Maximum Power Transfer
edit
in the notes, he says
P
=
Re
{
V
s
2
(
R
L
+
j
X
L
)
(
R
S
+
R
L
)
2
+
(
X
S
+
X
L
)
2
}
{\displaystyle P={\textrm {Re}}\left\{{\frac {V_{s}^{2}\left(R_{L}+jX_{L}\right)}{(R_{S}+R_{L})^{2}+(X_{S}+X_{L})^{2}}}\right\}}
but in the derivation the denominator is
[
(
R
S
+
R
L
)
+
j
(
X
S
+
X
L
)
]
2
{\displaystyle \left[(R_{S}+R_{L})+j(X_{S}+X_{L})\right]^{2}}
which doesn't expand out. I'm wondering if you could have a look at your notes to see if they match
Question 5a
edit
The Sum Thing
edit
∑
x
=
1
n
f
(
x
)
{\displaystyle \sum _{x=1}^{n}f(x)}
means that you take
f
(
1
)
{\displaystyle f(1)}
,
f
(
2
)
{\displaystyle f(2)}
,
f
(
3
)
{\displaystyle f(3)}
, etc. up to
f
(
n
)
{\displaystyle f(n)}
, and add them all together.
For example,
∑
x
=
1
5
2
x
=
2
(
1
)
+
2
(
2
)
+
2
(
3
)
+
2
(
4
)
+
2
(
5
)
=
30
{\displaystyle \sum _{x=1}^{5}2x=2(1)+2(2)+2(3)+2(4)+2(5)=30}
Note that:
∑
x
=
1
5
2
x
=
2
∑
x
=
1
5
x
{\displaystyle \sum _{x=1}^{5}2x=2\sum _{x=1}^{5}x}
and that a rule exists that
∑
x
=
1
n
x
=
n
(
n
+
1
)
2
{\displaystyle \sum _{x=1}^{n}x={\frac {n(n+1)}{2}}}
so
∑
x
=
1
5
2
x
=
2
∑
x
=
1
5
x
=
2
5
(
5
+
1
)
2
=
2
30
2
=
30
{\displaystyle \sum _{x=1}^{5}2x=2\sum _{x=1}^{5}x=2{\frac {5(5+1)}{2}}=2{\frac {30}{2}}=30}