Physics 212 Lecture 18 Slide 1 From the

Physics 212 Lecture 18, Slide 1

From the prelecture: Self Inductance Wrap a wire into a coil to make an “inductor”… e = -L d. I dt Physics 212 Lecture 18, Slide 2

What this really means: emf induced across L tries to keep I constant e. L = -L d. I dt L current I Inductors prevent discontinuous current changes ! It’s like inertia! Physics 212 Lecture 18, Slide 3

I(t) e = -L d. I dt Suppose d. I/dt > 0. Induced EMF tries to counteract this change (Lenz’s Law). e + I(t) - V 1 V 2 + - Voltage across inductor V 1 – V 2 = VL = + L d. I/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 (1/2)2 2 Compare the inductance of the two solenoids 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 i-clicker Physics 212 Lecture 18, Slide 6

How to think about RL circuits Episode 1: When no current is flowing initially: VL I=0 L I=V/R R L t = L/R R I VBATT At t = 0: I=0 VL = VBATT VR = 0 (L is like a giant resistor) VBATT t = L/R At t >> L/R: VL = 0 VR = VBATT I = VBATT/R (L is like a short circuit) Physics 212 Lecture 18, Slide 7

Checkpoint 2 a In the circuit, the switch has been open for a long time, and the current is zero everywhere. I At time t=0 the switch is closed. What is the current I through the vertical resistor immediately after the switch is closed? I IL=0 (+ is in the direction of the arrow) A) I = V/R B) I = V/2 R C) I = 0 D) I = -V/2 R E) I = -V/R Before: IL = 0 After: IL = 0 I = + V/2 R 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/2 R C) I = 0 D) I = -V/2 R E) I = -V/R After a long time in any static circuit: VL = 0 - + + - KVR: VL + IR = 0 Physics 212 Lecture 18, Slide 9

Checkpoint 2 b 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/2 R C) I = 0 D) I = -V/2 R E) I = -V/R Current through inductor cannot change DISCONTINUOUSLY circuit when switch opened L IL=V/R R Physics 212 Lecture 18, Slide 10

Why is there exponential behavior ? I 2 V 1 – V 2 = L d. I dt L 1 VL 3 R t = L/R V 3 -V 4 = IR 4 t = L/R where Physics 212 Lecture 18, Slide 11

I L VL R VBATT t = L/R Lecture: Prelecture: Did we mess up? ? No: The resistance is simply twice as big in one case. Physics 212 Lecture 18, Slide 12

Checkpoint 3 a 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 Physics 212 Lecture 18, Slide 13

Checkpoint 3 b 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 C) The same Before switch moved: Physics 212 Lecture 18, Slide 14

Checkpoint 3 c 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 C) The same Immediately after: Physics 212 Lecture 18, Slide 15

Calculation The switch in the circuit shown has been open for a long time. At t = 0, the switch is closed. What is d. IL/dt, the time rate of change of the current through the inductor immediately after switch is closed R 1 V R 2 L R 3 • 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 d. IL/dt immediately after switch is closed. Physics 212 Lecture 18, Slide 16

Calculation The switch in the circuit shown has been open for a long time. At t = 0, the switch is closed. R 1 V R 2 L IL = 0 R 3 What is IL, the current in the inductor, immediately after the switch is closed? (A) IL =V/R 1 up (B) IL =V/R 1 down (C) IL = 0 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 17

Calculation The switch in the circuit shown has been open for a long time. At t = 0, the switch is closed. R 1 V R 2 L R 3 IL(t=0+) = 0 What is the magnitude of I 2, the current in R 2, immediately after the switch is closed? (A) (B) (C) (D) We know IL = 0 immediately after switch is closed R 1 Immediately after switch is closed, V circuit looks like: I R 2 R 3 Physics 212 Lecture 18, Slide 18

Calculation The switch in the circuit shown has been open for a long time. At t = 0, the switch is closed. R 1 V IL(t=0+) = 0 R 2 L I 2 R 3 I 2(t=0+) = V/(R 1+R 2+R 3) What is the magnitude of VL, the voltage across the inductor, immediately after the switch is closed? (A) (B) (C) (D) (E) Kirchhoff’s Voltage Law, VL-I 2 R 2 -I 2 R 3 =0 VL = I 2 (R 2+R 3) Physics 212 Lecture 18, Slide 19

Calculation The switch in the circuit shown has been open for a long time. At t = 0, the switch is closed. What is d. IL/dt, the time rate of change of the current through the inductor immediately after switch is closed (A) (B) (C) R 1 V R 2 L R 3 VL(t=0+) = V(R 2+R 3)/(R 1+R 2+R 3) (D) The time rate of change of current through the inductor (d. I L /dt) = VL /L Physics 212 Lecture 18, 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) (A) (B) R 1 V R 2 L (C) R 3 (D) After a long time, d. I/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 !! Physics 212 Lecture 18, Slide 21

Follow Up 2 The switch in the circuit shown has been closed for a long time at which point, the switch is opened. R 1 (A) (B) (C) I 2 IL V What is I 2, the current through R 2 immediately after switch is opened ? (Positive values indicate current flows to the right) R 2 L (D) R 3 (E) 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/R 1 Immediately AFTER switch is opened: IL flows in right loop Therefore, IL = -V/R 1 Physics 212 Lecture 18, Slide 22