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.
Question 2
A transistor is used as a switch and the waveforms are shown in Figure 2. The parameters are
Vcc = 225 V, VBE(sat) = 3 V, IB = 8 A, VCE(sat) = 2 V, Ics = 90 A, td = 0.5 µs, tr = 1 µs, ts = 3 µs, tƒ
= 2 μs, and f
10 kHz. The duty cycle is k 50%. The collector- emitter leakage current is
ICEO = 2 mA. Determine the power loss due to the collector current:
=
=
=
(a) during turn-on ton = td + tr
VCE
Vcc
(b) during conduction period tn
V CE(sat)
0
toff"
ton
Ics
0.9 Ics
(c) during turn-off toff = ts + tf
(d) during off-time tot
(e) the total average power losses PT
ICEO
0
IBS
0
Figure 2
V BE(sat)
0
主
*
td
tr
In
Is
If
to
iB
VBE
T= 1/fs
Question 1:
The beta (B) of the bipolar transistor shown in Figure 1 varies from 12 to 60. The load resistance
is Rc = 5. The dc supply voltage is VCC = 40 V and the input voltage to the base circuit is
VB = 5 V. If VCE(sat) = 1.2 V, VBE(sat) = 1.6 V, and RB = 0.8 2, calculate:
(a) the overdrive factor ODF.
(b) the forced ẞ
(c) the power loss in the transistor PT.
IB
VB
RB
+
V BE
RC
Vcc'
Ic
+
IE
Figure 1
VCE
I need help in creating a matlab code to find the currents
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