Unveiling the Secrets of DC Circuits

This chapter focuses on direct current (DC) circuits containing resistors. You will learn how to analyze circuits where electric charge flows consistently in a single direction, and discover fundamental laws and techniques for understanding circuit behavior.

Learning Objectives:

• Apply Kirchhoff's Current and Voltage Laws to analyze DC circuits
• Calculate power supplied and absorbed in DC circuits
• Analyze series and parallel resistive circuits
• Use voltage divider principles to determine voltages in resistive circuits
• Understand how measuring instruments interact with circuits
• Apply circuit simplification techniques to solve complex problems

Kirchhoff's Laws: Currents' Crossroads and Voltage's Roundabout

Podcast: Kirchhoff Laws

Kirchhoff's Laws are fundamental tools in electrical circuit analysis that allow us to analyze complex circuits by relating currents and voltages. These laws are based on the principles of conservation of charge and energy, making them universally applicable to all electrical circuits.

Currents at Node (KCL)

Kirchhoff's Current Law (KCL) is based on the conservation of charge. It states that:

The sum of all currents entering a node equals the sum of all currents leaving that node.

A node is a point where multiple wires are connected. Since charge cannot accumulate at any point in the circuit, the net current flowing into any node must be zero. This can be expressed mathematically as:

$$\sum_{i=1}^n I_i = 0$$

where $I_i$ represents individual currents at the node, and $n$ is the total number of currents at that node. By convention, currents entering the node are considered positive, and currents leaving are considered negative.

Figure: KCL application at a node: Current $I_1$ enters the node, while currents $I_2$ and $I_3$ leave. By KCL, $I_1 = I_2 + I_3$ or equivalently $I_1 - I_2 - I_3 = 0$.

Applying KCL

To apply KCL in circuit analysis, follow these steps:

1. Identify the Nodes: Find all points where three or more circuit elements connect.
2. Assign Current Directions: For each branch connected to a node, assign a direction to the current. The actual direction does not need to be known beforehand. If the assumption is incorrect, the calculated current will simply be negative.
3. Apply KCL at Each Node: Write an equation based on KCL. You can express this as either:
4. Sum of currents entering equals sum of currents leaving: $\sum I_{\text{in}} = \sum I_{\text{out}}$
5. Algebraic sum of currents equals zero: $\sum I_i = 0$ (currents entering are positive, currents leaving are negative)
6. Solve the Equations: Apply KCL to all but one node in the circuit (the remaining equation would be redundant). Combine with other circuit relationships to solve for unknown currents.

Note: For a circuit with $n$ nodes, you can write at most $n-1$ independent KCL equations. The remaining equation can be derived from the others.

Example 1: Finding Currents in a Parallel Circuit

What are the currents $I_1$ and $I_2$ flowing through resistors $R_1$ and $R_2$ in this circuit?

Solution

Applying KCL at the top node, with both $I_1$ and $I_2$ leaving the node:

$$2 = I_1 + I_2$$

Since the resistors are connected in parallel, the voltage drop across them is the same. Applying Ohm's Law ($V = IR$) to each resistor:

$$3 I_1 = 4 I_2$$

Solving for $I_1$ in terms of $I_2$: $I_1 = \frac{4}{3}I_2$. Substituting into the KCL equation:

$$2 = \frac{4}{3}I_2 + I_2 = \frac{7}{3}I_2 \implies I_2 = \frac{6}{7}\,\text{A} \approx 0.857\,\text{A}$$

Then: $I_1 = \frac{8}{7}\,\text{A} \approx 1.143\,\text{A}$

Verification: $V_1 = I_1 R_1 = \frac{8}{7} \cdot 3 = \frac{24}{7}\,\text{V} \approx 3.43\,\text{V}$ and $V_2 = I_2 R_2 = \frac{6}{7} \cdot 4 = \frac{24}{7}\,\text{V} \approx 3.43\,\text{V}$. The voltages match, confirming the solution.

Example 2: Battery and Parallel Resistors

What are the currents $I_0$, $I_1$, and $I_2$ in this circuit?

Solution

At the node where $R_1$ and $R_2$ meet, KCL gives: $I_0 = I_1 + I_2$.

Since the resistors are in parallel, both see 12 V. Using Ohm's Law:

$$I_1 = \frac{12}{4} = 3\,\text{A}, \quad I_2 = \frac{12}{6} = 2\,\text{A}$$

Therefore: $I_0 = I_1 + I_2 = 3\,\text{A} + 2\,\text{A} = 5\,\text{A}$

Voltage's Voyage: A Loop de Loop (KVL)

Kirchhoff's Voltage Law (KVL) is based on the principle of energy conservation. It states that:

The sum of all voltage drops and rises around any closed loop in a circuit must be equal to zero.

