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## Lecture slides available at physics.ucdavis/physics7

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**Physics 7B - ABLecture 2April 10Recap + examplesSteady-State**Energy Density ModelApplied to Fluid Circuit & Electrical Circuit Lecture slides available athttp://physics.ucdavis.edu/physics7**Course Website**http://physics.ucdavis.edu/physics7 Click on Physics 7B-A/B Today Quiz 1! Next week Lecturer 3 Dr. Kevin Klapstein April 16 My office hr is cancelled.**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]. I vs v Fluid velocity is how fast fluid is moving. So it has the unit of distance per second. [velocity] = [m/s]**Recap on Fluids**Basic rules for looking at fluids: Energy (density) conservation a.k.a. Energy Density Model (P2– P1) + (1/2)(v22 –v12) + g( h2 –h1) + IR12 = 12 2. Current Conservation Current 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!**Current is shown flowing down the pipe shown. Choose the**correct statement. 1 2 (1) I1 = I2 (2) I1 > I2 (3) I1 < I2 (4) Cannot tell**Current is shown flowing down the pipe shown. Choose the**correct statement. 1 2 (1) I1 = I2 (2) I1 > I2 (3) I1 < I2 (4) Cannot tell**Current is shown flowing down the (R=0) pipe shown. Fluid is**flowing from 1 to 2.Choose the correct statement: 1 2 Pumping this way (1) I1 = I2 (2) I1 > I2 (3) I1 < I2 (4) Cannot tell**Current is shown flowing down the (R=0) pipe shown. Fluid is**flowing from 1 to 2.Choose the correct statement: 1 2 Pumping this way (1) I1 = I2 (2) I1 > I2 (3) I1 < I2 (4) Cannot tell**Current is shown flowing down the pipe shown. Choose the**correct statement. 1 2 (1) v1 = v2 (2) v1 > v2 (3) v1 < v2 (4) Cannot tell**Current is shown flowing down the pipe shown. Choose the**correct statement. 1 2 (1) v1 = v2 (2) v1 > v2 (3) v1 < v2 (4) Cannot tell**Current is shown flowing down the (R=0) pipe shown. Fluid is**flowing from 1 to 2.Choose the correct statement: 1 2 Pumping this way (1) v1 = v2 (2) v1 > v2 (3) v1 < v2 (4) Cannot tell**Current is shown flowing down the (R=0) pipe shown. Fluid is**flowing from 1 to 2.Choose the correct statement: 1 2 Pumping this way (1) v1 = v2 (2) v1 > v2 (3) v1 < v2 (4) Cannot tell What energy system changed ?**Pumping this way**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**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**(1) Side 1**(2) Side 2 (3) Need to know (4) Even knowing would not be enough info 2 The following shows a resistanceless pipe (R=0) with a pump attached. On which side is the pressure greater? 1 Pumping this way * (The change in height is 2 m from 1 to 2)**(1) Side 1**(2) Side 2 (3) Need to know (4) Even knowing would not be enough info 2 The following shows a resistanceless pipe (R=0) with a pump attached. On which side is the pressure greater? 1 Pumping this way * (The change in height is 2 m from 1 to 2)**1**3 2 Assume no dissipation, i.e., R = 0. How does P1, P2 , P3 compare?**1**3 2 Same height (approximately)1,3 have the same speed (same area, current conserved) 2 is faster (smaller area, current conserved) No pumps between 1 and 3Assume no resistance between 1 and 3.This implies P1 = P3 > P2.**1**3 2 Small amount of resistance modifies this. Pressures are P1 > P3 > P2**1**3 2 Large amount of resistance modifies this. Pressures are P1 > P2 > P3**Recap and Examples on**Electrical circuits**Analogies between Fluid and Electrical circuits**(P2– P1) + (1/2)(v22 –v12) + g( h2 –h1) + IR12 = 12**Analogies between Fluid and Electrical circuits**(V2– V1) + (1/2)(v22 –v12) + g( h2 –h1) + IR12 = 12**Analogies between Fluid and Electrical circuits**(V2– V1) + IR12 = 12 Or ∆V = – IR Energy Density Equation a.k.a. transport equation**Energy Density Equation applied to Electrical Circuitsa.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.**2**Current into A = I1 Current out of A = I2 + I3 Current in = current out I1 = I2 + I3 4 From rest of circuit To rest of circuit 1 3 Current into B = I2 + I3 Current out of B = I4 I2 + I3 = I4 = I1 Current (charge) conservationa.k.a. junction rule(everything that flows into a junction must be equal to everything that flows out) A B**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 5000 - 8000 Watts 40 - 120Watts**Electric PowerElecric energy is useful to us because it can**be easily transformed into other forms of energy, e.g. hair dryer, lightbulb… 120V Power = (∆V) x I = I2R Rbulb Rhair dryer 1200 Watts Rclothes dryer 5000 - 8000 Watts 40 - 120Watts Bright ness of a bulb**Bulbs 1 and 2 are connected with a battery as shown to the**right. The bulbs have different Resistances. Which statement must be true? R1 R2 1 2 The two bulbs have the same voltage different, ∆V, across them and that ∆V is half the battery voltage. The two bulbs are equally bright. The two bulbs and the battery have all the same current through them. All of the above must be true. None of the above must be true. Note : For electrical circuits, wires connecting circuit elements are assumed to have zero resistance.**Bulbs 1 and 2 are connected with a battery as shown to the**right. The bulbs have different Resistances. Which statement must be true? R1 R2 1 2 The two bulbs have the same voltage different, ∆V, across them and that ∆V is half the battery voltage. The two bulbs are equally bright. The two bulbs and the battery have all the same current through them. All of the above must be true. None of the above must be true. Note : For electrical circuits, wires connecting circuit elements are assumed to have zero resistance.**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) R R R 1 emits more light. The two emit the same amount of light. 2 emits more light. R **If the four bulbs in the figure are identical and the**batteries are identical, which circuit puts out more light? * For a review of circuit analysis, refer to P.21 of the course text (Method of Equivalent Reduction) (Remember brightness depends on the power transformed) R R R 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 R **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 emits more light. The two emit the same amount of light. 2 emits more light.**If the two bulbs in the circuit 1 are NOT identical,**which bulb puts out more light? (Remember brightness depends on the power transformed) RA Bulb A emits more light. The two emit the same amount of light. Bulb B emits more light. RA > RB RB Example Solve this for = 6V, RA= 3 Ohm, RB= 1 Ohm**If the two bulbs in the circuit 1 are NOT identical,**which bulb puts out more light? (Remember brightness depends on the power transformed) RA Bulb A emits more light. The two emit the same amount of light. Bulb B emits more light. RA > RB RB Example Solve this for = 6V, RA= 3 Ohm, RB= 1 Ohm You should find, IA = 2A, IB = 6A, and PA = 12Watts, PB = 36Watts