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Generators Textbook Sections 23-6 – 23-10 Physics 1161:

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Presentation on theme: "Generators Textbook Sections 23-6 – 23-10 Physics 1161:"— Presentation transcript:

1 Generators Textbook Sections 23-6 – 23-10 http://www.walter-fendt.de/ph14e/electricmotor.htm http://www.walter-fendt.de/ph14e/generator_e.htm Physics 1161: Lecture 16

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: curl along B field Palm: out of page. B I F + v +++ I

3 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:

4 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

5 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

6 At which time does the loop have the greatest emf (greatest  /  t)? 1.1 2.2 3.3

7 At which time does the loop have the greatest emf (greatest  /  t)? 1.1 2.2 3.3 1) Has greatest flux, but  = 0 so  = 0. 2) (Preflight example)   30 so    AB/2. 3) Flux is zero, but  = 90 so  =  AB.

8 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

9 Preflights 16.1, 16.2, 16.3  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!   

10 Preflights 16.1, 16.2, 16.3  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!   

11 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

12 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?

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(  )  = NI A B sin(  ) Note: Emf is maximum at  =90 Note: Torque is maximum at  =90 = (40) (0.2m) 2 (0.15T) (2.5 radians/s) = 0.6 Volts  = 40*I0.15A*(0.2m) 2 * 0.15 T* 1 = 0.036 Newton-meters 

14 Power Transmission, Preflight 16.5 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?

15 Power Transmission, Preflight 16.5 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 = 1.2  10 -6  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.  of Cu = 10 -8  -m 1 inch square copper wire has about 0.1 ohm resistance in 7 km

16 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

17 Preflight 13.6 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 Phy1152 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

18 Preflight 13.6 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 Phy1161 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

19 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

20 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? Transformers depend on a change in flux so they only work for alternating currents! 1.V s = 0 2.V s = 6 3.V s = 12 4.V s = 24

21 Transformers Key to Modern electrical system Starting with 120 volts AC – Produce arbitrarily small voltages. – Produce arbitrarily large voltages. Nearly 100% efficient

22 In a transformer the side with the most turns always has the larger peak voltage. (T/F) 1.True 2.False

23 In a transformer the side with the most turns always has the larger peak current. (T/F) 1.True 2.False

24 In a transformer the side with the most turns always dissipates the most power. (T/F) 1.True 2.False


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