1. 22-1 Current & Circuits

2. Potential Difference Charges can “lose” potential energy by moving from a location at high potential (voltage) to a location at low potential. Charges will continue to move as long as the potential difference (voltage) is maintained.

3. Producing Electric Currents When 2 conducting spheres touch, charges flow from the sphere with higher potential difference to the one at a lower difference  Flow of charged particles is called electric current  Flow of positive charges is called conventional current  Flow stops when this potential difference is equal  How could you keep the flow going?

4. Electric Current Keeping the Potential difference changing requires a pump that is powered by an external force  Galvanic cell  Chemical energy to electricity  Photovoltaic cell  Solar to electrical  Generators  Mechanical to electrical

5. Electric Circuits Closed loop where charges can flow  Includes a charge pump  increases PE from A-B  Device that reduces the PE from B back to A  Converts to some other form of energy  Motor converts electric energy to   Lamp converts electric energy to 

6. Generators Device that converts mechanical energy to electrical energy. Turn a loop of wire between magnets. Energy lost due to thermal energy ***Friction Generator

7. Rates of Charge Flow Power is the rate at which work is done  If a generator transfers 1 J of KE to electric energy per 1 sec  1 J / sec = 1 Watt  The energy carried by an electric current depends on the charge transferred and the potential difference which it crosses

8. Electric Current A sustained flow of electric charge past a point is called an electric current. Specifically, electric current is the rate that electric charge passes a point, so Current = Charge or I = q/t Time = 1 Ampere

9. Energy Transfer Finding power  Power is the amount of energy delivered to the motor per second  P = IV If the current through a motor is 3.0 A, and the potential difference is 120 J of energy, what is the power of the motor I = 3.0 AV = 120 J/C (3.0 A)(120 J/C) = 360 J/s = 360 Watts (W)

10. Problem A 6.0 V battery delivers a 0.50 A current to an electric motor that is connected across it terminals.  A) What is the power run by he motor?  B) If the motor runs for 5.0 minutes, how much energy is delivered? A) I = 0.50 AV = 6.0 V  = (0.50 A)(6.0 V) = 3.0 W B) Power = Energy / Time = P = E/t  E = Pt  (3.0 W)(5.0 min x 60s/1 min) = 9.0 x 10 2 J