Lecture Notes: Electric Current and Ohm's Law

Introduction

• Discussed basic equations and practice problems involving electric current and Ohm's Law.
• Conventional Current: Flows from high voltage (positive terminal) to low voltage (negative terminal).
• Electron Flow: Opposite to conventional current, electrons flow from negative to positive terminal.

Concepts and Formulas

Electric Current

• Definition: Rate of charge flow, mathematically ( I = \frac{\Delta Q}{\Delta T} )
  • ( Q ): Electric charge in coulombs
  • ( T ): Time in seconds
• Unit: Ampere (amp), where 1 Amp = 1 coulomb/second

Ohm's Law

• Formula: ( V = IR )
  • ( V ): Voltage
  • ( I ): Current
  • ( R ): Resistance in ohms
• Relationships:
  • Voltage and current are directly related.
  • Resistance and current are inversely related.

Electric Power

• Formula: ( P = VI ), alternate forms are ( P = I^2R ) and ( P = \frac{V^2}{R} )
  • Unit: Watt, where 1 watt = 1 joule/second

Practice Problems

Problem 1: Electric Charge

• Given: Current = 3.8 amps, Time = 12 minutes
• Solution:
  • Convert time to seconds: ( 12 \times 60 = 720 ) seconds
  • Calculate charge: ( Q = I \times T = 3.8 \times 720 = 2736 ) coulombs
  • Calculate the number of electrons: ( \frac{2736}{1.6 \times 10^{-19}} = 1.71 \times 10^{22} )

Problem 2: Current in a Resistor

• Given: 9-volt battery, 250-ohm resistor
• Solution:
  • Current: ( I = \frac{V}{R} = \frac{9}{250} = 0.036 ) amps
  • Power dissipated: ( P = I^2R = 0.324 ) watts
  • Power delivered: ( P = VI = 0.324 ) watts (balanced)

Problem 3: Resistance of a Light Bulb

• Given: 12-volt battery, 150 milliamps
• Solution:
  • Resistance: ( R = \frac{V}{I} = \frac{12}{0.15} = 80 ) ohms
  • Power consumed: ( P = VI = 1.8 ) watts
  • Energy cost: 14 cents/month at 11 cents/kWh

Problem 4: Voltage Across a Motor

• Given: 50 watts, 400 milliamps
• Solution:
  • Voltage: ( V = \frac{P}{I} = \frac{50}{0.4} = 125 ) volts
  • Internal resistance: ( R = \frac{V}{I} = 312.5 ) ohms

Problem 5: Current through a Resistor

• Given: 12.5 coulombs, 5 kiloohms, 8 minutes
• Solution:
  • Current: ( I = \frac{Q}{T} = \frac{12.5}{480} = 0.026 ) amps
  • Power consumed: ( P = I^2R = 3.38 ) watts
  • Voltage: ( V = IR = 130 ) volts

Conclusion

• These equations and concepts are fundamental in solving electrical problems related to current, voltage, resistance, and power.