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From Wikipedia, the free encyclopedia
Personal LaTeX testing area. Please do not edit.
P
l
o
s
s
=
I
a
2
∗
R
a
=
(
1.5
)
2
∗
R
a
{\displaystyle P_{loss}=I_{a}^{2}*R_{a}=(1.5)^{2}*R_{a}}
V
p
=
V
L
=
232
V
{\displaystyle V_{p}=V_{L}=232V}
I
p
=
I
L
3
{\displaystyle I_{p}={\frac {I_{L}}{\sqrt {3}}}}
I
p
=
0.62
A
3
=
0.357
A
{\displaystyle I_{p}={\frac {0.62A}{\sqrt {3}}}=0.357A}
P
p
=
P
3
p
3
=
5.83
3
=
1.94
W
{\displaystyle P_{p}={\frac {P_{3p}}{3}}={\frac {5.83}{3}}=1.94W}
R
c
=
V
o
c
2
P
o
c
=
232
2
1.94
=
27.744
k
Ω
{\displaystyle R_{c}={\frac {V_{oc}^{2}}{P_{oc}}}={\frac {232^{2}}{1.94}}=27.744k\Omega }
S
o
c
=
V
I
=
82.82
V
A
{\displaystyle S_{oc}=VI=82.82VA}
Q
=
S
2
−
P
2
=
82.82
2
−
1.94
2
=
82.8
V
A
R
{\displaystyle Q={\sqrt {S^{2}-P^{2}}}={\sqrt {82.82^{2}-1.94^{2}}}=82.8VAR}
X
m
=
V
2
Q
=
232
2
82.8
=
650.05
Ω
{\displaystyle X_{m}={\frac {V^{2}}{Q}}={\frac {232^{2}}{82.8}}=650.05\Omega }
{\displaystyle }
{\displaystyle }
|
Z
e
q
|
=
V
s
c
I
s
c
=
8
2.62
=
3.05
Ω
{\displaystyle |Z_{eq}|={\frac {V_{sc}}{I_{sc}}}={\frac {8}{2.62}}=3.05\Omega }
R
e
q
=
P
s
c
I
s
c
2
=
2.39
2.62
2
=
0.348
Ω
{\displaystyle R_{eq}={\frac {P_{sc}}{I_{sc}^{2}}}={\frac {2.39}{2.62^{2}}}=0.348\Omega }
X
e
q
=
|
Z
e
q
|
2
−
R
e
q
2
=
3.05
2
−
0.348
2
=
3.03
Ω
{\displaystyle X_{eq}={\sqrt {|Z_{e}q|^{2}-R_{eq}^{2}}}={\sqrt {3.05^{2}-0.348^{2}}}=3.03\Omega }
{\displaystyle }
{\displaystyle }
V
p
=
V
L
=
8
V
{\displaystyle V_{p}=V_{L}=8V}
I
p
=
I
L
3
{\displaystyle I_{p}={\frac {I_{L}}{\sqrt {3}}}}
I
p
=
4.54
A
3
=
2.62
A
{\displaystyle I_{p}={\frac {4.54A}{\sqrt {3}}}=2.62A}
P
p
=
P
3
p
3
=
7.15
W
3
=
2.39
W
{\displaystyle P_{p}={\frac {P_{3p}}{3}}={\frac {7.15W}{3}}=2.39W}
V
p
=
V
L
3
{\displaystyle V_{p}={\frac {V_{L}}{\sqrt {3}}}}
V
p
=
8
3
=
4.62
V
{\displaystyle V_{p}={\frac {8}{\sqrt {3}}}=4.62V}
S
3
=
A
3
′
(
A
2
+
A
1
+
A
0
)
+
A
3
A
2
′
A
1
′
A
0
′
=
A
3
⊕
(
A
2
+
A
1
+
A
0
)
{\displaystyle S_{3}=A_{3}'(A_{2}+A_{1}+A_{0})+A_{3}A_{2}'A_{1}'A_{0}'=A_{3}\oplus (A_{2}+A_{1}+A_{0})}
S
2
=
A
2
′
(
A
1
+
A
0
)
+
A
2
A
1
′
A
0
′
=
A
2
⊕
(
A
1
+
A
0
)
{\displaystyle S_{2}=A_{2}'(A_{1}+A_{0})+A_{2}A_{1}'A_{0}'=A_{2}\oplus (A_{1}+A_{0})}
S
1
=
A
1
A
0
′
+
A
1
′
A
0
=
A
1
⊕
A
0
{\displaystyle S_{1}=A_{1}A_{0}'+A_{1}'A_{0}=A_{1}\oplus A_{0}}
