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
- Label all currents: Assign a direction to the current in each branch of the circuit.
- 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).
- Choose a loop and direction: Select any closed loop in the circuit and assign a direction (clockwise or counterclockwise) to traverse the loop.
- Write down voltage drops across resistors:
- Loop rule: Use the first sign you encounter when traversing the element.
- List voltages: Account for voltage sources (EMFs) and voltage drops across resistors.
- Write down current equation: Apply Kirchhoff's Junction Rule at appropriate junctions.
- 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.