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Electric Potential

Electric Potential. 12 m. KE + PE g = E. 10 kg. Let’s review. N R,b = 100N. W E,b = 100N. 12 m. KE + PE g = E. 10 kg. Force by Hand?. Work ext ?. Work ext = F ext D x. = 100N (12 m ) = 1200 J. a = 0. D PE g = mgh. WORK. 12 m.

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Electric Potential

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  1. Electric Potential

  2. 12 m KE + PEg = E 10 kg Let’s review...

  3. NR,b = 100N WE,b = 100N 12 m KE + PEg = E 10 kg Force by Hand? Workext? Workext =FextDx =100N(12 m) =1200 J a = 0

  4. DPEg = mgh WORK 12 m KE + PEg = E KE + PEg = E 10 kg Which buckets were filled?

  5. 12 m KE + PEg = E KE + PEg = E 10 kg 10 kg Energy sloshes

  6. qtest=+1 C X A B

  7. Like Gravitational potential energy a charged object can have potential energy due to it’s location in an electric field. Just like work is required to lift an object, work is required to push a charged particle against the electric field of a charged body.

  8. Electrical potential = electric potential energy divided by charge V = PE/q 1 volt = 1 Joule/Coulomb PE = q V

  9. Current, voltage and resistance

  10. battery battery Consider a flashlight …..

  11. Battery Battery Resistor Switch Wiring Diagram or a Schematic

  12. Predict the relative brightness

  13. CURRENT: • Measure of the rate of flow of charge. • I = Dq/Dt • Measured in coulombs/second, which we define as the Ampere (A).

  14. I = ? A How do we measure current? • use an ammeter • How do we insert it? What drives the current?

  15. Electromotive Force (EMF) • The battery acts like a (charge) pump. • EMF is the energy that one coulomb of charge gains in passing through a battery. • Measured in J/C or volts. • We’ll just call this the “battery voltage” • Also given the symbol ebatt.

  16. B C A D EMF and Voltage ebatt = 12 volts Loses 12 J/C DVAB= 12 J/C DVCD= -12 J/C

  17. e = 12 volts Loses 12 J/C

  18. e = 12 volts Loses 12 J/C

  19. e = 12 volts Loses 12 J/C

  20. e = 12 volts Loses 12 J/C

  21. e = 12 volts Loses 12 J/C

  22. e = 12 volts Loses 12 J/C

  23. e = 12 volts Loses 12 J/C

  24. e = 12 volts Loses 12 J/C

  25. e = 12 volts Loses 12 J/C

  26. e = 12 volts Loses 12 J/C

  27. 10.000 A A 9.999 A .001 A V B How do we measure voltage? • Voltage is a measure of the electric potential. • use a voltmeter • How do we insert it? DVA,B

  28. Consider a simple circuit • What if we increase the push? • more push means more flow • What if we change the bulb? • not all identical pushes produce identical flows

  29. Resistance... …is futile

  30. Resistance V = I R • Represents the ability of a circuit element to impede the flow of current. • R = DV/I • volt/ampere = ohm W • Georg S. Ohm (1789-1854)

  31. Which has more resistance? steel wood

  32. Which has more resistance? steel steel

  33. Which has more resistance? steel steel

  34. Which has more resistance? Steel at 0oC Steel at 25oC

  35. - + AC Delco adjustable battery More flow more glow!

  36. Vbatt = IbattRcircuit Vbatt = IbattRcircuit R R

  37. indicator bulb A B Which box contains the greater resistance?

  38. R R R A B R B A Which is box A and which is box B?

  39. R R R A B R B A The bigger the R the smaller the current of the circuit, through the battery.

  40. Junction

  41. Vbatt = IbattRcircuit Vbatt = IbattRcircuit R R

  42. R R A B A B Which is box A and which is box B?

  43. Adding in Series • What happens to Ibatt if a bulb or network of bulbs is inserted between these two bulbs? A B • In all cases, it goes down. Why? • Clog up an existing path, • increase the resistance of the circuit, and • decrease the current through the battery.

  44. Summary • When a bulb or network of bulbs is added in series, the resistance of the circuit increases. A B • The important thing is not what is added but how it is added. • If you have to break the circuit, you are adding something in series.

  45. A B Adding in Parallel • What happens to Ibatt when a bulb (or network) is added in parallel to bulb B? A • It goes up. Why? • Add a new path (opportunity), • decrease the resistance of the circuit, and • increase the current through the battery.

  46. A A B Summary • When a bulb or network of bulbs is added in parallel, the resistance of the circuit decreases. • Again, the important thing is not what is added but how it is added. • If you don’t have to break the circuit, you are adding something in parallel.

  47. R R A B A B Which is box A and which is box B?

  48. R R A B B A Which is box A and which is box B?

  49. A B C Be the river • Rank all the bulbs below for brightness. Explain your reasoning. A>B=C A gets all of the current while B and C each get half of total the current. B and C share equally since each branch has equal resistance.

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