2.1 A transformer is made up of a 1200-turn primary coil and an open-circuited 75-turn
secondary coil wound around a closed core of cross-sectional area 42 cm2. The
core material can be considered to saturate when the rms applied flux density
reaches 1.45 T. What maximum 60-Hz rms primary voltage is possible without
reaching this saturation level? What is the corresponding secondary voltage?
How are these values modified if the applied frequency is lowered to 50 Hz?
Ans
1
1.50
E max = ωNφ max = 2πfNAc Bmax
E max, primary = 2πfNAc Bmax
= 2π (60)(1200)(42 × 10 −4 )(1.45)
= 2755.12 V
E max, primary
⇒ E rms , primary =
= 1948.16 V
2
E max,sec ondary = 2πfNAc Bmax
= 2π (60)(75)(42 × 10 −4 )(1.45)
= 172.19 V
E max,sec ondary
⇒ E rms ,sec ondary =
= 121.76 V
2
2
f = 60Hz → 50Hz
E max = ωNφ max = 2πfNAc Bmax
1.50
E max, primary = 2πfNAc Bmax
= 2π (50)(1200)(42 × 10 − 4 )(1.45)
= 2294.71 V
E max, primary
⇒ E rms , primary =
= 1622.61 V
2
E max,sec ondary = 2πfNAc Bmax
= 2π (50)(75)(42 × 10 −4 )(1.45)
= 143.42 V
E max,sec ondary
⇒ E rms ,sec ondary =
= 101.41 V
2
2.2 A magnetic circuit with a cross-sectional area of 15 cm2 is to be operated at 60 Hz
from a 120-V rms supply. Calculate the number of turns required to achieve a
peak magnetic flux density of 1.8 T in the core.
Ans
1.50
E max = ωNφ max = 2πfNAc Bmax
120 2 = 2π (60)N (15 × 10 −4 )(1.8)
⇒N=
120 2
= 167 turns
2π (60 )(15 × 10 − 4 )(1.8)
2.3 A transformer is to be used to transform the impedance of a 8- Ω resistor to an
impedance of 75 Ω . Calculate the required turns ratio, assuming the transformer to
be ideal.
Ans:
⎛N
Z1 = ⎜⎜ 1
⎝ N2
2
⎞
⎟⎟ Z 2
⎠
75
= 3.061
8
= 3 turns
:N =
2.4 A 100- Ω resister is connected to the secondary of an idea transformer with a turns
ratio of 1:4 (primary to secondary). A 10-V rms , 1-kHz voltage source is connected
to the primary . Calculate the primary current and the voltage across the 100- Ω
resister.
Ans:
10-V rms ,1-k Hz
1:4
100
⎛N
R1 = ⎜⎜ 1
⎝ N2
2
⎞
⎟⎟ R2
⎠
2
⎛1⎞
= ⎜ ⎟ × 100 = 6.25Ω
⎝4⎠
I1 =
V1
10
=
= 1.6 A
R1 6.25
⎛N ⎞
⎛4⎞
V2 = ⎜⎜ 2 ⎟⎟V1 = ⎜ ⎟ × 10 = 40V
⎝1⎠
⎝ N1 ⎠
2.7 A single-phase 60Hz transformer has a nameplate voltage rating of 7.97kv:266v ,
which is based on it winding turns ratio. The manufacturer calculates that the
primary (7.97kv) leakage inductance is 165mH and the primary magnetizing
inductance is 135H. For an applied primary voltage of 7970v at 60Hz,calculate the
resultant open-circuit secondary voltage.
Ans:
Ll1 =165mH
7.97kv : 266v
+
Vp=7970 V
Voc
Lm=135H
-
Voc = (
135 × 7970
266
)×(
) = 265.67V
135 + 0.165
7970
2.8 The manufacturer calculates that the transformer of Problem 2.7 has a secondary
leakage inductance of 0.225mH.
a. Calculate the magnetizing inductance as referred to the secondary side.
b. A voltage of 266v,60Hz is applied to the secondary. Calculate
(i) the resultant open-circuit primary voltage and
(ii) the secondary current which would result if the primary were short-circuited.
Ans:
part(a): Referred to the secondary
Lm ,2 = Lm ,12
N1
(
)
N2
135
=
(
7970 2
)
266
= 150.38mH
part(b): Referred to the secondary
Xm = ωLm , 2 = 2πf × 150.38mH = 56.7Ω
Xl 2 = ωLl 2 = 2πf × 0.225mH = 84.82mΩ
Xl1 = ωLl1 = 2πf × 165mH = 62.2mΩ
N1
Xm
(
)V 2
N 2 Xm + Xl 2
7970
56.7
=
×
× 265.67 = 7948 V
266 56.7 + 0.0848
(i ) V 1 =
7.97kv : 266v
Ll2 =0.225mH
V1
AC 265.67V
Lm=150.38mH
(ii ) Isc =
V2
V2
=(
)
Xsc
Xl 2 + Xl1 // Xm
265.5
=
= 1807.35 A
0.0848 + (56.7 // 0.0622)
Xl2 =84.82m
Xl1 // Xm
AC 265.67V
Xm=56.7