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Physics 102: Lecture 11, Slide 1 GeneratorsGenerators and Transformers Today’s lecture will cover Textbook Sections 20.2, 6 Physics 102: Lecture 11.

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Presentation on theme: "Physics 102: Lecture 11, Slide 1 GeneratorsGenerators and Transformers Today’s lecture will cover Textbook Sections 20.2, 6 Physics 102: Lecture 11."— Presentation transcript:

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2 Physics 102: Lecture 11, Slide 1 GeneratorsGenerators and Transformers Today’s lecture will cover Textbook Sections 20.2, 6 Physics 102: Lecture 11

3 Physics 102: Lecture 11, Slide 2 Review: Two uses of RHR’s Force on moving charge in Magnetic field –Thumb: v (or I) –Fingers: B –Palm: F on + charge Magnetic field produced by moving charges –Thumb: I (or v for + charges) –Fingers: where you want to know B –Palm: B Palm: out of page. B I F + v +++ Thumb: out Fingers: up Palm: left. x

4 Physics 102: Lecture 11, Slide 3 Review: Two uses of RHR’s Force on moving charge in Magnetic field –Thumb: v (or I) –Fingers: B –Palm: F on + charge Magnetic field produced by moving charges –Thumb: I (or v for + charges) –Fingers: curl along B field Palm: out of page. B I F + v +++ I

5 Physics 102: Lecture 11, Slide 4 Review: Induction Lenz’s Law –If the magnetic flux (  B ) through a loop changes, an EMF will be created in the loop to oppose the change in flux –EMF current (V=IR) additional B-field. Flux decreasing => B-field in same direction as original Flux increasing => B-field in opposite direction of original Faraday’s Law –Magnitude of induced EMF given by:

6 Physics 102: Lecture 11, Slide 5 Review: Rotation Variables v, , f, T Velocity (v): –How fast a point moves. –Units: usually m/s Angular Frequency (  ): –How fast something rotates. –Units: radians / sec  r v v v =  r f =  / 2  T = 1 / f = 2  /  Frequency ( f ): –How fast something rotates. –Units: rotations / sec = Hz Period (T): –How much time one full rotation takes. –Units: usually seconds

7 Physics 102: Lecture 11, Slide 6 Generators and EMF  rL = A  side 1 =  r B L sin(  )  side 2 =  r B L sin(  )  loop =  side 1 +  side 2  2  r B L sin(  )  loop =  A B sin(  )  loop =  A B sin(  t) v v x   r 1 2  t  AB  AB EMF is voltage!  side 1 = v B L sin(  ) v =  r

8 Physics 102: Lecture 11, Slide 7 ACT: Generators and EMF x  =  A B sin(  ) x x 123 At which time does the loop have the greatest emf (greatest  /  t)? 

9 Physics 102: Lecture 11, Slide 8 ACT: Generators and EMF x  =  A B sin(  ) x x 1 2 3 At which time does the loop have the greatest emf (greatest  /  t)? 1) Has greatest flux, but  = 0 so  = 0. 2) (Preflight example)   30 so    AB/2. 3) Flux is zero, but  = 90 so  =  AB. θ

10 Physics 102: Lecture 11, Slide 9 Comparison: Flux vs. EMF Flux is maximum –Most lines thru loop EMF is minimum –Just before: lines enter from left –Just after: lines enter from left –No change! Flux is minimum –Zero lines thru loop EMF is maximum –Just before: lines enter from top. –Just after: lines enter from bottom. –Big change! x x

11 Physics 102: Lecture 11, Slide 10 Preflights 11.1, 11.2, 11.3  v v x  r Flux is _________ at moment shown. Increasing decreasing not changing When  =30°, the EMF around the loop is: increasing decreasing not changing EMF is increasing!   

12 Physics 102: Lecture 11, Slide 11 Preflights 11.1, 11.2, 11.3  v v x  r Flux is decreasing at moment shown. When  =30°, the EMF around the loop is: increasing decreasing not changing EMF is increasing!   

13 Physics 102: Lecture 11, Slide 12 Generators and Torque v v x   r  =  A B sin(  ) Recall:  = A B I sin(  ) =  A 2 B 2 sin 2 (  )/R Torque, due to current and B field, tries to slow spinning loop down. Must supply external torque to keep it spinning at constant  Voltage! Connect loop to resistance R use I=V/R: I =  A B sin(  ) / R

14 Physics 102: Lecture 11, Slide 13 Generator v v x  A generator consists of a square coil of wire with 40 turns, each side is 0.2 meters long, and it is spinning with angular velocity  = 2.5 radians/second in a uniform magnetic field B=0.15 T. Determine the direction of the induced current at instant shown. Calculate the maximum emf and torque if the resistive load is 4 .  = NA B  sin(  ) Units?  = NI A B sin(  ) Units?

