AP Physics 2- 11.5 Compound Direct Current (DC) Circuits- Study Notes- New Syllabus
AP Physics 2- 11.5 Compound Direct Current (DC) Circuits – Study Notes
AP Physics 2- 11.5 Compound Direct Current (DC) Circuits – Study Notes – per latest Syllabus.
Key Concepts:
- Equivalent Resistance of Multiple Resistors
- Circuit with Resistive Wires and a Battery with Internal Resistance
- Measurement of Current and Potential Difference in a Circuit
Equivalent Resistance of Multiple Resistors
The equivalent resistance of a combination of resistors is the single resistance that can replace them in a circuit without changing the overall current-voltage relationship.
Resistors in Series:
- Current is the same through each resistor.
- The total (equivalent) resistance is the sum of all resistances:
\(\text{R}_{\text{eq}} = R_1 + R_2 + R_3 + \dots \)
- The equivalent resistance is always greater than the largest individual resistance.
- Voltage Rule: The total voltage is the sum of voltages across each resistor: \(\text{V}_{\text{total}} = V_1 + V_2 + V_3 + \dots\).
Resistors in Parallel:
- Voltage across each resistor is the same.
- The reciprocal of the total resistance is the sum of reciprocals of each resistance:
\(\dfrac{1}{\text{R}_{\text{eq}}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3} + \dots \)
- The equivalent resistance is always less than the smallest individual resistance.
- Current Rule: The total current divides among branches: \(\text{I}_{\text{total}} = I_1 + I_2 + I_3 + \dots\).
Combination Circuits:
- Circuits often contain both series and parallel parts.
- Apply the series rule first, then the parallel rule step by step, until the circuit is simplified.
Example:
Find the equivalent resistance of three resistors in series: \(R_1 = 2 \, \Omega\), \(R_2 = 3 \, \Omega\), and \(R_3 = 5 \, \Omega\).
▶️ Answer/Explanation
Step 1: Series formula: \(\text{R}_{\text{eq}} = R_1 + R_2 + R_3\).
Step 2: Substitute values: \(\text{R}_{\text{eq}} = 2 + 3 + 5 = 10 \, \Omega\).
Answer: Equivalent resistance = \(10 \, \Omega\).
Extra Insight: Current is the same through all three resistors, but each resistor has a different voltage drop.
Example:
Find the equivalent resistance of two resistors in parallel: \(R_1 = 6 \, \Omega\), \(R_2 = 3 \, \Omega\).
▶️ Answer/Explanation
Step 1: Parallel formula: \(\dfrac{1}{\text{R}_{\text{eq}}} = \dfrac{1}{R_1} + \dfrac{1}{R_2}\).
Step 2: Substitute values: \(\dfrac{1}{\text{R}_{\text{eq}}} = \dfrac{1}{6} + \dfrac{1}{3} = \dfrac{1}{6} + \dfrac{2}{6} = \dfrac{3}{6}\).
Step 3: Invert: \(\text{R}_{\text{eq}} = \dfrac{6}{3} = 2 \, \Omega\).
Answer: Equivalent resistance = \(2 \, \Omega\).
Extra Insight: Voltage across both resistors is the same, but current divides: \(I_1\) through \(R_1\) and \(I_2\) through \(R_2\).
Circuit with Resistive Wires and a Battery with Internal Resistance
Real Battery Model:
- A real battery can be modeled as an ideal emf source (\(\text{ε}\)) in series with an internal resistance (\(\text{r}\)).
- The emf represents the maximum potential difference the battery can provide when no current flows.
- The internal resistance accounts for energy loss due to the battery’s internal materials and chemical processes.
Potential Difference Across Battery Terminals:
When current flows, the voltage across the external circuit (terminal voltage) is:
\(\text{V}_{\text{terminal}} = \text{ε} – I\text{r}\)
- \(\text{ε}\): emf of the battery (ideal voltage)
- \(I\): current in the circuit
- \(\text{r}\): internal resistance of the battery
Effect of Resistive Wires:
- Resistive wires add resistance (\(\text{R}_{\text{wire}}\)) to the circuit, reducing the current.
- Total circuit resistance: \(\text{R}_{\text{total}} = R_{\text{load}} + R_{\text{wire}} + r\).
- Energy is dissipated in both the load resistor and the internal/battery resistance as heat.
Power Distribution:
- Power delivered by the battery: \(\text{P}_{\text{battery}} = \text{ε}I\).
- Useful power delivered to the load: \(\text{P}_{\text{load}} = I^2R_{\text{load}}\).
- Power lost inside the battery: \(\text{P}_{\text{internal}} = I^2r\).
Example :
A battery with emf \(\text{ε} = 12 \, V\) and internal resistance \(\text{r} = 0.5 \, \Omega\) is connected to a \(5.5 \, \Omega\) resistor. Find the current in the circuit and the terminal voltage.
▶️ Answer/Explanation
Step 1: Total resistance: \(\text{R}_{\text{total}} = R_{\text{load}} + r = 5.5 + 0.5 = 6.0 \, \Omega\).
Step 2: Current: \(I = \dfrac{\text{ε}}{\text{R}_{\text{total}}} = \dfrac{12}{6} = 2.0 \, A\).
Step 3: Terminal voltage: \(\text{V}_{\text{terminal}} = \text{ε} – I r = 12 – (2.0)(0.5) = 11 \, V\).
Answer: Current = \(2.0 \, A\), Terminal Voltage = \(11 \, V\).
Measurement of Current and Potential Difference in a Circuit
Ammeters:
An ammeter is a device used to measure the current at a specific point in a circuit.
- It must be connected in series with the element where the current is being measured, because the same current flows through elements in series.
- Ideal Ammeter: Has zero resistance so that it does not reduce the current in the circuit or change the voltage distribution.
- Real Ammeter: Has very small resistance, which slightly decreases the current and affects the circuit minimally.
Voltmeters:
A voltmeter is a device used to measure the potential difference (voltage) between two points in a circuit.
- It must be connected in parallel with the component across which the voltage is to be measured, since all elements in parallel share the same voltage.
- Ideal Voltmeter: Has infinite resistance so that no current flows through it, ensuring it does not change the current distribution in the circuit.
- Real Voltmeter: Has very high resistance, but not infinite, so a tiny current flows through it and slightly alters the measurement.
Practical Considerations:
- When measuring current, always connect the ammeter in series and ensure it can handle the expected current to avoid damage.
- When measuring voltage, always connect the voltmeter in parallel across the component and ensure the scale is suitable for the expected potential difference.
- Non ideal ammeters and voltmeters introduce slight changes in circuit behavior; therefore, high-quality instruments are designed to minimize these effects.
Example:
A resistor of \(10 \, \Omega\) is connected to a \(5 \, V\) battery. An ideal ammeter is used to measure the current. What current does the ammeter read?
▶️ Answer/Explanation
Step 1: Use Ohm’s law: \(\text{I} = \dfrac{\text{V}}{R} = \dfrac{5}{10} = 0.5 \, A\).
Step 2: Since the ammeter is ideal (zero resistance), it does not affect the current.
Answer: The ammeter reads \(0.5 \, A\).
Example:
A \(12 \, V\) battery is connected across a \(6 \, \Omega\) resistor. An ideal voltmeter is connected across the resistor. What voltage does the voltmeter read?
▶️ Answer/Explanation
Step 1: The potential difference across the resistor is equal to the battery emf (since only one resistor is connected): \(\text{V} = 12 \, V\).
Step 2: Since the voltmeter is ideal (infinite resistance), no current flows through it and it does not disturb the circuit.
Answer: The voltmeter reads \(12 \, V\).