Presentation on theme: "22-1 Current & Circuits. Potential Difference Charges can “lose” potential energy by moving from a location at high potential (voltage) to a location."— Presentation transcript:
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.
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?
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
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
Generators Device that converts mechanical energy to electrical energy. Turn a loop of wire between magnets. Energy lost due to thermal energy ***Friction Generator
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
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
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)
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