gonzales (pag757) – Homework 11 – Hoffmann – (56745)
1
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After the parts that are withdrawn there
are 20 problems.
001 (part 1 of 2) 10.0 points
A
smalldiameter
inductor
with
self
inductance
L
s
and a largediameter inductor
with selfinductance
L
ℓ
are connected in se
ries as shown in the figure. Consider the two
inductors to be far apart. Thus, the mutual
inductance between the two inductors is 0.
a
b
Determine the equivalent selfinductance
for the system.
1.
L
eq
=
radicalbig
(
L
s

L
ℓ
)
2
2
2.
L
eq
=
L
s
L
ℓ
L
s
+
L
ℓ
3.
L
eq
=
radicalBig
(
L
s

L
ℓ
)
2
4.
L
eq
=
L
s
+
L
ℓ
correct
5.
L
eq
=
L
s
L
ℓ
radicalBig
L
2
s
+
L
2
ℓ
6.
L
eq
=
L
s
L
ℓ
2 (
L
s
+
L
ℓ
)
7.
L
eq
=
L
s
L
ℓ
2
radicalBig
L
2
s
+
L
2
ℓ
8.
L
eq
=
radicalbig
(
L
s
+
L
ℓ
)
2
2
9.
L
eq
=
L
s
+
L
ℓ
2
10.
L
eq
=
radicalBig
(
L
s
+
L
ℓ
)
2
Explanation:
Let
E
ℓ
be
the
emf
across
the
larger
diameter inductor,
E
s
be the
emf
across the
smallerdiameter inductor, and
E
be the total
emf
across both inductors.
The current
I
entering terminal
a
and ex
iting terminal
b
of this series combination is
related to the total
emf
E
by
E
=

L
eq
d I
dt
,
(1)
so
L
eq
=

E
d I
dt
.
(2)
In general, the
emf
E
across the inductor
complex is due to selfinductance
and
mutual
inductance between the two inductors. How
ever, as
M
= 0 and as
I
ℓ
=
I
s
=
I
, we have
E
s
=

L
s
d I
s
dt
=

L
s
d I
dt
,
(3)
and similarly
E
ℓ
=

L
ℓ
d I
ℓ
dt
=

L
ℓ
d I
dt
.
(4)
Using
E
=
E
ℓ
+
E
s
together with Eqs. 1, 3, and
4, we have

L
eq
d I
dt
=
E
ℓ
+
E
s
=

L
ℓ
d I
dt

L
s
d I
dt
=

(
L
ℓ
+
L
s
)
d I
dt
.
(5)
Therefore for
L
eq
, we have
L
eq
=
L
ℓ
+
L
s
.
(6)
Note:
Series inductors combine like series
resistors (if no mutual inductance is present).
002 (part 2 of 2) 0.0 points
WITHDRAWN
003 (part 1 of 2) 10.0 points
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gonzales (pag757) – Homework 11 – Hoffmann – (56745)
2
A smalldiameter solenoidal inductor with
selfinductance 131 mH and a largediameter
solenoidal
inductor
with
selfinductance
54 mH are connected in parallel as shown
in the figure.
Consider the two inductors to
be far apart.
Thus, the mutual inductance
between the two inductors is 0 mH.
a
b
Determine the equivalent selfinductance
for the system.
Correct answer: 38
.
2378 mH.
Explanation:
Let :
L
s
= 131 mH
and
L
ℓ
= 54 mH
.
The source of the current through the smaller
diameter inductor
I
s
and the largerdiameter
inductor
I
ℓ
is from the same terminal. That
is, the total current
I
in this parallel inductor
complex is the sum of the currents in the two
inductors
I
=
I
s
+
I
ℓ
.
The
emf
E
is the same across both induc
tors because they are connected in parallel.
The
emf
is related to the total current
I
by
E
=

L
eq
d I
dt
.
(1)
Therefore
L
eq
=

E
d I
dt
.
(2)
In general the total induced
emf
E
across
an inductor is due to selfinductance with
itself
and
mutualinductance with the other
inductor. But with
M
= 0 mH, we only have
selfinductance, so
E
=

L
s
d I
s
dt
(3)
d I
s
dt
=

E
L
s
.
(4)
Similarly
E
=

L
ℓ
d I
ℓ
dt
(5)
d I
ℓ
dt
=

E
L
ℓ
.
(6)
Using Eqs. 4 and 6 to solve for
d I
dt
, where
I
=
I
s
+
I
ℓ
, we have
d I
dt
=
d I
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 Spring '08
 Turner
 Inductance, Work, Correct Answer, Inductor

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