1 Physics 7B - AB Lecture 2 April 10 Recap + examples Steady-State Energy Density Model Applied to Fluid Circuit & Electrical Circuit Lecture slides available.

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Physics 7B - ABLecture 2

April 10Recap + examples

Steady-State Energy Density ModelApplied to

Fluid Circuit & Electrical Circuit

Lecture slides available athttp://physics.ucdavis.edu/physics7

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Course Websitehttp://physics.ucdavis.edu/physics7

Click on Physics 7B-A/B

Next weekLecturer 3Dr. Kevin Klapstein

April 16

My office hr is cancelled.

Today Quiz 1!

Current I VS Fluid velocity v

Remember Energy Density Equation ?a.k.a. Fluid Transport Equation/Extended Bernoulli eq.

∆P + (1/2)∆(v2) + g∆h + IR = Epump/V

Current I is “ Volumetric flow rate”

How much water (in terms of volume) flows through a section of a pipe per unit time. So it has the unit of volume per second [m3/s].

Fluid velocity is how fast fluid is moving.

So it has the unit of distance per second. [velocity] = [m/s]

I vs v

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Recap on Fluids

Basic rules for looking at fluids:

1. Energy (density) conservation a.k.a. Energy Density Model

(P2– P1) + (1/2)(v22 – v1

2) + g( h2 – h1) + IR12 = 12

2. Current ConservationCurrent entering = current leaving

3. Pressures where two fluids systems touch are equal

4. Resistance is proportional to length

5. ALWAYS, pick two points within the SAME fluid system to apply 1 & 2!

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Current is shown flowing down the pipe shown. Choose the correct statement.

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(1) I1 = I2(2) I1 > I2(3) I1 < I2(4) Cannot tell

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Current is shown flowing down the pipe shown. Choose the correct statement.

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(1) I1 = I2(2) I1 > I2(3) I1 < I2(4) Cannot tell

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Current is shown flowing down the (R=0) pipe shown. Fluid is flowing from 1 to 2.Choose the correct statement: 1

2

(1) I1 = I2(2) I1 > I2(3) I1 < I2(4) Cannot tell

Pumping this way

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Current is shown flowing down the (R=0) pipe shown. Fluid is flowing from 1 to 2.Choose the correct statement: 1

2

(1) I1 = I2(2) I1 > I2(3) I1 < I2(4) Cannot tell

Pumping this way

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Current is shown flowing down the pipe shown. Choose the correct statement.

12(1) v1 = v2

(2) v1 > v2

(3) v1 < v2

(4) Cannot tell

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Current is shown flowing down the pipe shown. Choose the correct statement.

12(1) v1 = v2

(2) v1 > v2

(3) v1 < v2

(4) Cannot tell

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Current is shown flowing down the (R=0) pipe shown. Fluid is flowing from 1 to 2.Choose the correct statement: 1

2Pumping this way(1) v1 = v2

(2) v1 > v2

(3) v1 < v2

(4) Cannot tell

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Current is shown flowing down the (R=0) pipe shown. Fluid is flowing from 1 to 2.Choose the correct statement: 1

2Pumping this way(1) v1 = v2

(2) v1 > v2

(3) v1 < v2

(4) Cannot tell

What energy system changed ?

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The following shows a pipe without dissipation (R=0) with a pump attached. On which side is the pressure greater?

1 2(1) Side 1(2) Side 2(3) Not enough information

Pumping this way

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The following shows a pipe without dissipation (R=0) with a pump attached. On which side is the pressure greater?

1 2(1) Side 1(2) Side 2(3) Not enough information

Pumping this way

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The following shows a resistanceless pipe (R=0) with a pump attached. On which side is the pressure greater?

1

2

(1) Side 1(2) Side 2(3) Need to know (4) Even knowing would not be enough info

Pumping this way

* (The change in height is 2 m from 1 to 2)

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The following shows a resistanceless pipe (R=0) with a pump attached. On which side is the pressure greater?

1

2

(1) Side 1(2) Side 2(3) Need to know (4) Even knowing would not be enough info

Pumping this way

* (The change in height is 2 m from 1 to 2)

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How does P1, P2 , P3

compare?

Assume no dissipation, i.e., R = 0.

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Same height (approximately)1,3 have the same speed (same area, current conserved)

2 is faster (smaller area, current conserved)

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No pumps between 1 and 3Assume no resistance between 1 and 3.

This implies P1 = P3 > P2.

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1 32

Small amount of resistance modifies this.

Pressures are P1 > P3 > P2

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1 32

Large amount of resistance modifies this.