S
0
=
A
0
{\displaystyle S_{0}=A_{0}}
D
=
x
⊕
y
⊕
z
{\displaystyle D=x\oplus y\oplus z}
B
=
x
′
y
+
x
′
z
+
y
z
{\displaystyle B=x'y+x'z+yz}
S
=
x
′
y
+
x
y
′
=
x
⊕
y
{\displaystyle S=x'y+xy'=x\oplus y}
B
=
x
′
y
{\displaystyle B=x'y}
S
=
x
′
y
′
z
+
x
′
y
z
′
+
x
y
′
z
′
+
x
y
z
=
z
⊕
(
x
⊕
y
)
{\displaystyle S=x'y'z+x'yz'+xy'z'+xyz=z\oplus (x\oplus y)}
C
=
x
y
+
x
z
+
y
z
{\displaystyle C=xy+xz+yz}
C
=
x
(
y
+
z
)
+
y
z
{\displaystyle C=x(y+z)+yz}
R
p
a
r
a
l
l
e
l
=
E
p
a
r
a
l
l
e
l
r
E
p
a
r
a
l
l
e
l
i
=
n
2
c
o
s
θ
i
−
n
1
c
o
s
θ
t
n
2
c
o
s
θ
i
+
n
1
c
o
s
θ
t
{\displaystyle R_{parallel}={\frac {E_{parallel}^{r}}{E_{parallel}^{i}}}={\frac {n_{2}cos\theta _{i}-n_{1}cos\theta _{t}}{n_{2}cos\theta _{i}+n_{1}cos\theta _{t}}}}
R
p
e
r
p
e
n
d
i
c
u
l
a
r
=
E
p
e
r
p
e
n
d
i
c
u
l
a
r
r
E
p
e
r
p
e
n
d
i
c
u
l
a
r
i
=
n
1
c
o
s
θ
i
−
n
2
c
o
s
θ
t
n
1
c
o
s
θ
i
+
n
2
c
o
s
θ
t
{\displaystyle R_{perpendicular}={\frac {E_{perpendicular}^{r}}{E_{perpendicular}^{i}}}={\frac {n_{1}cos\theta _{i}-n_{2}cos\theta _{t}}{n_{1}cos\theta _{i}+n_{2}cos\theta _{t}}}}
α
=
R
F
E
T
‖
R
2
‖
R
3
R
1
+
R
F
E
T
‖
R
2
‖
R
3
{\displaystyle \alpha ={\frac {R_{FET}\lVert R_{2}\rVert R_{3}}{R_{1}+R_{FET}\lVert R_{2}\rVert R_{3}}}}
y
=
−
138815
x
+
333947
{\displaystyle y=-138815x+333947}
R
F
E
T
=
m
X
+
b
=
−
139
k
Ω
x
V
B
I
A
S
+
334
k
Ω
{\displaystyle R_{FET}=mX+b=-139k\Omega xV_{BIAS}+334k\Omega }
R
F
E
T
=
23.64
k
Ω
{\displaystyle R_{FET}=23.64k\Omega }
α
=
50
k
Ω
‖
23.64
k
Ω
47
k
Ω
+
50
k
Ω
‖
23.64
k
Ω
0.25
{\displaystyle \alpha ={\frac {50k\Omega \lVert 23.64k\Omega }{47k\Omega +50k\Omega \lVert 23.64k\Omega }}0.25}
R
p
e
r
p
e
n
d
i
c
u
l
a
r
=
s
i
n
(
θ
i
−
θ
t
)
/
s
i
n
(
θ
i
+
θ
t
)
{\displaystyle R_{perpendicular}=sin(\theta _{i}-\theta _{t})/sin(\theta _{i}+\theta _{t})}
θ
i
+
θ
t
=
90
O
{\displaystyle \theta _{i}+\theta _{t}=90^{O}}
θ
B
=
t
a
n
−
1
n
2
n
1
{\displaystyle \theta _{B}=tan^{-1}{\frac {n_{2}}{n_{1}}}}
I
(
θ
)
o
u
t
=
I
i
n
[
H
90
+
(
H
0
−
H
90
)
]
c
o
s
2
θ
{\displaystyle I(\theta )_{out}=I_{in}[H_{90}+(H_{0}-H_{90})]cos^{2}\theta }
H
0
=
1
2
(
k
1
2
+
k
2
2
)
{\displaystyle H_{0}={\frac {1}{2}}(k_{1}^{2}+k_{2}^{2})}
H
90
=
k
1
k
2
{\displaystyle H_{90}=k_{1}k_{2}}
I
(
θ
)
o
u
t
=
I
i
n
c
o
s
2
θ
{\displaystyle I(\theta )_{out}=I_{in}cos^{2}\theta }