Transcript Section 1-A

22-1 Current & Circuits
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
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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
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Galvanic cell  Chemical energy to electricity
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Photovoltaic cell  Solar to electrical
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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
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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
charge passes a point, so
Current = Charge
Time
= 1 Ampere
or
rate that electric
I = q/t
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 A
V = 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.
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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 A
V = 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 102 J