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Chapter E5. Driving Currents. Idealized Battery. -. +. -. -. +. +. -. +. -. +. -. +. -. -. +. +. -. +. Capacitor (two charged plates). Arrows show current. What is necessary to keep current flowing?.
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Chapter E5 Driving Currents
Idealized Battery - + - - + + - + - + - + - - + + - + Capacitor (two charged plates) Arrows show current What is necessary to keep current flowing? The arrows also show the direction of the electric field (the way positive charges would move). What direction are the electrons really moving?
Batteries • A battery is a device that maintains a charge between two plates by transporting a charge from one plate to the other against the electrical forces on that charge. • In the Van de Graaff, the belt carries the electrons away from the sphere. • In a battery this is done by chemical reactions.
Why electrons flow through a wire. • The instant the ends of a wire are connected across the battery, the charges adjust on the surface of the wire in a way so that the electric field inside the wire is exactly uniform from one end to the other. • On the previous slide the electric field would be in the direction of the arrows. • This is a dynamic equilibrium. • True in all wires that conduct electric current.
The way a battery works(Lead-Acid) • PbO2+HSO4-+3H3O++2e-→PbSO4+4H2O • This reaction removes two electrons from the positive electrode. • Pb+HSO4-+H2O→PbSO4 + H3O++ 2e- • This reaction puts two electrons on the negative electrode. • The net effect is a conveyor belt moving the electrons from the + to the – electrode against the electric field.
EMF-electromotive force = ξ The potential difference across the terminals of a battery is the characteristic electromotive force minus the electromotive force lost to thermal motion.
The electric field in a wire The potential difference across the ends of thin wire with a current flowing through it is the product of the electric field and the length of the wire.
Ohm’s law The unit of resistance is the ohm (Ω). One ohm is a volt per ampere.
Practical electricity problems • Will a 20 amp (120 V) circuit breaker support a 950 watt TV, a 200 w computer, a 1000 w iron and three 100 w light bulbs? • Total power required = 2450 watts • Power supplied by the circuit = 2400 w • The breaker will blow, one item must be turned off.
Discharging a capacitor V + R1 » R2 R - t
Problems due Monday • E5B.1, E5B.4, E5B.5