University of Rhode Island
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PHY 204: Elementary Physics II
Physics Course Materials
2015
18. RL Circuits
Gerhard Müller
University of Rhode Island, gmuller@uri.edu
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Abstract
Part eighteen of course materials for Elementary Physics II (PHY 204), taught by Gerhard Müller at
the University of Rhode Island. Documents will be updated periodically as more entries become
presentable.
Some of the slides contain figures from the textbook, Paul A. Tipler and Gene Mosca. Physics for
Scientists and Engineers, 5th/6th editions. The copyright to these figures is owned by W.H. Freeman.
We acknowledge permission from W.H. Freeman to use them on this course web page. The textbook
figures are not to be used or copied for any purpose outside this class without direct permission from
W.H. Freeman.
Recommended Citation
Müller, Gerhard, "18. RL Circuits" (2015). PHY 204: Elementary Physics II. Paper 8.
http://digitalcommons.uri.edu/elementary_physics_2/8
This Course Material is brought to you for free and open access by the Physics Course Materials at DigitalCommons@URI. It has been accepted for
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RL Circuit: Fundamentals
Specifications:
a
• E (emf)
• R (resistance)
• L (inductance)
ε
S
b
I(t)
R
L
Switch S:
• a: current buildup
• b: current shutdown
Time-dependent quantities:
• I(t): instantaneous current through inductor
dI
: rate of change of instantaneous current
dt
• VR (t) = I(t)R: instantaneous voltage across resistor
•
• VL (t) = L
dI
: instantaneous voltage across inductor
dt
4/11/2015
[tsl271 – 1/15]
RL Circuit: Current Buildup in Inductor
dI
=0
dt
dI
E/R − I
dI
= E − IR ⇒
=
• Differential equation: L
dt
dt
L/R
„
«
Z I
Z t
dI
E/R − I
dt
t
=
⇒ − ln
=
E/R
L/R
0 E/R − I
0 L/R
i
E h
−Rt/L
1−e
• Current through inductor: I(t) =
R
dI
E −Rt/L
• Rate of current change:
=
e
dt
L
• Loop rule: E − IR − L
I(t)
ε
ε
⇒
E/R − I
= e−Rt/L
E/R
dI/dt
L
R
t
t
4/11/2015
[tsl272 – 2/15]
RL Circuit: Current Shutdown in Inductor
dI
=0
dt
dI
dI
R
• Differential equation: L
+ IR = 0 ⇒
=− I
dtZ
dt
L
Z I
R t
R
I
I
dI
=−
=− t ⇒
= e−Rt/L
dt ⇒ ln
⇒
L 0
E/R
L
E/R
E/R I
• Loop rule: −IR − L
E −Rt/L
e
R
E
dI
= − e−Rt/L
• Rate of current change:
dt
L
• Current: I(t) =
I(t)
ε
ε
−dI/dt
L
R
t
t
4/11/2015
[tsl273 – 3/15]
RL Circuit: Energy Transfer During Current Buildup
Loop rule: IR + L
dI
=E
dt
(I > 0,
dI
> 0)
dt
• IE: rate at which EMF source delivers energy
• IVR = I 2 R: rate at which energy is dissipated in resistor
• IVL = LI
dI
: rate at which energy is stored in inductor
dt
Balance of energy transfer: I 2 R + LI
dI
= IE
dt
a
ε
b
S
I(t)
R
L
4/11/2015
[tsl274 – 4/15]
RL Circuit: Energy Transfer During Current Shutdown
Loop rule: IR + L
• IVL = LI
dI
=0
dt
(I > 0,
dI
< 0)
dt
dI
: rate at which inductor releases energy
dt
• IVR = I 2 R: rate at which energy is dissipated in resistor
Balance of energy transfer: I 2 R + LI
dI
=0
dt
a
ε
b
S
I(t)
R
L
4/11/2015
[tsl275 – 5/15]
RL Circuit: Some Physical Properties
Specification of RL circuit
by 3 device properties:
• E [V] (emf)
• R [Ω] (resistance)
ε
S
R
L
I(t)
• L [H] (inductance)
Physical properties of RL circuit during current buildup determined by 3 combinations of the
device properties:
˛
dI ˛˛
E
=
•
: initial rate at which current increases
L
dt ˛t=0
E
= I(t = ∞): final value of current
R
• L/R = τ : time it takes to build up 63% of the current through the circuit
[1 − e−1 = 0.632 . . .]
•
4/11/2015
[tsl276 – 6/15]
RL Circuit: Application (7)
In the circuit shown the switch S is closed at time t = 0.
