Compound DC circuits integrate both series and parallel connections to create complex arrangements of circuit elements. These circuits appear in virtually all electronic devices and understanding how to analyze them is essential for predicting current flow, voltage distribution, and power dissipation throughout the system.

Equivalent Resistance in Circuits

Series and Parallel Connections

When analyzing compound circuits, we first need to understand the fundamental ways components can be connected together.

Series connections form a single pathway for current, like a one-lane road where all traffic must follow the same route. In these connections:

• Current is identical through each component
• The total voltage is distributed across components
• There are no junction points between elements

Parallel connections provide multiple pathways for current, similar to a highway with multiple lanes. In these connections:

• Voltage is identical across each component
• Current divides among the available paths
• Junction points exist where current can split or recombine

Equivalent Resistance Calculations

To simplify analysis of complex circuits, we can replace groups of resistors with a single equivalent resistance that would have the same effect on the circuit.

For resistors connected in series, the equivalent resistance is the sum of all individual resistances:

$R_{\mathrm{eq}, s}=\sum_{i} R_{i} = R_1 + R_2 + R_3 + ... + R_n$

This makes intuitive sense because each resistor in series adds more opposition to current flow, increasing the total resistance.

For resistors connected in parallel, we calculate the equivalent resistance using:

$\frac{1}{R_{e q, p}}=\sum_{i} \frac{1}{R_{i}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + ... + \frac{1}{R_n}$

Parallel connections always result in an equivalent resistance smaller than any individual resistor in the arrangement. This occurs because adding parallel paths is like widening a road—it allows more current to flow with the same voltage applied.

Circuits with Resistive Wires and Batteries

Ideal vs Nonideal Components

In introductory circuit analysis, we often use idealized components to simplify calculations. However, real-world components have limitations that affect circuit behavior.

Ideal components have these characteristics:

• Wires have zero resistance
• Batteries maintain constant voltage regardless of current
• Components have no energy losses other than those explicitly defined

Real components deviate from these ideals:

• Wires have small but measurable resistance
• Batteries have internal resistance that causes voltage to drop under load
• Components may have capacitive or inductive effects not accounted for in simple models

Internal Resistance Effects

A real battery can be modeled as an ideal voltage source (emf) in series with an internal resistance. This internal resistance represents the opposition to current flow within the battery itself, caused by the electrolyte and electrodes.

When current flows through a battery, some voltage is "lost" across this internal resistance, resulting in a terminal voltage lower than the battery's emf:

$\Delta V_{\text {terminal }}=\mathcal{E}-I r$

Where:

• $\mathcal{E}$ is the battery's electromotive force (ideal voltage)
• $I$ is the current through the battery
• $r$ is the battery's internal resistance

This explains why batteries under heavy load (high current) deliver less voltage to the circuit than their rated value.

Terminal Voltage Equation

The terminal voltage equation helps us understand how a battery's performance changes under different load conditions:

$\Delta V_{\text {terminal }}=\mathcal{E}-I r$

This relationship shows that:

• At zero current (open circuit), terminal voltage equals the emf
• As current increases, terminal voltage decreases linearly
• The slope of this decrease is determined by the internal resistance

A battery with low internal resistance will maintain its voltage better under load, which is why high-quality batteries are designed to minimize internal resistance.

Current and Voltage Measurement

Ammeter Usage and Placement

Ammeters measure the rate of charge flow (current) through a specific point in a circuit. To properly use an ammeter:

1. Break the circuit at the point where you want to measure current
2. Insert the ammeter in series at this break point
3. Ensure the ammeter's positive terminal faces the direction of conventional current

Ammeters are designed to have very low resistance to minimize their impact on the circuit being measured. An ideal ammeter would have zero resistance, allowing it to measure current without altering the circuit.

Voltmeter Usage and Placement

Voltmeters measure the electric potential difference between two points in a circuit. To properly use a voltmeter:

1. Identify the two points between which you want to measure voltage
2. Connect the voltmeter in parallel across these points
3. Ensure the positive lead connects to the higher potential point

Voltmeters are designed with very high resistance to minimize current draw through the meter. An ideal voltmeter would have infinite resistance, drawing no current from the circuit being measured.

Effects of Nonideal Measuring Devices

Real measuring instruments are not ideal and can affect the circuits they measure:

A nonideal ammeter has some resistance, which:

• Adds to the total circuit resistance
• Reduces the current flowing in the circuit
• Creates a voltage drop across the meter itself

A nonideal voltmeter has finite resistance, which:

• Creates a parallel current path
• Draws some current from the circuit
• May significantly affect high-resistance circuits

For accurate measurements, the ammeter's resistance should be much smaller than the circuit being measured, and the voltmeter's resistance should be much larger than the circuit components across which it's connected.

Practice Problem 1: Equivalent Resistance

Solution: First, we need to find the equivalent resistance of R₁ and R₂ in parallel:

$\frac{1}{R_{eq,p}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{4Ω} + \frac{1}{6Ω} = \frac{6 + 4}{24Ω} = \frac{10}{24Ω} = \frac{5}{12Ω}$

$R_{eq,p} = \frac{12Ω}{5} = 2.4Ω$

Now, we find the total equivalent resistance by adding this value to R₃ (since they're in series):

$R_{eq,total} = R_{eq,p} + R_3 = 2.4Ω + 12Ω = 14.4Ω$

Therefore, the equivalent resistance of the entire circuit is 14.4Ω.

Practice Problem 2: Battery with Internal Resistance

Solution: (a) To find the current, we use Ohm's law with the total resistance (internal plus external):

$I = \frac{\mathcal{E}}{R_{total}} = \frac{\mathcal{E}}{r + R} = \frac{12V}{0.5Ω + 5.5Ω} = \frac{12V}{6Ω} = 2A$

(b) To find the terminal voltage, we use the terminal voltage equation:

$\Delta V_{terminal} = \mathcal{E} - Ir = 12V - (2A)(0.5Ω) = 12V - 1V = 11V$

The terminal voltage of the battery is 11V, which is less than the emf due to the voltage drop across the internal resistance.

Practice Problem 3: Measuring Instruments

Solution: First, let's calculate the voltage across R₂ without the voltmeter using the voltage divider principle:

$V_{R2} = \frac{R_2}{R_1 + R_2} \times 9V = \frac{2kΩ}{1kΩ + 2kΩ} \times 9V = \frac{2}{3} \times 9V = 6V$

When the voltmeter is connected across R₂, it creates a parallel combination with R₂:

$R_{parallel} = \frac{R_2 \times R_{voltmeter}}{R_2 + R_{voltmeter}} = \frac{2kΩ \times 100kΩ}{2kΩ + 100kΩ} = \frac{200,000kΩ}{102kΩ} \approx 1.96kΩ$

Now we recalculate the voltage using this new equivalent resistance:

$V_{measured} = \frac{R_{parallel}}{R_1 + R_{parallel}} \times 9V = \frac{1.96kΩ}{1kΩ + 1.96kΩ} \times 9V = \frac{1.96}{2.96} \times 9V \approx 5.96V$

The voltmeter will read approximately 5.96V, which is slightly less than the actual 6V that would exist without the voltmeter. This demonstrates how even a high-resistance voltmeter can slightly affect the circuit it's measuring.