Physics 212 Lecture 17, Slide 1 Physics 212 Lecture 17
Original:Mats Selen Modified:W.Geist Slide 2 Physics 212 Lecture
17, Slide 2 From the prelecture: Self Inductance Wrap a wire into a
coil to make an inductor dIdI dt = -L Slide 3 Physics 212 Lecture
17, Slide 3 What this really means: current I L dIdI dt L = -L emf
induced across L tries to keep I constant Inductors prevent abrupt
current changes ! Its like inertia! Slide 4 Physics 212 Lecture 17,
Slide 4 1) Two solenoids are made with the same cross sectional
area and total number of turns. Inductor B is twice as long as
inductor A Compare the inductance of the two solenoids A) L A = 4 L
B B) L A = 2 L B C) L A = L B D) L A = (1/2) L B E) L A = (1/4) L B
A) L A = 4 L B B) L A = 2 L B C) L A = L B D) L A = (1/2) L B E) L
A = (1/4) L B (1/2) 2 2 Slide 5 Physics 212 Lecture 17, Slide 5
Inside your i-clicker Slide 6 Physics 212 Lecture 17, Slide 6 How
to think about RL circuits Episode 1: When no current is flowing
initially: R L At t = 0: I = 0 V L = V BATT V R = 0 (L is like a
huge resistance) V BATT I=0 At t >> L/R: V L = 0 V R = V BATT
I = V BATT /R (L is like a short circuit) R I=V/R L V BATT VLVL I =
L/R Slide 7 Physics 212 Lecture 17, Slide 7 Circuit in Preflight (I
L = 0 before switch is closed) R L V R IRIRIRIR I L =0 Current
through L is 0 right after switch closed CD What is I R right after
closing switch? A) I = V/R B) I = V/2R C) I = 0 D) I = -V/2R E) I =
-V/R Remember current through inductor cant change abruptly Slide 8
Physics 212 Lecture 17, Slide 8 What is the current I through the
vertical resistor immediately after the switch is closed? (+ is in
the direction of the arrow) A) I = V/R B) I = V/2R C) I = 0 D) I =
-V/2R E) I = -V/R In the circuit above, the switch has been open
for a long time, and the current is zero everywhere. At time t=0
the switch is closed. Slide 9 Physics 212 Lecture 17, Slide 9 What
is the current I through the vertical resistor after the switch has
been closed for a long time? (+ is in the direction of the arrow)
A) I = V/R B) I = V/2R C) I = 0 D) I = -V/2R E) I = -V/R RL Circuit
(Long Time) After a long time in any static circuit: V L = 0 KVR: V
L + IR = 0 + + - - Slide 10 Physics 212 Lecture 17, Slide 10 How to
think about RL circuits Episode 2: When steady current is flowing
initially: R L I = V BATT /R V R = IR V L = V R (L is like a
battery) At t = 0: V BATT I=V/R At t >> L/R: I = 0 V L = 0 V
R = 0 R I=0 L VLVL = L/R Slide 11 Physics 212 Lecture 17, Slide 11
R L V R Current through inductor cant change abruptly. I L is the
same just before and just after switch opens = V/R. I L =V/R What
is the current I through the vertical resistor immediately after
the switch is opened? (+ is in the direction of the arrow) A) I =
V/R B) I = V/2R C) I = 0 D) I = -V/2R E) I = -V/R CD +ve Slide 12
Physics 212 Lecture 17, Slide 12 What is the current I through the
vertical resistor immediately after the switch is opened? (+ is in
the direction of the arrow) A) I = V/R B) I = V/2R C) I = 0 D) I =
-V/2R E) I = -V/R After a long time, the switch is opened, abruptly
disconnecting the battery from the circuit. Slide 13 Physics 212
Lecture 17, Slide 13 R I V = IR dIdI dt V = L I at t=0 L where + +
VLVL = L/R Why is there exponential behavior ? Slide 14 Physics 212
Lecture 17, Slide 14 Prelecture: R L I Lecture: VLVL = L/R V BATT
Did we mess up?? No: The resistance is simply twice as big in one
case. Slide 15 Physics 212 Lecture 17, Slide 15Calculation The
switch in the circuit shown has been open for a long time. At t =
0, the switch is closed. What is dI L /dt, the time rate of change
of the current through the inductor immediately after switch is
closed Conceptual Analysis Once switch is closed, currents will
flow through this 2-loop circuit. KVR and KCR can be used to
determine currents as a function of time. Strategic Analysis
Determine currents immediately after switch is closed. Determine
voltage across inductor immediately after switch is closed.
