Current and Resistance Lecture Notes

Current and Resistance

1. Simple Circuit

• A battery creates a potential difference of V when connected to a light bulb.

2. Circuit Diagram: Symbol for Battery

• The circuit diagram illustrates how to represent a battery.

• The battery is depicted as two parallel lines of unequal length:

  • Longer line: Positive terminal

  • Shorter line: Negative terminal

3. Circuit Diagram: Bulb and Battery

• The circuit diagram includes a bulb connected to a battery.

• Wires are represented as straight lines connecting the battery to the bulb.

4. Direction of Current

• The battery generates potential by separating charges chemically.

• Positive Terminal: Positively charged

• Negative Terminal: Negatively charged

• When connecting a wire to the battery:

  • Free electrons in the wire move towards the positive terminal, pass through the battery, and flow out of the negative terminal (counter-clockwise direction).

  • Historically, it was believed that positive charges moved, resulting in a clockwise movement.

• Definition of Current: The movement of charges, analogous to the flow of water, always goes from high potential (positive) to low potential (negative).

5. Quantifying Current

• Current Definition: Movement of charge through a circuit.

• Quantified by the equation:

  • I = \frac{\Delta Q}{\Delta t}

• Units of Current:

  • [I] = \frac{[\Delta Q]}{[\Delta t]} = 1 C/s = 1 \text{ Ampere} = 1 A

• More charge flows when current increases in a defined time interval.

6. Ohm’s Law

• Current can also be determined by potential difference.

• Analogy: Water flow in a stream based on gravitational difference in height.

  • Greater height difference leads to increased water flow (current).

• Proportionality:

  • I \propto V where I is current and V is potential difference.

• Electrical Resistance (R): Impurities in the wire impede the flow of electrons.

  • Shows that R \propto I is inversely proportional.

• Ohm's Law:

  • I = \frac{V}{R}

  • Rewrite as:

  • V = IR

  • Units for resistance:

  • [R] = \frac{[V]}{[I]} = 1 V/A = 1 \text{ Ohm} = 1 \Omega

7. Resistance of Wires

• Conductors typically have some resistance (including wires for circuits).

• Wire Characteristics:

  • Length (L): Longer wires increase chances of electron collisions, increasing resistance.

  • R \propto L

  • Cross-Sectional Area (A): Wider wires decrease resistance; electrons can avoid impurities more effectively.

  • R \propto \frac{1}{A}

• Business Equation:

  • R = \frac{\rho L}{A}

  • Where \rho (resistivity) is a constant dependent on the material. For example, copper has specific resistivity, as does silver.

• Units of Resistivity:

  • [\rho] = \frac{[R][A]}{[L]} = 1 \frac{\Omega m^2}{m} = 1 \Omega m

• Smaller \rho suggests better conductivity, allowing more current.

• Resistivity is temperature-dependent, as increased heat vibrates and collides with electrons, causing resistance to rise slightly with temperature, although not significantly worrying for practical classes.

8. Resistance of Wires

• Summary of wire characteristics:

  • Cylindrical wire with cross-sectional area of A and length of L.

9. Practical Application of Wire Resistance

• Generally, wire resistance is sufficiently low; however, its effects become relevant in specific scenarios.

• Examples include speaker wires in stereo systems.

  • Current affects speaker volume—greater current results in louder output.

  • For extended wires, wider wires or lower gauge wires are optimal for transferring maximum current to speakers.

• Wider wire translates to more material used, thus increasing cost.

10. Wire Gauges

• Visual representation of the cross-sectional sizes of various gauge wires.

• Gauge 10 is the widest and most expensive type.

11. Resistors

• Resistors are explicitly designed to have high resistance levels.

• Used for controlling the flow of current in circuits.

  • Similar to limiting water flow in a river with debris.

• Resistors possess significantly higher resistance compared to connecting wires, so the resistance of wires can often be ignored during calculations.

12. Revisiting Ohm’s Law

• When a light bulb (acting as a resistor) connects to a battery:

  • Impurities in the filament create resistance, limiting electron flow.

• Suppose a 10 V battery powers a 5 Ω light bulb:

  • Utilizing Ohm's Law:

  • V = I R

  • This becomes:

  • I = \frac{V}{R} = \frac{10 V}{5 \Omega} = 2 A

• Current remains consistent throughout the circuit; 2 A flows through battery, wires, and bulb correspondingly from the positive to negative terminal.

13. Current

• 2 A of current traverses the circuit from positive to negative terminal, resulting in a clockwise direction.

14. Electrical Power for a Battery

• As current flows from battery negative to positive, as charge passes through, it gains energy, denoting a voltage difference (Vb > Va).

• Energy gained by charge ΔQ when passing through battery:

  • ΔE_{PE} = q ΔV = +ΔQ V

• Power measure of energy change over time:

• P = \frac{ΔE_{PE}}{Δt} = +(\frac{ΔQ}{Δt}) V = I V

• Power defined in Watts (1 W = 1 J/s):

  • Units: [P] = [I][V],\ 1 W = 1 A imes V

• Positive current flow from positive battery end results in power:

  • P = +IV

• Current may also flow incorrectly out of the negative end:

  • In such cases, power is negative:

  • P = -IV

15. Dissipated Power by a Resistor

• Current flows from high potential to lower potential through resistors.