Pressures are P1 > P2 > P3

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ElectricalElectrical circuits circuits

Recap and Examples on

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Analogies between Fluid and Electrical circuits

FluidsFluids ElectricityElectricityDeals with Fluid flow Charge flow

CurrentWater flowing

through a cross-section of pipe per

unit time

Electric charge flowing through a wire per unit time

v Indicator of kinetic energy

(neglected)

hIndicator of

gravitational potential energy

(neglected)

R Relates current and thermal energy loss

Relates current and thermal energy loss

Energy stored /unit volume

or chargePressure Voltage

(P2– P1) + (1/2)(v22 – v1

2) + g( h2 – h1) + IR12 = 12

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Analogies between Fluid and Electrical circuits

FluidsFluids ElectricityElectricityDeals with Fluid flow Charge flow

CurrentWater flowing

through a cross-section of pipe per

unit time

Electric charge flowing through a wire per unit time

v Indicator of kinetic energy

(neglected)

hIndicator of

gravitational potential energy

(neglected)

R Relates current and thermal energy loss

Relates current and thermal energy loss

Energy stored /unit volume

or chargePressure Voltage

(V2– V1) + (1/2)(v22 – v1

2) + g( h2 – h1) + IR12 = 12

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Analogies between Fluid and Electrical circuits

FluidsFluids ElectricityElectricityDeals with Fluid flow Charge flow

CurrentWater flowing

through a cross-section of pipe per

unit time

Electric charge flowing through a wire per unit time

v Indicator of kinetic energy

(neglected)

hIndicator of

gravitational potential energy

(neglected)

R Relates current and thermal energy loss

Relates current and thermal energy loss

Energy stored /unit volume

or chargePressure Voltage

Energy Density Equationa.k.a. transport equation

(V2– V1) + IR12 = 12

Or ∆V = – IR

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Energy Density Equation applied to Electrical Circuits

a.k.a. transport equation, loop rule

(V2– V1) + IR12 = 12

Or ∆V = – IR

The change in the electrical potential energy per charge, (what we call voltage, or voltage drop), as we move from one point to another point will increase due to energy added by a battery or generator and will decrease due to the transfer of electric potential energy per charge to thermal energy system.

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Current in = current out

From rest of circuit

To rest of

circuit1

2

3

4AA BB

Current into A = I1Current out of A = I2 + I3

Current into B = I2 + I3Current out of B = I4

II11 = I = I22 + I + I33

II22 + I + I3 3 = I= I44 = = II11

Current (charge) conservation

a.k.a. junction rule (everything that flows into a junction must be equal to everything that flows out)

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Electric PowerElecric energy is useful to us because it can be easily

transformed into other forms of energy, e.g. hair dryer, lightbulb…

Power: rate at which energy is transformed by an electric device = Energy transformed/unit time

Ex. 120Watt = 120 Joule/sec

1200 Watts

40 - 120Watts 5000 - 8000 Watts

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Electric PowerElecric energy is useful to us because it can be easily

transformed into other forms of energy, e.g. hair dryer, lightbulb…

Power = (∆V) x I = I2R

1200 Watts

40 - 120Watts

Bright ness of a bulb

5000 - 8000 Watts

Rhair dryer

Rclothes dryer

Rbulb

120V

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1. Bulbs 1 and 2 are connected with a battery as shown to the right. The bulbs have different resistances (R 1 ? R2). Which

statement must ?be true

)A , The two bulbs have the same voltage difference? , V across them and that ? V is half the

battery voltage . B) The two bulbs are equally bright . )C The two bulbs and the battery all have the same

current through them . )D .All of the above must be true )E N one of the above must be true .

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R1 R2Bulbs 1 and 2 are connected with a battery as shown to the right. The bulbs have different Resistances. Which statement must be true?

(1) The two bulbs have the same voltage different, ∆V, across them and that ∆V is half the battery voltage.

(2) The two bulbs are equally bright.

(3) The two bulbs and the battery have all the same current through them.

(4) All of the above must be true.

(5) None of the above must be true.

Note : For electrical circuits, wires connecting circuit elements are assumed to have zero resistance.

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1. Bulbs 1 and 2 are connected with a battery as shown to the right. The bulbs have different resistances (R 1 ? R2). Which

statement must ?be true

)A , The two bulbs have the same voltage difference? , V across them and that ? V is half the

battery voltage . B) The two bulbs are equally bright . )C The two bulbs and the battery all have the same

current through them . )D .All of the above must be true )E N one of the above must be true .

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R1 R2Bulbs 1 and 2 are connected with a battery as shown to the right. The bulbs have different Resistances. Which statement must be true?

