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: $R<em>{eq} = R</em>1 + R<em>2 + R</em>3 + …$
• Parallel: Resistors connected side-by-side such that the voltage across each is the same.
  • Equivalent Resistance: $\frac{1}{R<em>{eq}} = \frac{1}{R</em>1} + \frac{1}{R<em>2} + \frac{1}{R</em>3} + …$

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.
• $\sum I<em>{in} = \sum I</em>{out}$

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 ($\epsilon<em>1 = 10V$ and $\epsilon</em>2 = 5V$) and two resistors ($R<em>1 = 10 \Omega$ and $R</em>2 = 10 \Omega$).
• Objective: Solve for the currents in all parts of the circuit ($I<em>1$, $I</em>2$, and $I_3$).
• Junction Equation: $I<em>1 = I</em>2 + I_3$
• Loop Equations:
  • Bottom Loop Clockwise from A: $I<em>2R</em>1 + \epsilon<em>2 - \epsilon</em>1 = 0$ (1)
  • Top Loop Clockwise from A: $I<em>3R</em>2 - \epsilon<em>2 - I</em>2R_1 = 0$ (2)
  • Outside Loop Clockwise from A: $I<em>3R</em>2 - \epsilon_1 = 0$ (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 $R<em>2$ in parallel with $R</em>3$ and $R<em>4$. This decreases the overall equivalent resistance of the parallel combination, leading to a lower voltage across $R</em>4$.

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: $I<em>1 + I</em>2 = I_3$

Clicker Question 4

• Question: According to Kirchhoff's Loop Rule, what does the left loop tell us about the circuit?
• Correct Answer: $12 V - (3\Omega)I<em>1 - (6\Omega)I</em>2 = 0$

Clicker Question 5

• Question: According to Kirchhoff's Loop Rule, what does the right loop tell us about the circuit?
• Correct Answer: $-12 V + (4\Omega)I<em>2 + (6\Omega)I</em>3 = 0$

Clicker Question 6

• Question: According to Kirchhoff's Loop Rule, what does the outside loop tell us about the circuit?
• Correct Answer: $24 V - (3\Omega)I<em>1 - (4\Omega)I</em>3 = 0$

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.