Unformatted text preview: Physics 212
Lecture 18 Physics 212 Lecture 18, Slide 1 From the prelecture: Self Inductance Wrap a wire into a coil to make an "inductor"... = L dI dt Physics 212 Lecture 18, Slide 2 What this really means:
e m f in d uc e d a c ro s s L trie s to ke e p I c o ns ta nt L = L dI dt L
current I
Ind uc to rs p re ve nt d is c o ntinu o us c u rre nt c h a ng e s ! It's like inertia!
Physics 212 Lecture 18, Slide 3 I(t) = L dI dt Suppose dI/dt > 0. Induced EMF tries to counteract this change (Lenz's Law). +  I(t) V1
+  V2 Voltage across inductor V1 V2 = VL = + L dI/dt > 0 Checkpoint 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 LB = 0 n 2r 2 z
(1/2)2 2 LB = 1 L A 2
C o m p a re th e ind uc ta nc e o f th e two s o le no id s A) LA = 4 LB B) LA = 2 LB C) LA = LB D) LA = (1/2) LB E) LA = (1/4) LB Physics 212 Lecture 18, Slide 5 WHAT ARE INDUCTORS AND CAPACITORS GOOD FOR? Inside your iclicker Physics 212 Lecture 18, Slide 6 How to think about RL circuits Episode 1: When no current is flowing initially: I=0 L R
VL I=V/R L R
I = L/R VBATT
I = 0 VL = VBATT VR = 0 (L is like a giant resistor) At t = 0: At t >> L/R:
VL = 0 VR = VBATT I = VBATT/R VBATT = L/R (L is like a short circuit)
Physics 212 Lecture 18, Slide 7 Checkpoint 2a
In the circuit, the switch has been open for a long time, and the current is zero everywhere. At time t=0 the switch is closed. I 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 Before: IL = 0 After: IL = 0 I = + V/2R I IL=0 Physics 212 Lecture 18, Slide 8 RL Circuit (Long Time)
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  + +  After a long time in any static circuit: VL = 0 KVR: VL + IR = 0
Physics 212 Lecture 18, Slide 9 VBATT How to think about RL circuits Episode 2: When steady current is flowing initially:
VL I=0 R L I=V/R R L R = L/R = L/R I = VBATT/R VR = IR VL = VR At t = 0: At t >> L/R:
I = 0 VL = 0 VR = 0
Physics 212 Lecture 18, Slide 10 Checkpoint 2b
After a long time, the switch is opened, abruptly disconnecting the battery from the circuit. 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 C urre nt th ro ug h ind uc to r c a nno t c h a ng e DIS C O NT INUO US LY circuit when switch opened L
IL=V/R R Physics 212 Lecture 18, Slide 11 Why is there exponential behavior ?
 I L R + VL = L/R V = L dI dt V = IR
 = L/R + dI L + IR = 0 dt I (t ) = I 0e tR / L = I 0e t / L where = R
Physics 212 Lecture 18, Slide 12 I L R VL VBATT = L/R Lecture: Prelecture: Did we mess up?? No: The resistance is simply twice as big in one case.
Physics 212 Lecture 18, Slide 13 Checkpoint 3a
After long time at 0, moved to 1 After long time at 0, moved to 2 After switch moved, which case has larger time constant? A) Case 1 B) Case 2 C) The same L 1 = 2R L 2 = 3R Physics 212 Lecture 18, Slide 14 Checkpoint 3b
After long time at 0, moved to 1 After long time at 0, moved to 2 Immediately after switch moved, in which case is the voltage across the inductor larger? A) Case 1 After switch moved: B) Case 2 V VL1 = 2 R C) The same R Before switch moved: I = V R VL 2 = V 3R R Physics 212 Lecture 18, Slide 15 Checkpoint 3c
After long time at 0, moved to 1 After long time at 0, moved to 2 After switch moved for finite time, in which case is the current through the inductor larger? A) Case 1 After awhile B) Case 2 I1 = Ie t / C) The same
1 Immediately after: I1 = I 2 I 2 = Iet / 2 1 > 2 Physics 212 Lecture 18, Slide 16 The switch in the circuit shown has been open for a long time. At t = 0, the switch is closed. What is dIL/dt, the time rate of change of the current through the inductor immediately after switch is closed Calculation
R1 V L R2 R3 Conceptual Analysis Once switch is closed, currents will flow through this 2loop 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. Physics 212 Lecture 18, Slide 17 The switch in the circuit shown has been open for a long time. At t = 0, the switch is closed. Calculation
R1 V L R2 IL = 0 R3 What is IL, the current in the inductor, immediately after the switch is closed? (A) IL =V/R1 up (B) IL =V/R1 down (C) IL = 0 (B) (C) 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: IL = 0 since no battery in loop Physics 212 Lecture 18, Slide 18 The switch in the circuit shown has been open for a long time. At t = 0, the switch is closed. Calculation
R1 V L R2 R3 IL(t=0+) = 0 What is the magnitude of I2, the current in R2, immediately after the switch is closed?
