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Published byLizbeth Carson Modified over 8 years ago
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e 10 cm 1cm e -e
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Calculate E y here.
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Cathode Ray Tube Conducting Paper +10 Volts 0 Volts A + B + C +
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ExEx EyEy + -
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+ V OUT V IN -
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V OUT V IN
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V=6 volts V=8 volts V=4 volts V=2 volts V=0 volts V=-2 volts = 1cm E=?
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6 V 5 V 4 V abc def ghi
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3 V A.B. C.
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3 C 4 C 3 cm -5 C
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3 C 4 C 3 cm -5 C q
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3 C 4 C 3 cm -5 C q
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3 V 6 V 1,000
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Baseball Diamond Heuristic of Electrostatics Equations
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(Usually easier to use Gauss’s Law.) (V POINT not really useful. V usually measured or otherwise derived.) (scalar) (vector) (Usually change in electrical potential is related to work on the particle.)
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3 V B. C.
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3 V 1 2 3
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V R1R1 R2R2
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V R DMM V DMM V R1R1 R DMM V DMM
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3 V 1 2 3 1
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100 200 100 10
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100 200 100 10 V
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V R1R1 R2R2
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3 V 1 2 3 1 2
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3 V 1 2 3 1 2 1
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9 V R 1 = 1 R 2 = 2 R 3 = 3 R 4 =4 I 4 =? V 4 =? I 1 =? V 1 =?I 2 =? V 2 =? I 3 =? V 3 =? I Battery =? R effective =?
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SCOPESCOPE tV
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t V motor t V resistor T t1t1 t2t2 V motor,on V resistor,on on off on off
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OSCOPEOSCOPE Voltage (0.5 volts per div) Time (1 second per div)
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R C + - red1 bottom ground red2
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R V source (t) = V MAX sin( t+ ) where V MAX = 5 Volts /(2 ) = 1,000 Hz
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V source (t) = V MAX sin( t+ ) where V MAX = 5 Volts /(2 ) = 1,000 Hz
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V amp =3 V 330 CH1CH2 red1 red2 bottom ground x-y mode
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200 100 red 1 red 2 (inverted) middle ground + - 200 100 red 1 bottom ground red 2 + -
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R C V source (t) = V MAX sin( t+ )
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SCOPESCOPE
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I I I Magnet B B Close is strong B Far is ~ zero Magnet B I L
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I I I I I I Beginning Position180 o Rotated Position current direction reversed (so is force on wire)
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I I I Alternating Contacts current direction always the same (so is force on wire) DC Power Supply + - these wires fixed brushes allow good contact as loop rotates
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I A. N I B. S I C. N S S NS N N SS N
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4 V 0.5 A B 1.5 2.5 4 4 V20 V
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12 V 1 2 3 1 2 1 2 BATTERY A B
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6 1 2 S 2 F 6 V
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Direction of I ? NS V velocity
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NS Direction of I ?
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NS V velocity
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SN Direction of I ?
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V receiver, amplitude f f maximum transmission
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I resistor,amplitude f drive f resonance
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R = 2,000 C = 15 F V source amplitude = 15 V L = 75 mH f drive = 750 Hz
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R [Ohm] C [Farad] V source L [Henry]
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10 L
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C L
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I resistor,amplitude f drive f resonance Same L and C with lower R
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L R red 1 red 2 ground C R red 1 ground red 2
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C R red 1 ground red 2 + - C R V source (t)=V source amp sin( D t) + -
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Pulses let through by the diode move speaker with frequency of desired audio wave. Quantum mechanical turn-on voltage of diode. Modulate Wave Transmitted by Diode to Speaker
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Function Generator RF Modulator INOUT Variable Capacitor Speaker Diode
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Speaker Diode Solenoid ASolenoid B
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Speaker Diode 3,600
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RF Modulator INGROUND Variable Capacitor Speaker Diode (This is just to provide a ground.) external antenna
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I2I2 I1I1 P d1d1 d2d2 I W H D
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2.0 Amp 1.0 Amp P 1.0 meter 2.0 meter
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Current carrying region 2. Current carrying region 1. Non-conducting material a b c
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6 1 2 S 2 6 V
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I I r a
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A. N B. S C. N S S NS N N SS N
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D1D1 D2D2
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(use more frames if necessary) Cartoon Frames
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30 V Ground 1000 V 2000 V 3000 V to ground constant voltage charge separation + + + + +- - - - - - -
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x V a (x) -200 100 -100 200
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+ - - + {upward} {outward} “{upward}” and “{outward}” describe which way the electron is deflected. - + {accelerated}
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+ - - + EaEa E d,v E d,h - +
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V d Volts 0 Volts d w v o,x x y coordinates z
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V d Volts 0 Volts d w x y coordinates v o,x v f,y yy
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V d,y Volts 0 Volts d w x y coordinates v o,x v f,y yy - - + + VaVa L y’ DyDy acceleration in x-direction acceleration in y-direction while crossing deflection plates constant motion while crossing remaining distance to screen
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S (magnet) B S B OR N (magnet) B N B OR Assessment #1Assessment #2
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I through V applied
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I LED V applied (many various applied V’s) (a non-Ohmic graph) V TURN ON ÷ R
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VRVR V applied (many various applied V’s)
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I through R
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V applied
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Energy (eV) Momentum Generic Plot of Energy Bands for Semiconductor conduction band (empty) valence band (filled with electrons) E is called Band Gap Energy
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C R V source Q Cap (0)=0 Q Cap (∞)= Q Max C Q Cap (0)=Q o R Q Cap (∞)= 0
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t V Cap (t) Delineate vertical scale:
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Algebraic EquationDifferential Equation y+3 = 2 (involves a function y(t) and it’s parameter t) (involves coordinate y) y = -1 (solution is a point/number) (solution is a function of t) (-1)+3 = 2…True! (check solution by plugging point into original algebraic equation) (check solution by plugging function into original differential equation) …True!
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