#  With a partner, get a battery, light bulb, and paper clip.  Find the two ways to light up the light bulb using just these three items.  Draw pictures.

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 With a partner, get a battery, light bulb, and paper clip.  Find the two ways to light up the light bulb using just these three items.  Draw pictures of what the set up looked like if you notes

 Charge flows from an item with a higher potential difference to something with a lower potential difference until the two have equal charges › The reason there is static discharge  Electric Current : the flow of charged particles

 If moving a charge against the electric field, work is done on the charge  The work done changes the charges potential energy to a higher value  The work is equal to the change in the potential energy  There is a difference in electric potential between the two locations  This difference is represented by ΔV and called electric potential difference

 By definition, electric potential difference is the difference in electric potential (V) between the final and initial location of the charge › And we drop the Δ as a historical convention (not a correct one) and just use V  Standard unit for electric potential difference is the volt (V), named in honor of Alessandra Volta

 An electric potential difference between two locations of 12 volts means that one coulomb of charge will gain 12 joules of potential energy when moved between those two locations  Because it is expressed in volts it is sometimes referred to as voltage

 Two models › Conventional current: assumes positive charges flow out of a positive terminal and travel to the negative terminal. › Electron Flow: what actually happens, is that electrons flow out of negative terminals and travel to the positive terminal

 Charge will always flow from one more charged body to another, but what makes it continue to flow? › You need a generator of some sort › Most common generator is several galvanic cells (or dry cells) connected together – they form a battery  Chemical energy generator  Other include hydro, steam, and wind generators  Photovoltaic cell: solar cell

 Electric Circuit: a closed loop in which current flows › Includes a charge pump (battery) which increases potential energy and a device to reduce potential energy (light bulb) › Think of a charge pump as the work done by a roller coaster to get the cars to the top of the hill, they go down the other side naturally  Once a positive charge gets from inside the battery to the positive terminal, it flows naturally to the negative terminal

 In a 9V battery, there is 9 volts of potential difference between the two terminals. That means that the battery must do 9 joules of work moving a positive charge from the negative wire to the positive wire. The positive charge then gains 9 joules of potential energy in which it can deliver to the light bulb and then come back with no energy and do it all again.

 When a charge moves through a circuit, the amount of potential energy it loses is qV (charge times potential difference)  So the generator/charge pump/battery needs to increase the charges potential and the energy required to do that is qV  The change in electric energy, E, is equal to qV › E = qV

 Power is the measure of the rate at which energy is transferred (P = E/t) › Transferring 1 joule per second is 1 watt  The rate of flow of electric charge, or electric current, I, is measured in coulombs per second › 1 coulomb per 1 second is 1 ampere, A › Ammeters measure amperes

 Suppose current is flowing at 3C/s (3A) and the potential difference is 120V, which means that each charge supplies the motor/light bulb/etc with 120J  To find the power delivered we multiply the current and the potential difference › P = IV  The power delivered in this situation would be 360 W

 A 6.0 V battery delivers a 0.50 A current to an electric motor that is connected across its terminals. › What power is consumed by the motor? › If the motor runs for 5 minutes, how much electric energy is delivered?

 The current though a lightbulb connected across the terminals of a 120- V outlet is 0.50 A. At what rate does the bulb convert electric energy to light?

 A car battery causes a current of 2.0 A through a lamp while 12 V is across it. What is the power used by the lamp?

 The current through the starter motor of a car is 210 A. If the battery keeps 12 V across the motor, what electric energy is delivered to the starter in 10.0 s?

 Resistance, R, is the ratio of the potential difference, V, to the current, I › R = V/I › Measured in volts per ampere, which is ohms  Named after Georg Simon Ohm › 1 Ω (ohm) is equal to 1 A of flow when 1 volt is applied  Ohm discovered that a devices resistance stays the same no matter the potential difference that is applied (Ohm’s Law)

 Most wires used in circuits have a very small resistance (they don’t reduce the potential difference much) › Factors of resistance – think garden hose  Small diameter v. large diameter  Short v. long  Resistors are devices use to have a specific resistance

 A 30.0 V batter is connected to a 10.0 Ω resistor. What is the current in the circuit?

 A lamp draws a current of 0.50 A when it is connected to a 120 V source. › What is the resistance of the lamp? › What is the power consumption of the lamp?

 P515  Redraw the circuits you made at the beginning of the hour with proper circuit diagrams

 Parallel › When two devices are connected so they are parallel to each other  Voltmeters need to be connected this way  Series › When two devices are connected so that the current that flows through one also flows through the other  Ammeters need to be connected this way

 Substituting new equations learned

 A heater has a resistance of 10.0 Ω. It operates on 120.0 V. › What is the current through the resistance? › What thermal energy is supplied by the heater in 10.0 s?

 A 30.0 Ω resistor is connected across a 60 V battery. › What is the current in the circuit? › How much energy is used by the resistor in 5 min?

 1 kWh is equal to 1000 watts delivered continuously for 3600 seconds (1 hour)  A television set draws 2.0 A when operated on 120 V. › How much power does the set use? › If the set is operated for an average of 7.0 h/day, what energy in kWh does it consume per month (30 days)? › At \$0.11 per kWh, what is the cost of operating the set per month?

 P527: 21, 23, 29, 32, 41, 44

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