(1) The two bulbs have the same voltage different, ∆V, across them and that ∆V is half the battery voltage.

(2) The two bulbs are equally bright.

(3) The two bulbs and the battery have all the same current through them.

(4) All of the above must be true.

(5) None of the above must be true.

Note : For electrical circuits, wires connecting circuit elements are assumed to have zero resistance.

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6. If the four light bulbs in the figure are identical and the batteries are identical, which circuit puts out more light? (Note: brightness depends on the power used so it is good enough to just determine which bulbs have the largest current?)

A) 1 emi ts more light. B) The two emit the same amount

of light. C) 2 emits more light. D) True E) False

1 2

If the four bulbs in the figure are identical and the batteries are identical, which circuit puts out more light?

(Remember brightness depends on the power transformed)

(1) 1 emits more light.

(2) The two emit the same amount of light.

(1) 2 emits more light.

R

R

R R

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6. If the four light bulbs in the figure are identical and the batteries are identical, which circuit puts out more light? (Note: brightness depends on the power used so it is good enough to just determine which bulbs have the largest current?)

A) 1 emi ts more light. B) The two emit the same amount

of light. C) 2 emits more light. D) True E) False

1 2

(Remember brightness depends on the power transformed)

From today’s lecture,

For circuit1: R1equivalent = R/2,

so I1 = /R1equi. = (2 )/R

As bulbs are identical, half of I1

flows through each bulb.

Power bulb in circuit 1 = ((I1/2)2R = { /R}2R = 2/R <= gives the measure of brightness of a bulb in circuit 1

For circuit2: R2equivalent = 2R,

so I2 = /R2equi. = /(2R) and this I2 flows through both bulbs.

Power bulb in circuit 2 = I22R = ( /2R)2R = 2/4R <= gives the measure of brightness of a bulb in circuit 2

If the four bulbs in the figure are identical and the batteries are identical, which circuit puts out more light?

R

R

R R

* For a review of circuit analysis, refer to P.21 of the course text (Method of Equivalent Reduction)

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6. If the four light bulbs in the figure are identical and the batteries are identical, which circuit puts out more light? (Note: brightness depends on the power used so it is good enough to just determine which bulbs have the largest current?)

A) 1 emi ts more light. B) The two emit the same amount

of light. C) 2 emits more light. D) True E) False

1 2

If the four bulbs in the figure are identical and the batteries are identical, which circuit puts out more light?

(Remember brightness depends on the power transformed)

(1) 1 emits more light.

(2) The two emit the same amount of light.

(1) 2 emits more light.

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6. If the four light bulbs in the figure are identical and the batteries are identical, which circuit puts out more light? (Note: brightness depends on the power used so it is good enough to just determine which bulbs have the largest current?)

A) 1 emi ts more light. B) The two emit the same amount

of light. C) 2 emits more light. D) True E) False

1 2

If the two bulbs in the circuit 1 are NOT identical,

which bulb puts out more light?

(1) Bulb A emits more light.

(2) The two emit the same amount of light.

(1) Bulb B emits more light.

RA

RB

(Remember brightness depends on the power transformed)

RA > RB

Example Solve this for = 6V, RA= 3 Ohm, RB= 1 Ohm

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6. If the four light bulbs in the figure are identical and the batteries are identical, which circuit puts out more light? (Note: brightness depends on the power used so it is good enough to just determine which bulbs have the largest current?)

A) 1 emi ts more light. B) The two emit the same amount

of light. C) 2 emits more light. D) True E) False

1 2

If the two bulbs in the circuit 1 are NOT identical,

which bulb puts out more light?

(1) Bulb A emits more light.

(2) The two emit the same amount of light.

(1) Bulb B emits more light.

RA

RB

(Remember brightness depends on the power transformed)

RA > RB

Example Solve this for = 6V, RA= 3 Ohm, RB= 1 Ohm

You should find, IA = 2A, IB = 6A, and

PA = 12Watts, PB = 36Watts

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Be sure to write your name, ID number & DL section!!!!!1 MR 10:30-12:50 Dan Phillips

2 TR 2:10-4:30 Abby Shockley

3 TR 4:40-7:00 John Mahoney

4 TR 7:10-9:30 Ryan James

5 TF 8:00-10:20 Ryan James

6 TF 10:30-12:50 John Mahoney

7 W 10:30-12:50 Brandon Bozek

7 F 2:10-4:30 Brandon Bozek

8 MW 8:00-10:20 Brandon Bozek

9 MW 2:10-4:30 Chris Miller

10 MW 4:40-7:00 Marshall Van Zijll

11 MW 7:10-9:30 Marshall Van Zijll