• Voltage drop across a resistor:

  • V = IR

• Dissipated energy manifests as heat; power across resistors often termed "dissipated power."

• Power remains valid across resistors:

  • P = IV

• Another corresponding equation using Ohm's Law:

  • P = I^2 R = \frac{V^2}{R}

• All versions yield identical results. Choosing the right formula depends on available information concerning resistors.

• Only P = IV is valid for battery power output calculations; other equations are exclusive to resistors.

16. Resistors In Series

• Connecting multiple resistors in sequence denotes a series connection.

• Current I through each resistor remains constant.

17. Finding Current in Series Circuits

• Ohm's Law, V = IR, governs series circuits.

• Current leads to voltage drops across resistors:

  • V = I R1 + I R2 + I R_3

• Resultant voltage equals initial battery voltage minus summed potential drops across resistors:

  • V = I (R1 + R2 + R_3)

18. Equivalent Resistance and Circuits for Series Connection

• Series circuit equivalent resistance defined as:

  • R{eqs} = R1 + R2 + R3

• Structural transformation to a single equivalent resistor with a battery:

  • V = I R_{eqs}

• Current is thus:

  • I = \frac{V}{R_{eqs}}

19. Notes and Summary for Series Connections

• Finding current in a series circuit involves calculating equivalent resistance:

  • R{eqs} = R1 + R2 + R3

  • Then, current found via:

  • I = \frac{V}{R_{eqs}}

• Voltage across individual resistors calculated using:

  • V1 = I R1

• Voltage across resistors equals battery voltage:

  • V = V1 + V2 + V_3

• Power calculation for individual resistor:

  • P = I^2 R

• Power dissipated highest through largest resistors in the series, and the smallest consumes least power.

20. Resistors In Parallel

• Multiple resistors connected parallelly confer individual connections to the battery.

• Voltage across all resistors equals battery voltage.

• Current splits from the battery into I1, I2, and I3:

  • Conclusively, I = I1 + I2 + I_3

21. Equivalent Resistance and Circuits for Parallel Connection

• Equivalent current expression:

  • I = \frac{V}{R1} + \frac{V}{R2} + \frac{V}{R3} = V(\frac{1}{R1} + \frac{1}{R2} + \frac{1}{R3})

• Rearranged:

  • V = I(\frac{1}{R1} + \frac{1}{R2} + \frac{1}{R_3})^{-1}

• Determines circuit using concept of equivalent parallel resistance (R_{eqp}).

• Equivalent current through this resistor equals total current exiting the battery.

22. Notes and Summary for Parallel Connections

• Multiple resistors in parallel yield total resistance:

  • R{eqp} = (\frac{1}{R1} + \frac{1}{R2} + \frac{1}{R3})^{-1}

• Voltage seen across all resistors equals battery voltage.

• Simplifies current calculation from battery:

  • I = \frac{V}{R_{eqp}}

• To find current through individual resistor:

  • I1 = \frac{V}{R1}

• Current sum held constant with battery current:

  • I = I1 + I2 + I_3

• Power for individual resistor based on voltage across it:

  • P = \frac{V^2}{R}

• Power inversely tied to resistance, smallest resistor dissipating most power.

23. Comparisons Between Series and Parallel Connection

• Series connections yield resistance:

  • R{eqs} = R1 + R2 + R3

• Parallel arrangements give:

  • R{eqp} = (\frac{1}{R1} + \frac{1}{R2} + \frac{1}{R3})^{-1}

• Adding resistors in series increases overall resistance.

• Adding resistors in parallel decreases overall resistance.

• Series circuits reduce current:

  • I = \frac{V}{R_{eq}}

• Adding parallel resistors increases operational current.

24. Comparisons: Removing a Resistor

• Removing a series resistor reduces total resistance.

  • Improving current flow through the battery and resistor increases the power output from the battery.

  • If light bulbs represent resistors, the brightness increases.

• Removing a parallel resistor retains voltage for others, achieving unchanged brightness for bulbs while reducing battery output.

25. Comparisons: Blown Out Light Bulb

• A blown bulb in a series circuit is akin to cutting the wire; thus, no current flows, extinguishing all bulbs.

• A blown bulb in parallel permits current to pass through other bulbs, preserving brightness.

26. Short Circuit

• Connecting a low resistance wire across a circuit restricts overall current through that circuit section.

• This nullifies resistor current, thus causing connected loads (e.g., light bulbs) to burn out.

27. Combination Circuits

• A combination circuit showcases series and parallel resistor configurations.

• Current I dispatches from the battery and bifurcates at the junction into paths I1 and I2.

  • Current traverses R2 and R3 for I1, while I2 moves through R4.

• Eventually, currents reunite at the next junction and flow through R5 back to the battery.

28. Combination Circuits

• Identify series resistors (R1 and R5) sharing current I.

