Massachusetts Institute of Technology
Department of Electrical Engineering and Computer Science
6.061/6.690 Introduction to Power Systems
Quiz 1 Solutions
March 16, 2011
Statistics: Quiz Mean was 58
Standard Deviation was 23
Median Grade was 60
Problem 1: You will recognize this as a delta-wye connected transformer with a +30◦ phase shift
from primary to secondary. Thus the voltages on the secondary side will be:
VA =
VB =
√
√
π
3N V ej 6
π
3N V e−j 2
The current in the resistor is:
π
3N V ej 3
VA − VB
IR =
=
R
R
Then currents on the primary of the transformer are:
Ia =
Ib =
Ic =
3N 2 V j π
e 3
R
6N 2 V −j 2π
e 3
R
3N 2 V j π
e 3
R
Noting that:
3N 2 V
3 × 100 × 100
=
=1
R
30, 000
Then:
π
Ia = 1ej 3
Ib = 2e−j
2π
3
π
Ic = 1ej 3
The currents are sketched on the template. Note that the units are not related to the voltage
units.
1
Ic
Vc
Ia
Va
Vb
Ib
Figure 1: Problem 1 Currents
To find the real and reactive power in each phase, multiply by voltage: Note that:
√
1
π
−j π3
e
= −j
2
2
(P + jQ)A = 100 ×
(P + jQ)B
(P + jQ)C
�
√ �
√
1
3
= 50 − j50 3 ≈ 50 − j86.6
−j
2
2
= 100 × 2 = 200
�
√ �
√
1
3
= 100 ×
+j
50 + j50 3 ≈ 50 + j86.6
2
2
Problem 2: Here, we can fall back on the simple relationships for the transmission line. If the
voltage magnitudes are the same at both ends of the line. At the sending end:
Ps =
Qs =
V2
sin δ
X
V2
(1 − cos δ)
X
And at the receiving end it is:
Pr =
V2
sin δ
X
2
Qr = −
Since δ =
π
6
V 2
(1 − cos δ)
X
= 30◦ , sin δ =
1
2
and cos δ =
6
(P + jQ)left = 500, 000 + j10
√
3
2 .
�
And
V2
X
= 106 . Then:
√ �
3
1−
≈ 500, 000 + j133, 975
2
(P + jQ)right = −500, 000 + j133, 975
To correct the power factor, we need
�
√ �
3
V2
6
= 1 × 10 1 −
Q=
XC
2
Or, in this case:
Xc =
1
1−
√
3
2
≈ 7.464Ω
Problem 3: The pulse launches a mode with I+ = VZ+0 and Vs = V+ + RI+ , or V+ = V2s . That
propagates until it hits the shorted right-hand end of the line and is inverted. When that pulse
gets back to the (matched) sending end, it generates no reflections. The resulting voltage in
the middle of the line is shown in Figure 2
Vo
1 kV
t
500 µ s
100 µ s
1500 µs
Figure 2: Problem 3 Solution
3
MIT OpenCourseWare
http://ocw.mit.edu
6.061 / 6.690 Introduction to Electric Power Systems
Spring 2011
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.