3020
Chapter 30
30.68.
I
DENTIFY
:
Apply the loop rule to the two loops. The current through the inductor doesn’t change abruptly.
S
ET
U
P
:
For the inductor
di
L
dt
=
E
and
E
is directed to oppose the change in current.
E
XECUTE
:
(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
v
L
dt
dt
=
⇒
=
=
=
The top circuit loop: 60.0
1
1
1
60.0 V
V
1.50 A.
40.0
i R
i
=
⇒
=
=
Ω
The bottom loop:
2
2
2
45.0 V
60 V
15.0 V
0
1.80 A.
25.0
i R
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
1
2
2.40 A.
i
i
=
=
E
VALUATE
:
The current through
1
R
changes abruptly when the switch is closed.
30.69.
I
DENTIFY
and
S
ET
U
P
:
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
E
XECUTE
:
(a)
At time
0
0
0
0,
0 so
0.
ac
t
i
v
i R
=
=
=
=
By the loop rule
0,
ac
cb
v
v
−
−
=
E
so
36.0 V.
cb
ac
v
v
=
−
=
=
E
E
(
0
0
i R
=
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
0
(
)
0.
i
R
R
−
+
=
E
0
0
36.0 V
0.180 A
50.0
150
i
R
R
=
=
=
+
Ω +
Ω
E
0
0
(0.180 A)(50.0
)
9.0 V
ac
v
i R
=
=
Ω =
Thus
0
0
(0.180 A)(150
)
0
27.0 V
cb
di
v
i R
L
dt
=
+
=
Ω +
=
(Note that
.
ac
cb
v
v
+
=
E
)
(c)
0
ac
cb
v
v
−
−
=
E
0
0
di
iR
iR
L
dt
−
−
−
=
E
0
0
0
(
) and
di
L
di
L
i R
R
i
dt
R
R
dt
R
R
⎛
⎞
=
−
+
= − +
⎜
⎟
+
+
⎝
⎠
E
E
0
0
/(
)
di
R
R
dt
i
R
R
L
+
⎛
⎞
=
⎜
⎟
− +
+
⎝
⎠
E
Integrate from
t
= 0, when
i
= 0, to
t
, when
0
:
i
i
=
0
0
0
0
0
0
0
0
0
ln
,
/(
)
i
i
t
di
R
R
R
R
dt
i
t
i
R
R
L
R
R
L
⎡
⎤
+
+
⎛
⎞
=
= −
− +
=
⎢
⎥
⎜
⎟
− +
+
+
⎝
⎠
⎣
⎦
∫
∫
E
E
so
0
0
0
0
ln
ln
R
R
i
t
R
R
R
R
L
⎛
⎞
⎛
⎞
+
⎛
⎞
−
+
−
= −
⎜
⎟
⎜
⎟
⎜
⎟
+
+
⎝
⎠
⎝
⎠
⎝
⎠
E
E
0
0
0
0
/(
)
ln
/(
)
i
R
R
R
R
t
R
R
L
⎛
⎞
−
+
+
+
⎛
⎞
= −
⎜
⎟
⎜
⎟
+
⎝
⎠
⎝
⎠
E
E
Taking exponentials of both sides gives
0
(
) /
0
0
0
/(
)
/(
)
R
R
t L
i
R
R
e
R
R
−
+
−
+
+
=
+
E
E
and
(
)
0
(
) /
0
0
1
R
R
t L
i
e
R
R
−
+
=
−
+
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
t
t
i
e
e
−
Ω
−
=
−
=
−
Ω +
Ω
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Inductance
3021
At
0
0,
(0.180 A)(1
1)
0
t
i
→
=
−
=
(agrees with part (a)). At
0
,
(0.180 A)(1
0)
0.180 A
t
i
→ ∞
=
−
=
(agrees with part (b)).
(
)
(
)
0
(
)
/
/0.020 s
0
0
0
0
1
9.0 V
1
R
R
t L
t
ac
R
v
i R
e
e
R
R
−
+
−
=
=
−
=
−
+
E
/0.020 s
/0.020 s
36.0 V
9.0 V(1
)
9.0 V(3.00
)
t
t
cb
ac
v
v
e
e
−
−
=
−
=
−
−
=
+
E
At
0,
0,
36.0 V
ac
cb
t
v
v
→
=
=
(agrees with part (a)). At
,
9.0 V,
27.0 V
ac
cb
t
v
v
→ ∞
=
=
(agrees with part (b)).
The graphs are given in Figure 30.69b.
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 Spring '06
 Buchler
 Physics, Current, long time, Inductor, dt, loop rule, R0

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