UNIT 3
Transformers
Q1>. A 25 KVA transformer has 500 turns on primary and 50 on secondary. The primary is connected to 2000V, 50 Hz supply. Find full load primary and secondary currents, the secondary emf and max flux in core. Neglect leakage drop and no load primary current.
Sol : K = N2/N1 = 50/500 = 1/10
I1 = 25,000/2000 = 12.5 A
I2 = I1/K = 10 x 12.5 A = 125 A
Emf/turn on primary side = 2000/500 = 4 V
E2 = KE1
E2 = 4 x 50 = 200 v
E1 = 4.44fN1 Φm
2000 = 4.44 x 50 x 500 x Φm
Φm = 18.02 m Wb
Q2>. A 2500/200 V transformer draws a no-load primary current of 0.5 A and absorbs 400 W. Find magnetising and loss currents.
Sol: Iron-loss current = No load input(W) / Primary voltage
= 400/2500 = 0.16 A
I20 = I2w + I2µ
Iµ = √I20 – I2w
= √ (0.5)2 – (0.16)2
Iµ = 0.473 A
Q3>. A 1- φ transformer with ratio of 440/110-V takes a no-load current of 6 A at 0.3 pf lagging. If secondary supplies 120 A at pf of 0.8 lagging. Find current taken by primary.
Sol>.
Cos φ2 = 0.8
Φ2 = 36.540
Cos φ0 = 0.3
Φ0 = 72.540
K = V2/V1 = 110/440 = ¼
I’2 = KI2 = 120 x ¼ = 30 A
I0 = 6A
Angle between I0 & I’2
= 72.54 – 36.54
= 35.670
From vector diagram,
I1 = √(62 + 302 + 2 x 6 x 30 cos 35.67)
I1 = 35.05 A
Q.4) The core of a 110 KVA, 10,000/500v, 50 Hz, 1-Φ core type transformer has a cross section of 18 cm x 18 cm. Find the number of HV and LV turns per phase and the emf per turn if the maximum core density does not exceed 1.3 tesla. Assume a stacking factor pf 0.9.
Sol: Bm= 1.3T
Area = (0.18 x 0.18) = 0.032m2
Emf induced in primary
E1 = 4.44 fN1BmA
10000=4.44 x 50 x N1 x 1.3 x 0.032
N1=1082.8
Emf induced in secondary
E2 = 4.44 fN2BmA
500= 4.44 x 50 x N2 x 1.3 x 0.032
N2= 54.14
i) The number of turns is N1=1082.8 and N2= 54.14
Ii) Emf per turn = E1/E2 = N1/N2 = K
=10000/1082=9.24V or 500/54.14=9.23V
Q.5) A 1-Φ transformer has 400 turns in primary and 110 turns in the secondary. The cross-sectional area of the secondary. The cross-sectional area of the core is 80cm2. If the primary winding is connected to the 50 Hz supply at 500V. Calculate peak flux density in core.
Sol: As we know Emf induced in primary
E1 = 4.44 fN1BmA
500 = 4.44 x 50 x 400 x Bm x (80x10‑4)
Bm=0.704Wb/m2
Q.6) A single phase transformer has 400 turns in primary and 1000 turns in secondary. The cross-sectional area is 80 cm2. If primary is connected to 50hz at 500V. Voltage induced in secondary?
Sol: As we know Emf induced in primary
E1 = 4.44 fN1BmA
500= 4.44x50x1000xBmx(80x10-4)
Bm= 0.28Wb/m2
The voltage induced in secondary is given as
E1/E2 = N1/N2 = K
E2=1000x500/400=1250V
Q7. A – 100 KVA transformer has 500 turns on primary and 80 turns on secondary. The primary and secondary resistances are 0.3 and 0.01 Ω respectively and the corresponding leakage reactances are 1.1 and 0.035 Ω. The supply voltage is 2400 V. Find
(i). Equivalent impedance referred to primary
(ii). Voltage regulation and the secondary terminal voltage for full load having pf 0.8 lagging?
