Electrical current

1. Chapter 23
Electric Current

2. Flow of Charge
• Charge flows when there is potential difference (voltage)
• Analogy: water flows from high pressure to low pressure, as
shown below, until equilibrium occurs
• To continue water flow, a pump is needed

3. Electric Current
• Electric current = flow of electric charge
• Meaning, the flow of electrons
• What is the charge of an electron?
– Answer: 1.60 × 10-19 coulombs
• How many electrons would you need to have a charge of 1
coulomb?
– Answer: 6.25 x 1018 electrons!
• The rate at which electrons flow is measured in amperes, or
amps for short
• 1 amp = 1 coulomb / second (That’s a lot of electrons!)
The bulb on this slide requires 2.4 amps!

4. Voltage Sources
• Batteries and electric generators remove negative charges
from positively charged objects
• The difference in charge results in the creation of a potential
difference, measured in Volts
• In the analogy below, the pump and battery create the
potential difference
• The valve and switch control the flow (on/off)
• Water / Electrons flow from high pressure to low pressure

5. Electrical Resistance
• Voltage pushes charge.
• To control how much charge flows, electrical resistance is
applied to the flow
• Consider a water hose
– Resistance is less when the hose is shorter and…
– Resistance is less when the hose is wider
– Example: closing the end of the water hose controls how
fast the water comes out
• In electric currents
– Resistance is less when the wire is shorter
– Resistance is less when the wire is thicker
• Electrical resistance is measured in Ohms, the symbol Ω
• In drawings, it is represented by

6. Ohm’s Law
• Voltage = current x resistance (V = IR)
• Voltage (volts), current (amperes), resistance (ohms)
• Volts = amps / ohms
Question
1. How much current will flow through a lamp that as
a resistance of 60Ω when 12V are impressed
across it?
I = V/R = 12V/60Ω = 0.20A
2. What is the resistance of an electric frying pan that
draws 12A when corrected to a 120V circuit?
R = V/I = 120V/12A = 10Ω

7. End of Part 1