nguyen (jmn727) – oldhomework 27 – Turner – (59070)
1
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001
(part 1 of 3) 10.0 points
One application of an RL circuit is the gen
eration of timevarying highvoltage from a
lowvoltage source, as shown in the figure.
3 H
14 Ω
1154 Ω
10
.
2 V
S
b
a
What is the current in the circuit a long
time after the switch has been in position
“
a
”?
Correct answer: 0
.
728571 A.
Explanation:
L
R
2
R
1
E
S
b
a
Let :
R
2
= 14 Ω
and
E
= 10
.
2 V
.
When the switch is at “
a
”, the circuit com
prises the battery, the inductor
L
, and the
resistor
R
2
. A long time after the switch has
been in position “
a
”, the current is steady.
This means that the inductor has no response
to the current. In other words, the circuit can
be considered as consisting of
E
and
R
2
only,
with the inductor reduced to a wire. In this
case, the current is simply found by Ohm’s
Law
I
0
=
E
R
2
=
10
.
2 V
14 Ω
=
0
.
728571 A
.
002
(part 2 of 3) 10.0 points
Now the switch is thrown quickly from “
a
” to
“
b
”.
Compute the initial voltage across the in
ductor.
Correct answer: 850
.
971 V.
Explanation:
Let :
R
1
= 1154 Ω
and
I
0
= 0
.
728571 A
.
When the switch is thrown from “
a
” to “
b
”,
the current in the circuit is the current passing
through
R
2
, which was found in Part 1 to be
0
.
728571 A
.
From Kirchhoff’s Loop Law, the
initial voltage across the inductor is equal to
the initial voltage across
R
1
and
R
2
. So, we
have
V
L
=
V
R
1
+
V
R
2
=
I
0
R
1
+
I
0
R
2
= (0
.
728571 A) (1154 Ω) + (0
.
728571 A) (14 Ω)
=
850
.
971 V
.
003
(part 3 of 3) 10.0 points
How much time elapses before the voltage
across the inductor drops to 11 V?
Correct answer: 11
.
169 ms.
Explanation:
Let :
L
= 3 H
and
V
L
= 11 V
.
The voltage across an inductor is
V
L
=

L
dI
dt
.
When the switch is at “
b
”, we are dealing with
an RL circuit with an initial current
I
0
that
decays as
I
=
I
0
e

t / τ
.
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nguyen (jmn727) – oldhomework 27 – Turner – (59070)
2
The time constant is
τ
=
L
R
t
=
L
R
1
+
R
2
=
3 H
1154 Ω + 14 Ω
= 0
.
00256849 s
.
Therefore, the voltage across the inductor
is
V
L
=

L
d
dt
I
0
e

t / τ
=
L I
0
τ
e

t / τ
.
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 Spring '08
 Turner
 Inductance, Work, Inductor, RL circuit, LC circuit

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