Sol. Equivalent impedance referred to primary
Z01 = √R201 + X201 = R01 + jX01
R01 = R1 + R2/K2 = 0.3 + 0.01/K2 = 0.69 Ω
K = 80/500 = 4/25
X01 = X1 + X2/K2 = 1.1 + 0.035/(0.16)2 = 2.467 Ω
Z01 = 0.69 + j2.46
(ii). Secondary terminal voltage Z02 = K2 Z01
Z02 = 0.018 + j 0.063
= 0.065 ( 74.050
No-load secondary voltage = KV1
= 0.16 x 2400 = 384 V
I2 = 100 x 103/384 = 260.42 A
Full load voltage drop referred to secondary
= I2 (R02 cosφ – X02 Sinφ)
Cosφ = 0.8
Φ = 36.860
Sinφ = 0.6
= 260.42(0.018 x 0.8 – 0.063 x 0.6)
= - 6.094 V
% regn = -6.094/384 x 100
= -1.587
Secondary terminal voltage on-load
= 384 – (-6.094)
= 390.09 V
Q8). In a 50 KVA, 2200/200 V, 1-φ transformer, the iron and full-load copper losses are 400 W and 450 W respectively. Calculate n at unity power factor on (i). Full load (ii). Half-full load?
Sol. (i). Total loss = 400 + 450 = 850 W
F.L output at unity power factor = 50 x 1
= 50 KVA
n = 50 / 50 + .850 = 50/50.850 = 0.98 = 98%
(ii). Half full load, unity pf
= 50 KVA/2 = 25 KVA
Cu loss = 400 x (1/2)2 = 100 W
Iron loss is same = 450 W
Total loss = 100 + 450 = 550 W
n = 25/25 + 0.55 = 25/25.55 = 0.978 = 97.8 %
Q9. A 40 KVA 440/220 V, 1- φ, 50 Hz transformer has iron loss of 300 W. The cu loss is found to be 100 W when delivering half full-load current. Determine (i) n when delivering full load current at 0.8 lagging pf (ii) the percentage of full-load when the efficiency will be max.
Sol. Full load efficiency at 0.8 pf
= 40 x 0.8/(40 x 0.8) + losses
Full load cu loss = (440/220)2 x 100
= 400 W
Iron loss = 400 + 300
= 700 W
n = 40 x 0.8/(40 x 0.8) + 0.7 = 97.8 %
(ii). KVA for maximum / F.L KVA = √ iron loss / F.L cu loss
= √300/400 = 0.866
Q10. An auto transformer suppliers load of 4KW at 100v at unity pf. IF the applied primary voltage is 220v. Calculating power transferred to load (a) Inductively (b) conductively.
Soln, Power transferred inductively = Input(1-k)
Power transferred conductively = K* Input
K= 
=
Input= Output =4KW
Inductively transferred power =4(
)
=3.82KW
Conductively transferred power = 
*4
= 0.182 kW
Q11 A 3-Ø, 50Hz transformer has delta connected primary and star-connected secondary, the line voltage being 20,000 V and 500 V respectively. The secondary has a star connected balanced load at 0.8 lagging p.f. The line current of primary side is 5A. Determine the current in each coil of the primary and secondary line. Also find the output of the transformer?
→Phase voltage on primary side = 20,000 V
Phase voltage on secondary = 
V
K = 
Primary phase current = 
Secondary phase current = 
= 
= 200 A
Output = 
VLILcosØ
= 
= 138.56 kW
Que 12. A 100 kVA, 3Ø, 50 Hz , 3500/500 V transformer is Δ – connected on the h. V side and Y-connected on the L.V. Side. The resistance of h.V. Winding is 3.5 Ω / phase and that of L.V is 0.02 Ω per phase. Calculate the iron losses of the transformer at normal voltage and frequency if its full load efficiently be 95.1 % at 0.8 p.f. ?
→Full load output = 100 × 0.8 = 80 kW
Input = 
kW
Total loss = Input – Output
= 83.2 – 80
= 3200 kW
K = 
K = 
= 
RO2 = R2 + k2R1
= 
= 0.044 Ωs
Full-load secondary phase current I2
= 
= 115.4 A
Total Cu Loss = 3I22RO2
= 3 × 0.044 × (115.4)2
= 1760 W
Iron Loss = Total loss – Cu Loss
= 3200 – 1760
= 1440 W
UNIT 3
Transformers
Q1>. A 25 KVA transformer has 500 turns on primary and 50 on secondary. The primary is connected to 2000V, 50 Hz supply. Find full load primary and secondary currents, the secondary emf and max flux in core. Neglect leakage drop and no load primary current.