This implies that the net change in energy for a charge traversing a closed loop is zero. Consider the circuit below, which contains four circuit elements. Applying KVL to this loop:

$$V_1 + V_2 + V_3 + V_4 = 0$$

where $V_i$ represents the voltage across each component.

Figure: KVL: The sum of voltages around a closed loop is zero.

Applying KVL

To apply KVL in circuit analysis, follow these steps:

1. Identify Loops: Find closed paths in the circuit where you start and end at the same node.
2. Choose a Direction: For each loop, choose either clockwise or counterclockwise as your traversal direction.
3. Assign Voltage Signs: As you move around the loop:
4. For voltage sources: if you move from negative to positive terminal, assign a positive sign (voltage rise). If from positive to negative, assign a negative sign (voltage drop).
5. For resistors or other passive elements: if the assumed current direction matches your loop direction, assign a negative sign (voltage drop). If directions are opposite, assign a positive sign.
6. Write the KVL Equation: Sum all voltages around the loop with proper signs and set the sum equal to zero.
7. Solve: Combine with other circuit relationships to find unknown voltages and currents.

Example 3: Finding a Voltage Using KVL

Consider a circuit with a 10 V voltage source and two passive elements #1 and #2. What is the voltage $V_{AB}$ across element #1?

Solution

Since the node voltages (relative to ground) at both ends of element #1 are known, $V_{AB}$ can be found directly:

$$V_{AB} = V_A - V_B = (10 - 2)\,\text{V} = 8\,\text{V}$$

Alternatively, a KVL equation for the loop gives: $10\,\text{V} - V_{AB} - V_{BC} = 0$, so $V_{AB} = 10 - V_{BC} = (10 - 2)\,\text{V} = 8\,\text{V}$.

Example 4: Using KVL to Find Current

What is the current $I$ in this circuit?

Solution

1. Choose a Loop: The entire circuit forms a single loop.
2. Assign a Direction: Clockwise.
3. Determine Voltage Changes:
4. Battery: $+12\,\text{V}$ (negative to positive terminal)
5. Resistor $R_1$: voltage drop $-IR_1 = -4I$
6. Resistor $R_2$: voltage drop $-IR_2 = -8I$
7. Apply KVL: $12 - 4I - 8I = 0$
8. Solve: $12 - 12I = 0 \implies I = 1\,\text{A}$

Example 5: KVL with Multiple Sources

What is the current $I$ in this circuit with two voltage sources?

Solution

1. Choose a loop: Single loop, clockwise.
2. Identify voltage drops and rises:
3. $V_1$ (11 V battery): $+11\,\text{V}$
4. $R_1$ (2 Ω): $-2I$
5. $V_2$ (3 V battery): $-3\,\text{V}$ (traversing positive to negative)
6. $R_2$ (6 Ω): $-6I$
7. Apply KVL: $11 - 2I - 3 - 6I = 0 \implies 8 - 8I = 0 \implies I = 1\,\text{A}$

Because of KCL (the current is the same through all components in a single loop), the circuit of Example 5 can also be drawn as:

This further reduces to the circuit below, because by KVL the 3 V battery subtracts from the 11 V battery to give an effective 8 V source:

Additional Resources

• Kirchhoff's Rules (Laws) — Introduction (Video)
• Kirchhoff's Circuit Laws (Wikipedia)

Power and Energy in DC Circuits

Podcast: Power in Electrical Circuits

In Chapter 1, electric potential was described as analogous to height: a charge at higher potential holds stored energy, just as a rock on a hill holds gravitational potential energy. When charges move through a circuit, that stored energy is transferred to the components they pass through. Power is the rate at which this transfer occurs.

Energy Transfer in Circuits

A common misconception about electrical circuits is that current gets consumed as it flows through components. Electrical charge is conserved. What actually happens is that energy is transferred and transformed as charges move through the circuit. The charges themselves continue to circulate; only their energy changes form.

Calculating Power

Power is the rate at which energy is transferred. In electrical systems, the power exchanged by any element is the product of the voltage across it and the current through it:

$$P = V \times I$$

where $P$ is power in watts (W), $V$ is voltage in volts (V), and $I$ is current in amperes (A). The sign of the result distinguishes two cases. When $P$ is positive, the element absorbs energy and converts it to another form, as a resistor does when it produces heat. When $P$ is negative, the element supplies energy to the rest of the circuit, as a battery does when it discharges.

Power in Resistors

For a resistor, the relationship $V = IR$ allows the power equation to be written in two additional equivalent forms:

$$P = I^2 R = \frac{V^2}{R}$$

Both forms are useful: the first when current is known, the second when voltage is known. Because $R$ is always positive, power absorbed by a resistor is always positive. A resistor never supplies energy; it only absorbs it.

