A single-phase, 50-kVA, 2400/240-V, 60-Hz distribution transformer has the following parameters: Resistance of the 2400-V winding: R 1 = 0.75 Ω Resistance of the 240-V winding: R 2 = 0.0075 Ω Leakage reactance of the 2400-V winding: X 1 = 1.0 Ω Leakage reactance of the 240-V winding: X 2 = 0.01 Ω Exciting admittance on the 240-V side = 0.00 3 − j 0.0 2 S (a) Draw the equivalent circuit referred to the high-voltage side of the transformer. (b) Draw the equivalent circuit referred to the low-voltage side of the transformer. Show the numerical values of impedances on the equivalent circuits.
A single-phase, 50-kVA, 2400/240-V, 60-Hz distribution transformer has the following parameters: Resistance of the 2400-V winding: R 1 = 0.75 Ω Resistance of the 240-V winding: R 2 = 0.0075 Ω Leakage reactance of the 2400-V winding: X 1 = 1.0 Ω Leakage reactance of the 240-V winding: X 2 = 0.01 Ω Exciting admittance on the 240-V side = 0.00 3 − j 0.0 2 S (a) Draw the equivalent circuit referred to the high-voltage side of the transformer. (b) Draw the equivalent circuit referred to the low-voltage side of the transformer. Show the numerical values of impedances on the equivalent circuits.
Solution Summary: The author illustrates the circuit referred to the high voltage side of transformer. The value of exciting admittance is given as l
A single-phase,
50-kVA, 2400/240-V, 60-Hz
distribution transformer has the following parameters:
Resistance of the 2400-V winding:
R
1
=
0.75
Ω
Resistance of the 240-V winding:
R
2
=
0.0075
Ω
Leakage reactance of the 2400-V winding:
X
1
=
1.0
Ω
Leakage reactance of the 240-V winding:
X
2
=
0.01
Ω
Exciting admittance on the 240-V side
=
0.00
3
−
j
0.0
2 S
(a) Draw the equivalent circuit referred to the high-voltage side of the transformer.
(b) Draw the equivalent circuit referred to the low-voltage side of the transformer. Show the numerical values of impedances on the equivalent circuits.
Calculate the current magnitude in the coils e1, e2 of theMagnetic circuit, if:ɸa = 3.00 x 10^-3 Wb, φb = 0.80 x 10^-3 Wb, ɸc = 2.20 x 10^-3 Wb
L ab = 0.10 m,A ab = 5.0 cm^2L afeb = L acdb = 0.40 mA afeb = A acdb = 20 cm^2
MATERIAL CHARACTERISTICSH (At/m) 240 350 530 1300 5000 9000B (T) 0.7 0.9 1.1 1.3 1.5 1.6
A toroid magnetic circuit is composed of three sections A, B and C, thesection C has an air gap, section A has an 850 round coil thatconsumes a current of 1.2 A. the physical and magnetic properties of each sectionare:
Section A: Length = 80 mm, Cross section = 120 mm^2, μr = 400
Section B: Length = 60 mm, Cross section = 40 mm^2, μr = 250
Section C: Length = 50 mm, Cross section = 200 mm^2, μr = 600
Gap: Length = 1 mm, Cross section = 40 mm^2, μr = 1
Calculate:The magnetic field density in each of the sections
3) Compute the input impedance of Fig. 3. (10 points)
Rin
R1
R₂
Figure 3
T
Vcc
Chapter 3 Solutions
Power System Analysis and Design (MindTap Course List)
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