Simple circuits

• Notes
  • Parallel resistors
    • ΔV MUST be same for all
    • To reduce: R_eq = 1/(1/R1 + 1/R2 + ...)
    • NOTE: R_eq < smallest of R1, R2, ...
    • To un-reduce: all ΔVs are same as R_eq ΔV. I's are unknown.
  • Series resistors
    • I MUST be the same for all
    • To reduce: R_eq = R1 + R2 + ...
    • To un-reduce: All I's are the same as R_eq. ΔVs are unknown.
• Lab instructions: For each circuit
  • ΔV = 1.5V, R₁ = 0.1kΩ, R₂ = 0.22kΩ, R₃ = 0.33kΩ, R₄ = 0.1kΩ, R₅ = 0.22kΩ
  • 1 - perform one reduction at a time, redraw the reduced circuit, fill in all information you know and leave blanks for information you don't know, keep reducing until circuit is power supply and one resistor, solve for I_total, then work backwards by un-reducing until you find the desired ΔV.
  • 2 - build the circuit using either cables and clamps, or a breadboard.
  • 3 - get me: I will test your circuit and prediction.
  • You must compute / build circuits in order (7, 8, 9, 10)
  • Grade is determined by the last circuit you successfuly compute and build (80, 85, 90, 100)
• Notes to me:
  • 6: I = 02.3mA, ΔV₂ = 0.508V
  • 7: I₄ = ΔV / (100Ω + 330Ω) = 03.5mA
    ΔV₄ = 03.5mA × 1000Ω = 0.35V
  • 8: R_eq = 466Ω, I = 03.2mA
    ΔV_{3 and 4} = 1.5V - I ×(200Ω) = 0.47V
    I_{3 and 4} = 0.47V / 430Ω = 01.1mA
    ΔV₃ = 01.1mA × 330Ω = 0.36V
  • 9: R₄₅ = 68.8Ω, R₃₄₅ = 398.8Ω, R₂₃₄₅ = 141.8Ω
    R_eq = 241.8Ω
    I₁ = 06.2mA
    ΔV₂ = 1.5V - 06.2mA×100Ω = 0.880V
    I₃ = 0.880V / 398.8Ω = 02.2mA
    ΔV₃ = 02.2mA × 330Ω = 0.728V
  • 10: