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3020
Chapter 30
30.68.
IDENTIFY:
Apply the loop rule to the two loops. The current through the inductor doesn’t change abruptly.
SET UP:
For the inductor
di
L
dt
=
E
and
E
is directed to oppose the change in current.
EXECUTE:
(a)
Switch is closed, then at some later time
50.0 A/s
(0.300 H) (50.0 A/s)
15.0 V.
cd
di
di
vL
dt
dt
=⇒
=
=
=
The top circuit loop: 60.0
11
1
60.0 V
V1
.
5
0
A
.
40.0
iR
i
=
=
Ω
The bottom loop:
22
2
45.0 V
60 V
15.0 V
0
1.80 A.
25.0
i
−−
=
⇒
=
=
Ω
(b)
After a long time:
2
60.0 V
2.40 A,
25.0
i
==
Ω
and immediately when the switch is opened, the inductor maintains
this current, so
12
2.40 A.
ii
EVALUATE:
The current through
1
R
changes abruptly when the switch is closed.
30.69.
IDENTIFY
and
SET UP:
The circuit is sketched in Figure 30.69a. Apply the loop rule. Just after
1
S is closed,
i
= 0.
After a long time
i
has reached its final value and
di
/
dt
= 0. The voltage across a resistor depends on
i
and the
voltage across an inductor depends on
di
/
dt.
Figure 30.69a
EXECUTE: (a)
At time
00
0
0,
0 so
0.
ac
ti
v
i
R
By the loop rule
0,
ac
cb
vv
−
−=
E
so
36.0 V.
cb
ac
=− ==
EE
(
0
0
=
so this potential difference of 36.0 V is across the inductor and is an induced emf produced by the
changing current.)
(b)
After a long time
0
0
di
dt
→
so the potential
0
di
L
dt
−
across the inductor becomes zero. The loop rule gives
()
0
.
iR R
−+
=
E
0
0
36.0 V
0.180 A
50.0
150
i
RR
=
+Ω
+
Ω
E
(0.180 A)(50.0
)
9.0 V
ac
vi
R
Ω
=
Thus
0
0
(0.180 A)(150
) 0
27.0 V
cb
di
R
L
dt
=+
=
Ω
+
=
(Note that
.
ac
cb
+
=
E
)
(c)
0
ac
cb
−−=
E
0
0
di
iR
iR
L
dt
−−− =
E
0
a
n
d
di
L
di
Li
R
R
i
dt
R
R
dt
R
R
⎛⎞
=−
+
=
⎜⎟
++
⎝⎠
E
E
0
0
/(
)
di
R
R
dt
R
L
+
=
+
E
Integrate from
t
= 0, when
i
= 0, to
t
, when
0
:
=
0
0
0
ln
,
/(
)
i
it
di
R
R
R
R
dt
i
t
R
L
R
R
L
⎡⎤
−
−
+
=
⎢⎥
+
+
⎣⎦
∫∫
E
E
so
0
0
ln
ln
L
⎛
⎞
+
−
⎜
⎟
⎝
⎠
0
0
/(
)
ln
/(
)
R
R
R
t
L
+
+
+
E
E
Taking exponentials of both sides gives
0
/
0
/(
)
/(
)
RRtL
R
e
−
=
+
E
E
and
0
/
0
0
1
ie
+
E
Substituting in the numerical values gives
(
)
(200
/ 4.00 H)
/0.020 s
0
36.0 V
1
(0.180 A) 1
50
150
tt
e
−Ω
−
=
−
Ω+
Ω
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3021
At
0
0,
(0.180 A)(1 1)
0
ti
→=
−
=
(agrees with part (a)). At
0
,
(0.180 A)(1 0)
0.180 A
→∞
=
−
=
(agrees with part (b)).
()
0
/
/0.020 s
0
00
0
1
9.0 V 1
RRtL
t
ac
R
vi
R
e
e
RR
−+
−
==
−
=
−
+
E
/0.020 s
/0.020 s
36.0 V 9.0 V(1
)
9.0 V(3.00
)
tt
cb
ac
vv
e
e
−−
=− =
−
−
=
+
E
At
0,
0,
36.0 V
ac
cb
tv
v
→==
(agrees with part (a)). At
9.0 V,
27.0 V
ac
cb
v
=
=
(agrees with part (b)).
The graphs are given in Figure 30.69b.
Figure 30.69b
EVALUATE:
The expression for
i
(
t
) we derived becomes Eq.(30.14) if the two resistors
0
R
and
R
in series are
replaced by a single equivalent resistance
0
.
R
R
+
30.70.
IDENTIFY:
Apply the loop rule. The current through the inductor doesn’t change abruptly.
SET UP:
With
2
S
closed,
cb
v
must be zero.
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 Spring '06
 Buchler
 Physics, Current

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