  • Their equivalent resistance combines as:

  • R_{15} = 2 \Omega + 6 \Omega = 8 \Omega

• R2 and R3 also have the same current (I1):

  • R_{23} = 5 \Omega + 10 \Omega = 15 \Omega

29. Combination Circuits

• Formulate changes into an equivalent circuit post-resistor combination.

• Identify parallel resistors with the same voltage and calculate resistances.

• Combine R23 and R4 (both in parallel):

  • R_{234} = (\frac{1}{15} + \frac{1}{30})^{-1} \Omega

  • Resulting in calculated: R_{234} = 10 \Omega

30. Combination Circuits

• Delivering the equivalent circuit yields R15 in series with R234:

  • Combined resistance forms: R_{12345} = 10 \Omega + 8 \Omega = 18 \Omega

  • Current, I, from a 20 V battery becomes:

  • I = \frac{20}{18} Amps = 1.11 A

31. Combination Circuits

• Expand re-formed circuit model to derive remaining currents.

• Voltage across R23 and R4 not equivalent to battery since R15 contributes voltage drop.

• Potential loss through R15:

  • V{15} = I \times R{15} = (\frac{20}{18} A)(8 \Omega) = \frac{80}{9} V

  • Potential remaining: 20 V – 8.89 V = 11.11 V across R23 and R4.

• Resulting currents for R23 and R4:

  • I_1 = (11.11 V)/(15 \Omega) = 0.74 A

  • I_2 = (11.11 V)/(30 \Omega) = 0.37 A

  • Confirm mathematical consistency: I1 + I2 = I

32. Traversing Loops

• Series circuit consists of three resistances, equating to:

  • 1 \Omega + 4 \Omega + 5 \Omega = 10 \Omega

• Current flow:

  • I = \frac{10 V}{10 \Omega} = 1 A

• Traverse the loop noting potential changes starting prior to battery:

  • Initial voltage loss from the battery changes to 10 V before losses through each resistor by: I R

  • Sequential losses yield voltage results across the circuit path.

33. Traversing Loops

• Counter-clockwise traversal captures potential gains.

• Each resistor contributes voltage gains when moving against current, leading back to a potential of 10 V.

34. Traversing Loops

• Transition through a loop without known current can still establish values.

• Voltage computations yield equivalent current, regardless of current direction established for further calculations.

35. Traversing Loops

• Alternative traversal confirms correct potential differences lead to equivalent current calculations despite inversion of current direction maintaining absolute values.

36. Traversing Loops

• Misinterpretation of current egress direction may lead to erroneous values while validating flows through resistance-based voltage drops.

37. Kirchhoff’s Loop Rules

• Apply Kirchhoff’s rules for complex circuits with multiple loops and batteries, facilitating analysis.

38. Kirchhoff’s Loop Rules

• Select one battery; devise current direction ignoring others and map all circuit elements sequentially.

39. Kirchhoff’s Loop Rules

• Develop equations based on elements traversed through loops, ensuring total sums of gains and losses resolve to zero.

40. Kirchhoff’s Loop Rules

• Equations derive from each loop, amassed applying Ohm's law under current influences.

41. Kirchhoff’s Loop Rules

• From these relations, segment currents accordingly, applying junction equations to form sequence in total circuit dynamics.

42. Kirchhoff’s Loop Rules

• Analysis consolidates total power output calculations: each battery adjusts for losses and gains while accounting for voltage directionality.

43. Kirchhoff’s Loop Rules

• Confirm re-evaluations and direction changes after deriving all meaningful outputs in responses to identified sources.

44. Kirchhoff’s Loop Rules: Summary

• Revisit battery assumptions; gather equations covering junction and loops while guaranteeing constraints yield suitable reversibility of negative results to notations indicating flow.

45. Capacitors in Parallel

• Capacitors in parallel share voltage identical to that of the battery while allowing charge computations corresponding to each capacitor’s capacitance.

• Total charge across capacitors integrates their capacitances.

46. Capacitors in Parallel

• Equivalent capacitance summarized by additive properties:

  • C{eqp} = C1 + C2 + C3

47. Capacitors in Series

• Series capacitors share equal charge and add voltage across the arrangement.

48. Capacitors in Series

• Integrate voicings for voltages respectively across each capacitor to derive cumulative effects:

49. RC Circuits - Charging

• Introducing RC circuits modifies charging dynamics impacting responsive characteristics accordingly.

50. RC Circuits - Charging

• Depot-charge development behaviors showcase initial transitions before stabilizing into the battery’s charge voltage at a fixed rate.

51. RC Circuits - Charging

• Capacitor charging reflected through exponential relationships noting time constants’ influence on final accumulated charge states.

52. RC Circuits - Charging

• Values integrate logically with established behaviors through both immediate and long-term temporal references defining the capacitor’s charge behavior.

53. RC Circuits - Discharging

• Capacitors exhibit reverse discharges under battery isolation, altering flowing behavior to adjustable resistor movements.

54. RC Circuits - Discharging

• Continued application of mathematical representations correspondingly assesses charge reduction through defined time constants established under designated initial states.

55. RC Circuits - Discharging

• Discharging outcomes modelled on traction deviations towards zero charge over established time constants while noting resultant equations corresponding to elapsed intervals.