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Physics 1308: General Physics II - Professor Jodi Cooley Gustav Robert Kirchhoff 12 March 1824 – 17 October 1887 Circuits Welcome Back to Physics 1308

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Physics 1308: General Physics II - Professor Jodi Cooley

Gustav Robert Kirchhoff 12 March 1824 – 17 October 1887

Circuits

Welcome Back to Physics 1308

Physics 1308: General Physics II - Professor Jodi Cooley

Announcements• Assignments for Thursday, October 4th:

- Reading: Chapter 27.2

- Watch Video: https://youtu.be/4Eh-mrHkUpY — Lecture 12 - Simple and Complex Circuits

• Homework 7 Assigned - due before class on Thursday, October 11th.

• No office hours or class during Fall break Monday, October 8th and Tuesday, October 9th due to Fall break.

• Midterm Exam 2 will be in class on Tuesday, October 16th. It will explicitly cover chapters 24 - 26. However, these chapters build on the previous material and you will be expected to apply all concepts/information from the beginning of class to this point to any problem or question that you encounter on the exam.

Physics 1308: General Physics II - Professor Jodi Cooley

Physics 1308: General Physics II - Professor Jodi Cooley

Optical Tweezers• Arthur Ashkin invented optical tweezers that grab particles, atoms, viruses and

other living cells with their laser beam fingers.

• This new tool allowed Ashkin to realise an old dream of science fiction – using the radiation pressure of light to move physical objects. He succeeded in getting laser light to push small particles towards the centre of the beam and to hold them there. Optical tweezers had been invented.

• A major breakthrough came in 1987, when Ashkin used the tweezers to capture living bacteria without harming them. He immediately began studying biological systems and optical tweezers are now widely used to investigate the machinery of life.

Physics 1308: General Physics II - Professor Jodi Cooley

Chirp Pulse Amplification• Gérard Mourou and Donna Strickland paved the way towards the shortest and most intense laser pulses

ever created by mankind. • Their revolutionary article was published in 1985 and was the foundation of Strickland’s doctoral thesis. • Using an ingenious approach, they succeeded in creating ultrashort high-intensity laser pulses without

destroying the amplifying material. First they stretched the laser pulses in time to reduce their peak power, then amplified them, and finally compressed them. If a pulse is compressed in time and becomes shorter, then more light is packed together in the same tiny space – the intensity of the pulse increases dramatically.

• Strickland and Mourou’s newly invented technique, called chirped pulse amplification, CPA, soon became standard for subsequent high-intensity lasers. Its uses include the millions of corrective eye surgeries that are conducted every year using the sharpest of laser beams.

• The innumerable areas of application have not yet been completely explored. However, even now these celebrated inventions allow us to rummage around in the microworld in the best spirit of Alfred Nobel – for the greatest benefit to humankind

Physics 1308: General Physics II - Professor Jodi Cooley

Physics 1308: General Physics II - Professor Jodi Cooley

Review Question 1Consider a capacitor made of two parallel metallic plates separated by a distance d. The top plate has a surface charge density +σ, the bottom -σ. Α slab of metal of thickness l < d is inserted between the plates, not connected to either one. Upon insertion of the metal slab, the potential difference between the plates

A) increases B) decreases C) remains the same

Because the upper face of the slab becomes negatively charged and the lower face positively charged, they are attracted to the capacitor plates. We must do work on the system if we wish to remove the slab. Thus, the system has a lower potential energy when the slab is in place, and the potential difference between the plates is lower.

Physics 1308: General Physics II - Professor Jodi Cooley

Review Key Concepts• Electric charge exists in two kinds, positive and negative. These exert influence on

each other via a field of force.

• The language of energy applies equally to the electric force and its field, which are conservative (energy is conserved in a closed system and there is a potential energy associated with the field).

• An electric potential difference (a voltage) causes charges to accelerate, and can be used to establish electric currents.

• Currents are not free to move without resistance in the medium in which they travel, due to collisions with atoms. A material that obeys “Ohm's Law” has a proportional relationship between the applied voltage and the established current.

Physics 1308: General Physics II - Professor Jodi Cooley

Key Conceptsemf device:

a device that—by doing work on the charge carriers—maintains a potential difference between a pair of terminals

- the positive terminal (+) is at a higher electric potential than the negative terminal (-).

- The emf of a device is represented with an arrow that points from the negative terminal toward the positive terminal.

- The arrow indicates the flow of current in the circuit.

