Lecture on Electric Current, Ohm's Law, and Practice Problems

Key Concepts

Conventional Current vs. Electron Flow

• Conventional Current: Flows from positive to negative terminal (high voltage to low voltage).
• Electron Flow: Actual electron movement is from negative to positive terminal.

Definitions

• Current (I): Rate of charge flow, measured in amperes (A). Defined as ( I = \frac{\Delta Q}{\Delta t} ).
  • ( Q ) is charge in coulombs (C).
  • ( t ) is time in seconds (s).
  • 1 ampere is equal to 1 coulomb/second.

Ohm's Law

• Equation: ( V = IR )
  • ( V ) is voltage, ( I ) is current, ( R ) is resistance.
  • Voltage and current are directly related; resistance and current are inversely related.
  • Resistance is measured in ohms (Ω).

Electric Power

• Power (P): Rate of energy transfer, measured in watts (W).
  • ( P = VI = I^2R = \frac{V^2}{R} )
  • 1 watt is equal to 1 joule/second.

Practice Problems

Problem 1: Charge and Current

• Given:
  • Current: 3.8 A
  • Time: 12 minutes (convert to 720 seconds)
• Find: Electric charge ( Q = I \times t = 3.8 \times 720 = 2736 ) C
• Convert: Charge to number of electrons:
  • One electron charge: ( 1.6 \times 10^{-19} ) C
  • Number of electrons: ( \frac{2736}{1.6 \times 10^{-19}} = 1.71 \times 10^{22} )

Problem 2: Ohm's Law Application

• Given:
  • Voltage: 9 V
  • Resistance: 250 Ω
• Current Calculation: ( I = \frac{V}{R} = \frac{9}{250} = 0.036 ) A (36 mA)
• Power Dissipated: ( P = I^2 \times R = 0.036^2 \times 250 = 0.324 ) W (324 mW)
• Battery Power: ( P = V \times I = 9 \times 0.036 = 0.324 ) W

Problem 3: Light Bulb Resistance

• Given:
  • Voltage: 12 V
  • Current: 150 mA (0.15 A)
• Resistance: ( R = \frac{V}{I} = \frac{12}{0.15} = 80 ) Ω
• Power Consumption: ( P = VI = 12 \times 0.15 = 1.8 ) W

Problem 4: Motor Voltage and Resistance

• Given:
  • Power: 50 W
  • Current: 400 mA (0.4 A)
• Voltage: ( V = \frac{P}{I} = \frac{50}{0.4} = 125 ) V
• Resistance: ( R = \frac{V}{I} = \frac{125}{0.4} = 312.5 ) Ω

Problem 5: Charge Flow and Power

• Given:
  • Charge: 12.5 C
  • Resistance: 5 kΩ (5000 Ω)
  • Time: 8 minutes (480 seconds)
• Current: ( I = \frac{Q}{t} = \frac{12.5}{480} = 0.026 ) A (26 mA)
• Power: ( P = I^2 \times R = 0.026^2 \times 5000 = 3.38 ) W
• Voltage: ( V = I \times R = 0.026 \times 5000 = 130 ) V

Summary

• Understanding and applying Ohm's Law and power equations is crucial for solving electrical circuit problems.
• Practice problems help reinforce concepts of charge, current, voltage, resistance, and power.