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Physics 2B: Lecture 17 - Electric Circuits and Kirchhoff's Laws

Resistance, Resistors, and Ohm's Law

  • Resistance is a fundamental property of materials that opposes the flow of electric current.
  • Resistors are circuit components designed to provide a specific resistance.
  • Ohm's Law: V = IR, where:
    • V is the voltage across the resistor.
    • I is the current flowing through the resistor.
    • R is the resistance of the resistor.

Series and Parallel Resistors

  • Series: Resistors connected end-to-end such that the same current flows through each.
    • Equivalent Resistance: R{eq} = R1 + R2 + R3 + …
  • Parallel: Resistors connected side-by-side such that the voltage across each is the same.
    • Equivalent Resistance: \frac{1}{R{eq}} = \frac{1}{R1} + \frac{1}{R2} + \frac{1}{R3} + …

Electric Circuits and Current

  • Electric circuits provide a closed path for current to flow.
  • Current (I) is the rate of flow of electric charge, measured in amperes (A).

Kirchhoff's Laws

  • Tools for analyzing complex circuits.

Kirchhoff’s Junction Law (Current Law)

  • States that the total current entering a junction (node) must equal the total current leaving the junction.
  • Based on the conservation of charge.
  • \sum I{in} = \sum I{out}

Kirchhoff’s Loop Law (Voltage Law)

  • States that the algebraic sum of the potential differences (voltages) around any closed loop in a circuit must be zero.
  • Based on the conservation of energy.
  • \sum V = 0

Energy and Power

  • Power Dissipated by a Resistor: The rate at which electrical energy is converted to heat in a resistor.
    • Formula: P = I^2R
    • Alternative Formula: P = \frac{V^2}{R}
  • Power is a key metric to determine brightness of a bulb.

Steps for Using Kirchhoff’s Rules

  1. Label all currents: Assign a direction to the current in each branch of the circuit.
  2. Label +/- for all elements:
    • Emf (Electromotive Force) sources have a regular + and - terminal.
    • For resistors, the current goes from + to – (representing a decrease in voltage).
  3. Choose a loop and direction: Select any closed loop in the circuit and assign a direction (clockwise or counterclockwise) to traverse the loop.
  4. Write down voltage drops across resistors:
    • Loop rule: Use the first sign you encounter when traversing the element.
  5. List voltages: Account for voltage sources (EMFs) and voltage drops across resistors.
  6. Write down current equation: Apply Kirchhoff's Junction Rule at appropriate junctions.
  7. Junction rule: Express the relationship between currents entering and leaving a junction.

Example Problem and Solution

  • Circuit Description: A circuit with two voltage sources (\epsilon1 = 10V and \epsilon2 = 5V) and two resistors (R1 = 10 \Omega and R2 = 10 \Omega).
  • Objective: Solve for the currents in all parts of the circuit (I1, I2, and I_3).
  • Junction Equation: I1 = I2 + I_3
  • Loop Equations:
    • Bottom Loop Clockwise from A: I2R1 + \epsilon2 - \epsilon1 = 0 (1)
    • Top Loop Clockwise from A: I3R2 - \epsilon2 - I2R_1 = 0 (2)
    • Outside Loop Clockwise from A: I3R2 - \epsilon_1 = 0 (3)
  • Solutions:
    • I_3 = 1 A
    • I_2 = 0.5 A
    • I_1 = 1.5 A

Clicker Questions and Answers

Clicker Question 1

  • Question: What happens to the voltage across resistor R_4 when the switch is closed in the given circuit?
  • Correct Answer: The voltage across R_4 decreases.
  • Explanation: Closing the switch adds R2 in parallel with R3 and R4. This decreases the overall equivalent resistance of the parallel combination, leading to a lower voltage across R4.

Clicker Question 2

  • Question: Which resistor has the greatest current going through it, assuming all resistors are equal?
  • Answer: R_1

Clicker Question 3

  • Question: According to Kirchhoff's Current Rule, which equation is correct for the given circuit?
  • Correct Answer: I1 + I2 = I_3

Clicker Question 4

  • Question: According to Kirchhoff's Loop Rule, what does the left loop tell us about the circuit?
  • Correct Answer: 12 V - (3\Omega)I1 - (6\Omega)I2 = 0

Clicker Question 5

  • Question: According to Kirchhoff's Loop Rule, what does the right loop tell us about the circuit?
  • Correct Answer: -12 V + (4\Omega)I2 + (6\Omega)I3 = 0

Clicker Question 6

  • Question: According to Kirchhoff's Loop Rule, what does the outside loop tell us about the circuit?
  • Correct Answer: 24 V - (3\Omega)I1 - (4\Omega)I3 = 0

Additional Notes

  • Current can only change at a junction.
  • In a segment without a junction, current must remain constant.
  • Loops do not need to go in the same direction as currents.