Sol : K = N2/N1 = 50/500 = 1/10
I1 = 25,000/2000 = 12.5 A
I2 = I1/K = 10 x 12.5 A = 125 A
Emf/turn on primary side = 2000/500 = 4 V
E2 = KE1
E2 = 4 x 50 = 200 v
E1 = 4.44fN1 Φm
2000 = 4.44 x 50 x 500 x Φm
Φm = 18.02 m Wb
Q2>. A 2500/200 V transformer draws a no-load primary current of 0.5 A and absorbs 400 W. Find magnetising and loss currents.
Sol: Iron-loss current = No load input(W) / Primary voltage
= 400/2500 = 0.16 A
I20 = I2w + I2µ
Iµ = √I20 – I2w
= √ (0.5)2 – (0.16)2
Iµ = 0.473 A
Q3>. A 1- φ transformer with ratio of 440/110-V takes a no-load current of 6 A at 0.3 pf lagging. If secondary supplies 120 A at pf of 0.8 lagging. Find current taken by primary.
Sol>.
Cos φ2 = 0.8
Φ2 = 36.540
Cos φ0 = 0.3
Φ0 = 72.540
K = V2/V1 = 110/440 = ¼
I’2 = KI2 = 120 x ¼ = 30 A
I0 = 6A
Angle between I0 & I’2
= 72.54 – 36.54
= 35.670
From vector diagram,
I1 = √(62 + 302 + 2 x 6 x 30 cos 35.67)
I1 = 35.05 A
Q.4) The core of a 110 KVA, 10,000/500v, 50 Hz, 1-Φ core type transformer has a cross section of 18 cm x 18 cm. Find the number of HV and LV turns per phase and the emf per turn if the maximum core density does not exceed 1.3 tesla. Assume a stacking factor pf 0.9.
Sol: Bm= 1.3T
Area = (0.18 x 0.18) = 0.032m2
Emf induced in primary
E1 = 4.44 fN1BmA
10000=4.44 x 50 x N1 x 1.3 x 0.032
N1=1082.8
Emf induced in secondary
E2 = 4.44 fN2BmA
500= 4.44 x 50 x N2 x 1.3 x 0.032
N2= 54.14
i) The number of turns is N1=1082.8 and N2= 54.14
Ii) Emf per turn = E1/E2 = N1/N2 = K
=10000/1082=9.24V or 500/54.14=9.23V
Q.5) A 1-Φ transformer has 400 turns in primary and 110 turns in the secondary. The cross-sectional area of the secondary. The cross-sectional area of the core is 80cm2. If the primary winding is connected to the 50 Hz supply at 500V. Calculate peak flux density in core.
Sol: As we know Emf induced in primary
E1 = 4.44 fN1BmA
500 = 4.44 x 50 x 400 x Bm x (80x10‑4)
Bm=0.704Wb/m2
Q.6) A single phase transformer has 400 turns in primary and 1000 turns in secondary. The cross-sectional area is 80 cm2. If primary is connected to 50hz at 500V. Voltage induced in secondary?
Sol: As we know Emf induced in primary
E1 = 4.44 fN1BmA
500= 4.44x50x1000xBmx(80x10-4)
Bm= 0.28Wb/m2
The voltage induced in secondary is given as
E1/E2 = N1/N2 = K
E2=1000x500/400=1250V
Q7. A – 100 KVA transformer has 500 turns on primary and 80 turns on secondary. The primary and secondary resistances are 0.3 and 0.01 Ω respectively and the corresponding leakage reactances are 1.1 and 0.035 Ω. The supply voltage is 2400 V. Find
(i). Equivalent impedance referred to primary
(ii). Voltage regulation and the secondary terminal voltage for full load having pf 0.8 lagging?