Determine dIL/dt immediately after switch is closed. C R1R1R1R1 L V
R2R2R2R2 R3R3R3R3 Slide 16 Physics 212 Lecture 17, Slide
16Calculation The switch in the circuit shown has been open for a
long time. At t = 0, the switch is closed. What is the direction of
I L, the current in the inductor, immediately after the switch is
closed? (B) (C) (A) I L flows up (B) I L flows down (C) I L = 0
R1R1R1R1 L V R2R2R2R2 R3R3R3R3 INDUCTORS: Current cannot change
discontinuously ! Current through inductor immediately AFTER switch
is closed IS THE SAME AS the current through inductor immediately
BEFORE switch is closed Immediately before switch is closed: I L =
0 since no battery in loop I L = 0 Slide 17 Physics 212 Lecture 17,
Slide 17Calculation The switch in the circuit shown has been open
for a long time. At t = 0, the switch is closed. What is the
magnitude of I 2, the current in R 2, immediately after the switch
is closed? (B) (C) (D) (A) (B) (C) (D) R1R1R1R1 L V R2R2R2R2
R3R3R3R3 We know I L = 0 immediately after switch is closed
Immediately after switch is closed, circuit looks like: R1R1R1R1V
R2R2R2R2 R3R3R3R3 I I L (t=0+) = 0 Slide 18 Physics 212 Lecture 17,
Slide 18Calculation The switch in the circuit shown has been open
for a long time. At t = 0, the switch is closed. R1R1R1R1 L V
R2R2R2R2 R3R3R3R3 I2I2I2I2 I L (t=0+) = 0 I 2 (t=0+) = V/(R 1 +R 2
+R 3 ) What is the magnitude of V L, the voltage across the
inductor, immediately after the switch is closed? (B) (C) (D) (E)
(A) (B) (C) (D) (E) The voltage across the inductor, V L, is also
equal to the voltage across R 2 + R 3 The voltage across R 2 + R 3
= I 2 (R 2 + R 3 ) Therefore: Slide 19 Physics 212 Lecture 17,
Slide 19Calculation The switch in the circuit shown has been open
for a long time. At t = 0, the switch is closed. What is dI L /dt,
the time rate of change of the current through the inductor
immediately after switch is closed (B) (C) (D) (A) (B) (C) (D)
R1R1R1R1 L V R2R2R2R2 R3R3R3R3 The time rate of change of current
through the inductor (dI L /dt) = V L /L V L (t=0+) = V(R 2 +R 3
)/(R 1 +R 2 +R 3 ) Hint: Slide 20 Physics 212 Lecture 17, Slide 20
Follow Up The switch in the circuit shown has been closed for a
long time. What is I 2, the current through R 2 ? (Positive values
indicate current flows to the right) (B) (C) (D) (A) (B) (C) (D)
R1R1R1R1 L V R2R2R2R2 R3R3R3R3 After a long time, dI/dt = 0
Therefore, the voltage across L = 0 Therefore the voltage across R
2 + R 3 = 0 Therefore the current through R 2 + R 3 must be zero !!
Slide 21 Physics 212 Lecture 17, Slide 21 R L V R IRIRIRIR ILILILIL
A long time later the current in the circuit is no-longer changing.
V L = 0 Inductor looks like a short circuit = 0 (current all going
through the short circuit inductor)
