lec15_021207_-_Current_II

# lec15_021207_-_Current_II - Current II Current Q I= = n q...

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Unformatted text preview: Current II Current: Q I= = n q Avd t Resistivity Experimentally, for conducting wires, we find: R= L A Conventional Current: The direction of the flow of positive charges! charges! Current Density: Va E J= Resistance: L Vb I e- I = n q vd A V R= I J = nqvd L A is the length of the wire is the cross-sectional area of the material ( cross- is the resistivity of the material ( m ) is a property of the material, and is defined by: to the flow of charge) = E J For most conductors: 110-8 For glass (insulator): 1010 - 1014 Resistivities of materials can change with temperature: E = J When the resistance does not depend on the voltage drop or the current, it is current, said to be an ohmic material! material! (T ) = o [1 + (T - To )] R(T ) = Ro [1 + (T - To )] Ohm's Law: V = IR Ohm' Monday, 12 February 2007 R is constant! 1 For most materials is positive! 2 Monday, 12 February 2007 Resistance of Non-Uniform Objects NonIf the area or resistivity is not constant over the length, then S = 4 +3x S Resistors in Parallel Typical R is big (k, (k M, G) ... G So V-drop across R >> V-drop between junction and top of each resistor! Must be @ same V E must be zero!?! dR = dL A of conductors ~ 10-8 m, so we need 108 m of wire for 1 resistance! 1 Current goes through the square crosscrosssection of this wire. Wire has a total length of L, and starts are x = 0. 0. Monday, 12 February 2007 R = dR = 0 L dx (4 + 3 x ) 2 3 Monday, 12 February 2007 4 1 Resistors in Series An object that resists current flow is a resistor Energy in Circuits When charge travels from point a to point b, the energy lost is given by: "Equivalence" relationships for Resistors are exact opposite Equivalence" of those for Capacitors! Capacitors! Parallel: Series: Const. V The rate of energy lost is given by: Ceq, parallel = i Ci 1 1 Ceq, series Monday, 12 February 2007 Const. Q = i 1 Ci Const. V Const. I Req, parallel = i 1 Ri In general, Power Lost / Gain: For Ohmic resistors specifically: 5 Monday, 12 February 2007 6 Req, series = i Ri Question Question When I add a 2nd bulb in parallel with the first, compared with the single bulb in the circuit, A. The brightness of both bulbs will be less than the single bulb B. The brightness of the first bulbs will be the same, but the 2nd bulb will be dimmer C. Both bulb will have the same brightness as the original D. None of the above Monday, 12 February 2007 7 Monday, 12 February 2007 8 2 Question 2nd Question When I add a bulb in series with the first, compared with the single bulb in the circuit, A. The brightness of both bulbs will be less than the single bulb B. The brightness of the first bulbs will be the same, but the 2nd bulb will be dimmer C. Both bulb will have the same brightness as the original D. None of the above Monday, 12 February 2007 9 Monday, 12 February 2007 10 Batteries Question The power company bills you based on the number of kW*hr used. This is a unit of A. Power B. Energy Power (W ) = U energy t time C. Charge D. Speed kW hr = energy time = energy time Monday, 12 February 2007 11 Monday, 12 February 2007 12 3 ...
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