Sol. Equivalent impedance referred to primary
Z01 = √R201 + X201 = R01 + jX01
R01 = R1 + R2/K2 = 0.3 + 0.01/K2 = 0.69 Ω
K = 80/500 = 4/25
X01 = X1 + X2/K2 = 1.1 + 0.035/(0.16)2 = 2.467 Ω
Z01 = 0.69 + j2.46
(ii). Secondary terminal voltage Z02 = K2 Z01
Z02 = 0.018 + j 0.063
= 0.065 ( 74.050
No-load secondary voltage = KV1
= 0.16 x 2400 = 384 V
I2 = 100 x 103/384 = 260.42 A
Full load voltage drop referred to secondary
= I2 (R02 cosφ – X02 Sinφ)
Cosφ = 0.8
Φ = 36.860
Sinφ = 0.6
= 260.42(0.018 x 0.8 – 0.063 x 0.6)
= - 6.094 V
% regn = -6.094/384 x 100
= -1.587
Secondary terminal voltage on-load
= 384 – (-6.094)
= 390.09 V
Q8). In a 50 KVA, 2200/200 V, 1-φ transformer, the iron and full-load copper losses are 400 W and 450 W respectively. Calculate n at unity power factor on (i). Full load (ii). Half-full load?
Sol. (i). Total loss = 400 + 450 = 850 W
F.L output at unity power factor = 50 x 1
= 50 KVA
n = 50 / 50 + .850 = 50/50.850 = 0.98 = 98%
(ii). Half full load, unity pf
= 50 KVA/2 = 25 KVA
Cu loss = 400 x (1/2)2 = 100 W
Iron loss is same = 450 W
Total loss = 100 + 450 = 550 W
n = 25/25 + 0.55 = 25/25.55 = 0.978 = 97.8 %
Q9. A 40 KVA 440/220 V, 1- φ, 50 Hz transformer has iron loss of 300 W. The cu loss is found to be 100 W when delivering half full-load current. Determine (i) n when delivering full load current at 0.8 lagging pf (ii) the percentage of full-load when the efficiency will be max.
Sol. Full load efficiency at 0.8 pf
= 40 x 0.8/(40 x 0.8) + losses
Full load cu loss = (440/220)2 x 100
= 400 W
Iron loss = 400 + 300
= 700 W
n = 40 x 0.8/(40 x 0.8) + 0.7 = 97.8 %
(ii). KVA for maximum / F.L KVA = √ iron loss / F.L cu loss
= √300/400 = 0.866
Q10. An auto transformer suppliers load of 4KW at 100v at unity pf. IF the applied primary voltage is 220v. Calculating power transferred to load (a) Inductively (b) conductively.
Soln, Power transferred inductively = Input(1-k)
Power transferred conductively = K* Input
K= 
=
Input= Output =4KW
Inductively transferred power =4(
)
=3.82KW
Conductively transferred power = 
*4
= 0.182 kW
Q11 A 3-Ø, 50Hz transformer has delta connected primary and star-connected secondary, the line voltage being 20,000 V and 500 V respectively. The secondary has a star connected balanced load at 0.8 lagging p.f. The line current of primary side is 5A. Determine the current in each coil of the primary and secondary line. Also find the output of the transformer?
→Phase voltage on primary side = 20,000 V
Phase voltage on secondary = 
V
K = 
Primary phase current = 
Secondary phase current = 
= 
= 200 A
Output = 
VLILcosØ
= 
= 138.56 kW
Que 12. A 100 kVA, 3Ø, 50 Hz , 3500/500 V transformer is Δ – connected on the h. V side and Y-connected on the L.V. Side. The resistance of h.V. Winding is 3.5 Ω / phase and that of L.V is 0.02 Ω per phase. Calculate the iron losses of the transformer at normal voltage and frequency if its full load efficiently be 95.1 % at 0.8 p.f. ?
→Full load output = 100 × 0.8 = 80 kW
Input = 
kW
Total loss = Input – Output
= 83.2 – 80
= 3200 kW
K = 
K = 
= 
RO2 = R2 + k2R1
= 
= 0.044 Ωs
Full-load secondary phase current I2
= 
= 115.4 A
Total Cu Loss = 3I22RO2
= 3 × 0.044 × (115.4)2
= 1760 W
Iron Loss = Total loss – Cu Loss
= 3200 – 1760
= 1440 W