15 Physics 102: Lecture 11, Slide 14 Generator v v x  A generator consists of a square coil of wire with 40 turns, each side is 0.2 meters long, and it is spinning with angular velocity  = 2.5 radians/second in a uniform magnetic field B=0.15 T. Determine the direction of the induced current at instant shown. Calculate the maximum emf and torque if the resistive load is 4 .  = NA B  sin(  )  = NI A B sin(  ) Note: Emf is maximum at  =90 Note: Torque is maximum at  =90 = (40) (0.2) 2 (0.15) (2.5) = 0.6 Volts = N 2  A 2 B 2 sin 2 (  )/R = (40) 2 (2.5) (0.2) 4 (0.15) 2 /4 = 0.036 Newton-meters 

16 Physics 102: Lecture 11, Slide 15 Power Transmission, Preflight 11.5 A generator produces 1.2 Giga watts of power, which it transmits to a town 10 km away through copper power lines. How low does the line resistance need to be in order to consume less than 10% of the power transmitted from the generator at 120 Volts? I = Current leaving/returning to the generator Find I? R = Line resistance for 12 Megawatt loss in lines So why use high voltage lines?

17 Physics 102: Lecture 11, Slide 16 Power Transmission, Preflight 11.5 A generator produces 1.2 Giga watts of power, which it transmits to a town 10 km away through copper power lines. How low does the line resistance need to be in order to consume less than 10% of the power transmitted from the generator at 120 Volts? I = 10 7 P = I V so 1.2  10 9 = 120 I or I = 10 7 amps R = P = I 2 R so 1.2  10 8 = (10 7 ) 2 R or R = 1.2  10 -6  This would require a cable more than 40 feet in diameter!! Large current is the problem. Since P=IV, use high voltage and low current to deliver power.

18 Physics 102: Lecture 11, Slide 17 Transformers Increasing current in primary creates an increase in flux through primary and secondary.   iron VsVs VpVp Same  t Energy conservation! I p V p = I s V s R (primary) (secondary) NSNS NPNP Key to efficient power distribution

19 Physics 102: Lecture 11, Slide 18 Preflight 11.6 The good news is you are going on a trip to France. The bad news is that in France the outlets have 240 volts. You remember from P102 that you need a transformer, so you wrap 100 turns around the primary. How many turns should you wrap around the secondary if you need 120 volts out to run your hair dryer?   iron VsVs VpVp R 1) 502) 1003) 200 (primary) (secondary) NSNS NPNP

20 Physics 102: Lecture 11, Slide 19 Preflight 11.6 The good news is you are going on a trip to France. The bad news is that in France the outlets have 240 volts. You remember from P102 that you need a transformer, so you wrap 100 turns around the primary. How many turns should you wrap around the secondary if you need 120 volts out to run your hair dryer?   iron VsVs VpVp R 1) 502) 1003) 200 (primary) (secondary) NSNS NPNP

21 Physics 102: Lecture 11, Slide 20 ACT: Transformers iron VsVs VpVp R A 12 Volt battery is connected to a transformer that has a 100 turn primary coil, and 200 turn secondary coil. What is the voltage across the secondary after the battery has been connected for a long time? 1) V s = 0 2) V s = 6 3) V s = 12 4) V s = 24 (primary) (secondary) NSNS NPNP

22 Physics 102: Lecture 11, Slide 21 ACT: Transformers iron VsVs VpVp R A 12 Volt battery is connected to a transformer that has a 100 turn primary coil, and 200 turn secondary coil. What is the voltage across the secondary after the battery has been connected for a long time? 1) V s = 0 2) V s = 6 3) V s = 12 4) V s = 24 Transformers depend on a change in flux so they only work for alternating currents! (primary) (secondary) NSNS NPNP

23 Physics 102: Lecture 11, Slide 22 Transformers Key to Modern electrical system Starting with 120 volts AC –Produce arbitrarily small voltages. –Produce arbitrarily large voltages. Nearly 100% efficient !!!Volt!!

24 Physics 102: Lecture 11, Slide 23 In a transformer the side with the most turns always has the larger peak voltage. (T/F) In a transformer the side with the most turns always has the larger peak current. (T/F) In a transformer the side with the most turns always dissipates the most power. (T/F) Which of the following changes will increase the peak voltage delivered by a generator –Increase the speed it is spinning. –Increase the area of the loop. –Increase the strength of the magnetic field. Exam Prep Questions

25 Physics 102: Lecture 11, Slide 24 See you next time! Read Sections 20.9, 21.1, 3-4


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