW)
16. A 415V, 3-phase a.c. motor has a power output of 12.75kW and operates at a power factor of
0.77 lagging and with an efficiency of 85 per cent. If the motor is delta-connected, determine (a)
the power input, (b) the line current and (c) the phase current.
(a) (85/100)=12.750 power input from which,
= 12. 750 × 10085
= 15 000W or 15Kw
(b) P=√3 VLIL cos φ,
(c) IL = P/ √3 (415) (0.77)
= 15 000/ √3 (415) (0.77)
= 27.10A
(d) For a delta connection, IL =√3 Ip,
Ip = IL/√3
= 27.10 /√3
= 15.65A
17. A 400V, 3-phase star connected alternator supplies a delta-connected load, each phase of which
has a resistance of 30_ and inductive reactance 40_. Calculate (a) the current supplied by the
alternator and (b) the output power and the kVA of the alternator, neglecting losses in the line
between the alternator and load.
(a) Considering the load
Phase current, Ip =Vp/Zp
Vp =VL for a delta connection,
Vp =400V.
Phase impedance,
Zp =√ (R2+XL2)
=√ (302 +402) =50
For a delta-connection,
Line current, IL =√3 Ip =√3 (8) =13.86 A.
=13.86A is the current supplied by the alternator.
(b) Alternator output power is equal to the power Dissipated by the load
P =√3 (400) (13.86) (0.6) = 5.76kW.
Alternator output kVA,
S =√3 VLIL =√3 (400) (13.86)
9.60 kVA.
18. In an A.C circuit, the resistance R=0.2Ω. At a certain instant, VA−VB=0.5V, I=0.5A, and the
current is increasing at the rate of ΔtΔI=8A/s. The inductance of the coil is:
19. A series RLC circuit is made as shown in the figure with an AC source of 60 V, 20 Hz.
20. An AC source is connected in parallel with an L-C-R circuit as shown. Let IS, IL, ICand IR
denote the currents through and VS, VL, VCand VRvoltages across the corresponding
components.
21. A series combination of resistor (R) and capacitor (C) is connected to an A.C. source of angular
frequency 'ω'. Keeping the voltage same, if the frequency is changed to ω/3, the current becomes
half of the original current. Then the ratio of the capacitive reactance and resistance at the former
frequency is:
22. A three-phase line has an impedance of 0.1 + j0.8 Ω/ϕ. The line feeds two balanced three-phase
loads connected in parallel. The first load is absorbing a total of 630 kW and absorbing 840
kVAR magnetizing vars. The second load is Y-connected and has an impedance of 15.36 - j4.48
Ω/ϕ. The line-to-neutral voltage at the load end of the line is 4000 V. What is the magnitude of
the line voltage at the source end of the line?
23. The impedance Z in the balanced three-phase circuit is 100 - j75 Ω. Find a) IAB, IBC, and ICA,
b) IaA, IbB, and IcC, c) Iba, Icb, and Iac.
24. A 12-phase star-connected transformer bank supplies power to a balanced load rated 10 MW,
at 4160 V at 85% lagging. Calculate the current in each line.
𝑃 = 𝑛𝑉𝑝 𝐼𝑝 cos 𝜃
𝑉
𝐿
𝑃 = 𝑛(2 sin(180°
)𝐼𝐿 cos 𝜃
10𝑥106 = 12 (
4160
2 sin(180°
)(𝐼𝐿 )(0.85)
12
𝐼𝐿 = 122 𝐴
25. Three identical impedances of 8 + j6 ohms are connected in delta across a balanced 208 V, 3phase, 60 Hz system. Determine the line current drawn.
𝑍 = 8 + 𝑗6 = 10 < −36.86° 𝑜ℎ𝑚𝑠
𝑉
208
𝐼𝐿 = √3𝐼𝑝 = √3 ( 𝑍𝐿 ) = √3 ( 10 ) = 36.03 𝐴
26. A balanced three-phased load is wye-connected and has an impedance Zp =4-j3 ohms per
phase. Find the line current if this load is connected across a 220 V three phase source.
𝑍 = 4 − 𝑗3 = 5 < −36.869°
𝐼𝐿 = 𝐼𝑃 =
𝑉𝑃
√3𝑍
=
220
√3(5)
= 25.4 𝐴
27. The primary of a 34.5 Kv/13.8 Kv, 50MVA transformer is wye-grounded. The three primary
currents are: Ia= 720 cis -30 degrees; Ib= 695 cis 209 degrees; Ic= cis 88 degrees. Determine
the neutral current.
𝐼𝑁 = 𝐼𝐴 + 𝐼𝐵 + 𝐼𝐶
𝐼𝑁 = 720 < −30° + 695 < 209° + 710 < 88°
𝐼𝑁 = 42.363 < 17.253° 𝐴
28. A generator supplies three-phase power to balanced load. The voltage is 230 volts, the current
is 18 A and the power factor is 85%. What is the power?
𝑃 = √3 𝑉𝐿 𝐼𝐿 cos 𝜃 = √3(230)(18)(0.85) = 6,095 𝑊
29. Three resistor 10, 15 and 30 ohmic values are connected in wye-configuration to a balanced
208-volt three phase supply. Calculate the total power of the system.
𝑅𝐴 =
𝑅𝑏 =
𝑅𝑐 =
15(30)+(30)(10)+(10)(15)
=
10
15(30)+(30)(10)+(10)(15)
=
15
15(30)+(30)(10)+(10)(15)
=
30
900
=
10
900
=
15
900
=
30
90 𝑂ℎ𝑚𝑠
60 𝑂ℎ𝑚𝑠
30 𝑂ℎ𝑚𝑠
𝑃=
2082
90
+
2082
60
+
2082
30
= 2644 𝑊
30. The phase b voltage and the phase b current of a balance three-phase system are V= 220 sin
(wt=210°) and I= 10 sin (wt-180°)
𝑃 = 3𝑉𝑃 𝐼𝑃 cos 𝜃
𝑃 = 3(220)(10)(cos 30°)
𝑃 = 5715.76 𝑊