V V V I2 = (A) (B) (C) (D) I2 = I2 = R2 + R3 R1 + R2 + R3 R1
We know IL = 0 immediately after switch is closed Immediately after switch is closed, circuit looks V like: R1
I= I2 = VR2 R3 R2 + R3
R2 R3 I V R1 + R2 + R3 Physics 212 Lecture 18, Slide 19 The switch in the circuit shown has been open for a long time. At t = 0, the switch is closed. Calculation
R1 V L R2 I2 R3 What is the magnitude of VL, the voltage across the inductor, immediately after the switch is closed?
3 VL = VL = 0 VL = V 2 VL = V (A) (B) (C) (D) (E)V R ( R2 +3 R ) R1 1 2 3 IL(t=0+) = 0 I2(t=0+) = V/(R1+R2+R3) R +R RR VL = V R2 + R3 R1 + R2 + R3 Kirchhoff's Voltage Law, VLI2 R2 I2 R3 =0 VL = I2 (R2+R3)
VL = V ( R2 + R3 ) R1 + R2 + R3 Physics 212 Lecture 18, Slide 20 The switch in the circuit shown has been open for a long time. At t = 0, the switch is closed. What is dIL/dt, the time rate of change of the current through the inductor immediately after switch is closed Calculation
R1 V L R2 R3 VL(t=0+) = V(R2+R3)/(R1+R2+R3) dI L V R2 + R3 dI L dI L V R + R3 = = (A) (B) (C) (D) 2 =0 dt L R1 dt L R1 + R2 + R3 dt
The time rate of change of current through the inductor (dIL /dt) = VL /L dI L V = dt L dI L V R2 + R3 = dt L R1 + R2 + R3
Physics 212 Lecture 18, Slide 21 The switch in the circuit shown has been closed for a long time. What is I2, the current through R2 ? (Positive values indicate current flows to the right) Follow Up
R1 V L R2 R3 V V ( R2 R3 ) I2 = + (A) (B) (C) (D) I2 = + I2 = 0 R2 + R3 R + R2 + R3
1 I2 =  V R2 + R3 After a long time, dI/dt = 0 Therefore, the voltage across L = 0 Therefore the voltage across R2 + R3 = 0 Therefore the current through R2 + R3 must be zero !! Physics 212 Lecture 18, Slide 22 The switch in the circuit shown has been closed for a long time at which point, the switch is opened. What is I2, the current through R2 immediately after switch is opened ? (Positive values indicate current flows to the right) Follow Up 2
R1 V IL L R3 R2 I2 (A) V V I2 = + I2 = 0 I2 = + (B) (C) (D) (E) R1 R1 + R2 + R3 I2 =  V R1 I2 =  V R1 + R2 + R3 Current through inductor immediately AFTER switch is opened IS THE SAME AS the current through inductor immediately BEFORE switch is opened Immediately BEFORE switch is opened: IL = V/R1 Immediately AFTER switch is opened: IL flows in right loop Therefore, IL = V/R1
Physics 212 Lecture 18, Slide 23 ...
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 Fall '12
 Inductance, Trigraph, long time, Inductor, RC circuit, iL, RL circuit

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