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Electric currents Physics 100 Chapt 13 (Motion of electric charges)

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Presentation on theme: "Electric currents Physics 100 Chapt 13 (Motion of electric charges)"— Presentation transcript:

1 Electric currents Physics 100 Chapt 13 (Motion of electric charges)

2 Alessandro Volta

3 Positive Ions _ _ _ _ + + + + + Atoms with one or more electrons removed ”net” charge = + 2q e _ _ _ _

4 Battery C Zn Zn ++ - Zn ++ - Zn ++ - + Zn ++ - + Zn - + acid

5 “Voltage” Anode + Cathode - - - - - - E Zn ++ F + + - - - W = Fd dF = 2q e E W = 2q e Ed  W 0 = 2q e E 0 d  =E 0 d W 0 2q e V “Voltage”

6 Anode + Cathode - - - - - - F=QE 0 + + - - - W = Fd E0E0 Q F=QE 0 Q =Q E 0 d = QV d Zn ++ Energy gained by the charge

7 Units again! W = Q V V = WQWQ joules coulombs  joules coulomb = Volt 1 V = 1 joule coulomb

8 Continuous charge flow = “electric current” Anode + Cathode - - - - - - + + - - - Zn ++ QQ Electrical “conductor” connected between anode & cathode

9 electric current Anode + Cathode - - - - - - + + - - - Zn ++ QQ I = Q t Units: Coulombs second =Amperes

10 The conductor can be a piece of wire Anode + Cathode - - - - - - + + - - - Zn ++ I = Q t + + +

11 The energy can be used to run a gadget Anode + Cathode - - - - - - + + - - - Zn ++ P= Energy time + + + QV t == I V I I I

12 Electric light 60 Watts I=? T Power = P = I V I = PVPV = 60 W 100V = 0.6 J/s J/C = 0.6 1/s 1/C = 0.6 CsCs = 0.6 A V=100V

13 General circuit 12V + - Appliance + - Energy source (device that separates + & - charge) I I

14 analogy Height ~ voltage Pump ~ battery Amt of water flow ~ current appliance pond pump


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