This preview shows pages 1–2. Sign up to view the full content.
dP
dx
Rr xR
rx
R
avg
=
−
+
ε
2
3
bg
.
This is zero for
xrR
==
=
//
1000
10
100
ΩΩ
b
g
. We note that for small
x, P
avg
increases linearly with
x
, and for large
x
it decreases in proportion to 1/
x
. Thus
x = r
/
R
is
indeed a maximum, not a minimum. Recalling
x
= (
N
p
/
N
s
)
2
, we conclude that the
maximum power is achieved for
/1
0
ps
NN x
=
=
.
88. (a) The amplifier is connected across the primary windings of a transformer and the
resistor
R
is connected across the secondary windings.
(b) If
I
s
is the rms current in the secondary coil then the average power delivered to
R
is
PI
R
s
avg
=
2
. Using
s
I
=
()
/
p
NNI
, we obtain
P
IN
N
R
pp
s
avg
=
F
H
G
I
K
J
2
Next, we find the current in the primary circuit. This is effectively a circuit consisting of
a generator and two resistors in series. One resistance is that of the amplifier (
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '08
 Any
 Physics, Current, Power

Click to edit the document details