(165) Electric Current & Circuits Explained, Ohm's Law, Charge, Power, Physics Problems, Basic Electricity

Electric Current and Ohm's Law

  • Conventional Current

    • Flows from positive terminal to negative terminal (high voltage to low voltage).

    • Water analogy: Like water flowing from high to low ground level.

  • Electron Flow

    • Flows from negative terminal to positive terminal (actual flow of electrons).

  • Current Definition

    • Current (I) = Rate of charge flow = Δq / Δt.

      • Q: Electric charge in coulombs (C).

      • t: Time in seconds (s).

    • Unit of Current: Ampere (A); 1 A = 1 C/s.

    • Charge of an electron = 1.6 x 10^-19 C (negative).

Ohm's Law

  • Ohm's Law Equation: V = IR

    • V: Voltage (volts), I: Current (amps), R: Resistance (ohms).

    • Direct and Inverse Relationships:

      • Increasing voltage → increases current (if resistance is constant).

      • Increasing resistance → decreases current (if voltage is constant).

  • Analogy:

    • Highway example: More lanes (lower resistance) allow more cars (current) to pass.

Electric Power

  • Power Equation: P = VI

    • V: Voltage, I: Current.

  • Alternative Power Forms:

    • P = I²R

    • P = V²/R

  • Unit of Power: Watts (W); 1 W = 1 Joule/s.

Practice Problems

Problem 1: Charge Flow Calculation

  • Given: 3.8 A current for 12 minutes.

    • Convert time: 12 min = 720 s.

    • Calculate charge (Q): Q = I × t = 3.8 A × 720 s = 2736 C.

  • Electrons Calculation:

    • Number of electrons = Q / charge of electron = 2736 C / (1.6 × 10^-19 C) ≈ 1.71 × 10^22 electrons.

Problem 2: Current through a Resistor

  • Given: 9 V battery across 250 ohm resistor.

    • Use Ohm's Law: I = V / R = 9 V / 250 ohms = 0.036 A (or 36 mA).

  • Power Dissipation:

    • P = I²R = (0.036 A)² × 250 ohms = 0.324 W (or 324 mW).

  • Power Delivered by Battery:

    • Confirm: P = VI = 9 V × 0.036 A = 0.324 W (balance confirmed).

Problem 3: Light Bulb Resistance

  • Given: 12 V battery, 150 mA through light bulb.

    • Convert current: 150 mA = 0.15 A.

    • Use Ohm's Law to find resistance: R = V / I = 12 V / 0.15 A = 80 ohms.

  • Power Consumption:

    • P = VI = 12 V × 0.15 A = 1.8 W.

Problem 4: Operating Cost of Light Bulb

  • Given: Power of 1.8 W, cost of electricity is 11 cents per kWh.

    • Energy per month = Power × Time.

    • Time = 30 days × 24 hours = 720 hours.

    • Convert to kW: 1.8 W = 0.0018 kW.

    • Energy = 0.0018 kW × 720 hours = 1.296 kWh.

    • Cost = 1.296 kWh × 0.11 dollars/kWh = 0.14256 dollars (≈ 14 cents).

Problem 5: Motor Voltage Calculation

  • Given: 50 W motor, 400 mA current.

    • Convert current: 400 mA = 0.4 A.

    • Voltage: V = P / I = 50 W / 0.4 A = 125 V.

  • Internal Resistance Calculation:

    • Use Ohm's Law: R = V / I = 125 V / 0.4 A = 312.5 ohms.

Problem 6: Charge Flow and Current through 5 kOhm Resistor

  • Given: 12.5 C charge, 5 kOhm resistor, 8 minutes.

    • Convert time: 8 min = 480 s.

    • Current: I = Q / t = 12.5 C / 480 s = 0.026 A (or 26 mA).

  • Power Consumption:

    • Calculate: P = I²R, where R = 5000 ohms.

    • P = (0.026 A)² × 5000 ohms = 3.38 W.

    • Voltage = I × R = 0.026 A × 5000 ohms = 130 V.