Physics 1308: General Physics II - Professor Jodi Cooley

Key ConceptsCurrent in a Single Loop Circuit:

Physics 1308: General Physics II - Professor Jodi Cooley

Key ConceptsResistors in Series:

or

Physics 1308: General Physics II - Professor Jodi Cooley

Question 1What should be the direction of the emf arrow across the “battery” in this circuit?

A) upward B) downward C) rightward D) leftward E) not enough information is given

Physics 1308: General Physics II - Professor Jodi Cooley

Question 1If R1 > R2 > R3, rank the three resistances according to the current through them, greatest first.

The current is the same for all resistances connected in series. Thus, i1 = i2 = i3.

Physics 1308: General Physics II - Professor Jodi Cooley

Question 2If R1 > R2 > R3, rank the potential difference across them, greatest first.

V1 > V2 > V3

Remember, the battery drives current through the resistor from high potential to low potential. (In the battery, current goes in the opposite direction from low potential to high potential).

Physics 1308: General Physics II - Professor Jodi Cooley

Instructor Problem: Double Battery CircuitIn the figure the ideal batteries have emfs ε1 = 14 V and ε2 = 9.0 V and the resistors have resistances R1 = 4.4 Ω and R2 = 8.3 Ω.

a) What is the current in resistors 1 and 2? b) What is the dissipation rate in resistor 1? c) What is the dissipation rate in resistor 2? d) What is the energy transfer rate in

battery 1? e) What is the energy transfer rate in

battery 2? f) Is energy being supplied or absorbed

battery 1? g) Is energy being supplied or absorbed

by battery 2?

Physics 1308: General Physics II - Professor Jodi Cooley

Part A: What is the current in resistors 1 and 2?

First, make an assumption about direction for the current.

ii =

emf

R

We know that the current is given by the equation

Req = R1 +R2

Since the the resistors are in series

Putting it together, we have,

i =✏2 � ✏1R1 +R2

=9 V � 14 V

4.4 ⌦+ 8.3 ⌦= �0.39 A

The negative sign means the current is actually flowing in the opposite direction then what we assumed.

i = 0.39 A CCW

Physics 1308: General Physics II - Professor Jodi Cooley

Part B: What is the dissipation rate in resistor 1?

The dissipation rate refers to the power dissipated and it is given by the equation

P = i2R = (0.39 A)2(4.4 ⌦)

P1 = 0.68 W

Part C: What is the dissipation rate in resistor 2?

Similar to part B we have

P = i2R = (0.39 A)2(8.3 ⌦)

P2 = 1.3 W

Physics 1308: General Physics II - Professor Jodi Cooley

Part D: What is the energy transfer rate in battery 1?

In the battery, the energy transfer rate is given by the formula

P = i✏

Note that if current and emf are in the same direction, then energy is supplied by the battery at a rate P. However, if current and emf are in opposite directions then energy is absorbed by the battery at a rate P.

Applying this to our problem —

P1 = i✏1

Part E: What is the energy transfer rate in battery 2?

P2 = i✏2

Similar to part D

�! P1 = 5.5 W

�! P2 = 3.5 W

= (0.39 A)(14 V )

= (0.39 A)(9 V )

Physics 1308: General Physics II - Professor Jodi Cooley

Part F: Is energy being supplied or absorbed battery 1?

In battery 1 the current is in the same direction as the emf. Therefore, this battery supplies energy to the circuit.

Part G: Is energy being supplied or absorbed battery 1?

In battery 1 the current is in the opposite direction as the emf. Therefore, this battery absorbs energy from the circuit.

Physics 1308: General Physics II - Professor Jodi Cooley

Student Problem

A) I n t h e f i g u r e w h a t v a l u e must R have if the current in the circuit is to be 0.93 mA? Take ε1 = 2.5 V, ε2 = 3.8 V, and r1 = r2 = 3.8 Ω.

B) What is the rate at which thermal energy appears in R?

Physics 1308: General Physics II - Professor Jodi Cooley

Part A:

First, make an assumption about direction for the current.

ii =

emf

R

We know that the current is given by the equation

Since the the resistors are in seriesReq = r1 + r2 +R

Apply to our problem

i =emf

Req=

✏2 � ✏1r1 + r2 +R

�! R =✏2 � ✏1

i� r1 � r2

=3.8 V � 2.5 V

0.93⇥ 10�3 A� 3.8 ⌦� 3.8 ⌦

R = 1400 ⌦

Physics 1308: General Physics II - Professor Jodi Cooley

Part B:

Thermal energy dissipated is given by the formula

P = i2R = (0.93⇥ 103 A)2(1400 ⌦)

P = 0.0012 W

Physics 1308: General Physics II - Professor Jodi Cooley

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