(a) Find the current I as a function of time for 0 < t < tF , where tF marks the instant the fuse
breaks.
(b) Find the current I as a function of time for t > tF .
4A fuse
15 Ω
12V
I
5H
S
4/11/2015
[tsl284 – 7/15]
RL Circuit: Application (8)
In the circuit shown the switch has been open for a long time.
Find the currents I1 and I2
• just after the switch has been closed,
• a long time later,
• as functions of time for 0 < t < ∞.
S
R 2 = 10 Ω
ε = 12V
Ι1
R1 = 5 Ω
Ι2
L = 5H
4/11/2015
[tsl285 – 8/15]
RL Circuit: Application (6)
In the RL circuit shown the switch has been at position a for a long time and is thrown to position b
at time t = 0. At that instant the current has the value I0 = 0.7A and decreases at the rate
dI/dt = −360A/s.
(a) Find the EMF E of the battery.
(b) Find the resistance R of the resistor.
(c) At what time t1 has the current decreased to the value I1 = 0.2A?
(d) Find the voltage across the inductor at time t1 .
a
ε
b
S
R
L = 0.3H
4/11/2015
[tsl283 – 9/15]
RL Circuit: Application (5)
Each RL circuit contains a 2A fuse. The switches are closed at t = 0.
• In what sequence are the fuses blown?
(1)
6V
S
1H
(2)
1Ω
1H
(3)
6V
S
2A
1H
2A
1H
1Ω
1H
1Ω
1Ω
(4)
2A
1Ω
1H
6V
S
1Ω
6V
S
2A
1H
1Ω
1Ω
1H
4/11/2015
[tsl282 – 10/15]
RL Circuit: Application (3)
The switch is closed at t = 0. Find the current I
(a) immediately after the switch has been closed,
(b) immediately before the fuse breaks,
(c) immediately after the fuse has broken,
(d) a very long time later.
4A fuse
15 Ω
12V
I
5H
S
4/11/2015
[tsl280 – 11/15]
RL Circuit: Application (1)
Each branch in the circuit shown contains a 3A fuse. The switch is closed at time t = 0.
(a) Which fuse is blown in the shortest time?
(b) Which fuse lasts the longest time?
R=6 Ω
S
12V
1
L=6H
2
C=6F
3
4/11/2015
[tsl278 – 12/15]
RL Circuit: Application (4)
Find the magnitude (in amps) and the direction (↑, ↓) of the current I
(a) right after the switch has been closed,
(b) a very long time later.
S
12V
1Ω
1F
1Ω
1H
I
1Ω
1Ω
4/11/2015
[tsl281 – 13/15]
RL Circuit: Application (2)
The switch in each RL circuit is closed at t = 0.
Rank the circuits according to three criteria:
(a) magnitude of current at t =1ms,
(b) magnitude of current at t = ∞,
(c) time it takes I to reach 63% of its ultimate value.
(1)
S
1H
(2)
6V
1Ω
1H
(3)
1H
1Ω
1H
1Ω
(4)
6V
S
1Ω
1H
1H
1Ω
6V
S
6V
S
1Ω
1H
1Ω
1Ω
1H
4/11/2015
[tsl279 – 14/15]
Intermediate Exam III: Problem #2 (Spring ’05)
In the circuit shown we close the switch S at time t = 0. Find the current I through the battery and
the voltage VL across the inductor
(a) immediately after the switch has been closed,
5H
(b) a very long time later.
4Ω
2Ω
2Ω
12V
S
Ι
4/11/2015
[tsl343 – 15/15]
Intermediate Exam III: Problem #2 (Spring ’05)
In the circuit shown we close the switch S at time t = 0. Find the current I through the battery and
the voltage VL across the inductor
(a) immediately after the switch has been closed,
5H
(b) a very long time later.
4Ω
2Ω
2Ω
12V
Solution:
12V
= 1.5A,
(a) I =
2Ω + 4Ω + 2Ω
S
Ι
VL = (4Ω)(1.5A) = 6V.
4/11/2015
[tsl343 – 15/15]
Intermediate Exam III: Problem #2 (Spring ’05)
In the circuit shown we close the switch S at time t = 0. Find the current I through the battery and
the voltage VL across the inductor
(a) immediately after the switch has been closed,
5H
(b) a very long time later.
4Ω
2Ω
2Ω
12V
Solution:
12V
= 1.5A,
VL = (4Ω)(1.5A) = 6V.
(a) I =
2Ω + 4Ω + 2Ω
12V
= 3A,
VL = 0.
(b) I =
2Ω + 2Ω
S
Ι
4/11/2015
[tsl343 – 15/15]