Lecture on Electric Current and Ohm's Law

Basics of Electric Current

• Conventional current: Flows from the positive terminal to the negative terminal of a battery, similar to water flowing from high to low.
• Electron flow: Opposite of conventional current; electrons move from the negative terminal to the positive terminal.

Definitions

• Current (I): Rate of charge flow, ( \frac{\Delta Q}{\Delta T} ), where ( Q ) is electric charge (coulombs) and ( T ) is time (seconds).
  • Unit: Ampere (A); 1 A = 1 coulomb/second.
• Electron charge: (-1.6 \times 10^{-19}) coulombs.

Ohm's Law

• Equation: ( V = IR ), where
  • ( V ) is voltage,
  • ( I ) is current,
  • ( R ) is resistance.
• Relationships:
  • Increasing voltage increases current.
  • Increasing resistance decreases current.
  • Voltage and current are directly related; resistance and current are inversely related.

Analogy

• Highways as circuits: More lanes (less resistance) allows more cars (current) to pass.

Electric Power

• Equations:
  • ( P = VI )
  • ( P = I^2R )
  • ( P = \frac{V^2}{R} )
• Unit: Watt (W); 1 W = 1 Joule/second.

Practice Problems

Problem 1

• Given: 3.8 A current for 12 minutes.
• Calculate charge: ( Q = I \cdot T )
  • Convert 12 minutes to seconds: 720 seconds.
  • ( Q = 3.8 \times 720 = 2736 ) coulombs.
• Calculate electrons:
  • One electron = (1.6 \times 10^{-19}) coulombs.
  • ( 2736 / 1.6 \times 10^{-19} \approx 1.71 \times 10^{22} ) electrons.

Problem 2

• Given: 9 V battery, 250 ohm resistor.
• Find current: ( I = \frac{V}{R} = \frac{9}{250} = 0.036 ) A (36 mA).
• Find power dissipated:
  • ( P = I^2R = 0.036^2 \times 250 = 0.324 ) W.
• Power delivery by battery:
  • ( P = VI = 9 \times 0.036 = 0.324 ) W.

Problem 3

• Given: 12 V battery, 150 mA current.
• Find resistance:
  • Convert 150 mA to 0.15 A.
  • ( R = \frac{V}{I} = \frac{12}{0.15} = 80 ) ohms.
• Power consumption:
  • ( P = VI = 12 \times 0.15 = 1.8 ) W.

Problem 4

• Given: 50 W power, 400 mA current.
• Find voltage:
  • Convert 400 mA to 0.4 A.
  • ( V = \frac{P}{I} = \frac{50}{0.4} = 125 ) V.
• Find resistance:
  • ( R = \frac{V}{I} = \frac{125}{0.4} \approx 312.5 ) ohms.

Problem 5

• Given: 12.5 C charge, 5 kohm resistor, 8 minutes.
• Find current:
  • Convert minutes to seconds: 480 seconds.
  • ( I = \frac{Q}{T} = \frac{12.5}{480} = 0.026 ) A (26 mA).
• Power consumption:
  • ( P = I^2R = 0.026^2 \times 5000 = 3.38 ) W.
• Voltage across resistor:
  • ( V = IR = 0.026 \times 5000 = 130 ) V.