Note: Every resistor has a maximum power rating, expressed in watts. Exceeding this rating causes the resistor to overheat. Common discrete resistors are rated at 1/4 W or 1/2 W. When designing a circuit, always verify that the expected power dissipation stays within the component's rating.

Power Conservation

Because energy is conserved, the total power supplied in a circuit must equal the total power absorbed:

$$\sum P_{\text{supplied}} = \sum P_{\text{absorbed}}$$

This identity provides a useful check: after computing all currents and voltages in a circuit, the power balance should hold exactly.

Energy vs. Power

Power is the rate of energy transfer, while energy is the total amount of work done over a period of time:

$$E = P \times t$$

where $E$ is energy in joules (J) and $t$ is time in seconds (s). In practical contexts, energy is often expressed in watt-hours (Wh): a 60 W bulb running for one hour consumes 60 Wh, or 216,000 J.

Example 6: Power in a Series Circuit

Using the circuit from Example 4 (a 12 V source with $R_1 = 4\,\Omega$ and $R_2 = 8\,\Omega$ in series), calculate the power supplied by the source and the power absorbed by each resistor. Verify that power is conserved.

Solution

From Example 4, the current in the circuit is $I = 1\,\text{A}$.

$$P_{\text{source}} = V \times I = 12 \times 1 = 12\,\text{W}$$

$$P_{R_1} = I^2 R_1 = (1)^2 \times 4 = 4\,\text{W}$$

$$P_{R_2} = I^2 R_2 = (1)^2 \times 8 = 8\,\text{W}$$

Checking conservation:

$$P_{R_1} + P_{R_2} = 4\,\text{W} + 8\,\text{W} = 12\,\text{W} = P_{\text{source}} \checkmark$$

The power delivered by the source is completely absorbed by the two resistors.

Example 7: Power Rating Check in a Voltage Divider

A voltage divider consists of $R_1 = 4.7\,\text{k}\Omega$ and $R_2 = 3.3\,\text{k}\Omega$ connected in series across a 12 V source (this is the circuit from Example 12 in the Voltage Divider section). Standard 1/4 W resistors are available. Determine whether they are adequate for this application.

Solution

The total series resistance is $R_{\text{total}} = 4.7\,\text{k}\Omega + 3.3\,\text{k}\Omega = 8\,\text{k}\Omega$.

$$I = \frac{V}{R_{\text{total}}} = \frac{12}{8000} = 1.5\,\text{mA}$$

$$\begin{aligned} > P_{R_1} &= I^2 R_1 = (1.5 \times 10^{-3})^2 \times 4700 \approx 10.6\,\text{mW} \\ > P_{R_2} &= I^2 R_2 = (1.5 \times 10^{-3})^2 \times 3300 \approx 7.4\,\text{mW} > \end{aligned}$$

Both values are well below 250 mW (the 1/4 W rating), so standard resistors are adequate. The total power drawn from the source is approximately 18 mW.

This calculation illustrates why voltage dividers intended only as reference networks are built from high-value resistors: high resistance reduces current, which reduces power dissipation.

Series and Parallel Connections

Podcast: Series and Parallel Connections

Two fundamental ways to connect circuit elements are in series and in parallel. Understanding these connection types simplifies the analysis of complex circuits by allowing us to reduce them to simpler equivalent circuits.

Defining Series and Parallel Connections

Series Connection

Two circuit elements are connected in series when they share a single node that connects only these two devices. This means:

• The elements are connected end-to-end
• The same current flows through both elements
• Every charge leaving one element must pass through the other

Figure: Components #1 and #2 are connected in series.

Parallel Connection

Two elements are connected in parallel when they are connected across the same two nodes. This means:

• Both elements share the same two connection points
• The voltage across both elements is identical
• Current divides between the elements

Figure: Components #1 and #2 are connected in parallel.

Combinations of Series and Parallel

Many circuits contain combinations of series and parallel connections. In such cases, it is often helpful to identify which components are in series and which are in parallel, and then simplify the circuit step by step.

Figure: A complex circuit: Components #1 and #2 are in series. The series combination of #1 and #2 is in parallel with the series combination of #3 and #4.

Resistors in Series

When resistors are connected in series, the total resistance is the sum of the individual resistances:

$$R_{total} = R_1 + R_2 + \ldots + R_n$$

Figure: Resistors in series. The total resistance is $R_{total} = R_1 + R_2$.

Some key properties of series-connected resistors:

• The same current flows through each resistor
• The voltage divides across the resistors proportionally to their resistance values
• If $R_1 \gg R_2$ ($R_1$ is much larger than $R_2$), then $R_{total} \approx R_1$
• If $R_1 = R_2$, then $R_{total} = 2R_1$

Example 8: Resistors in Series

Calculate the total resistance of a 330 Ω resistor in series with a 470 Ω resistor.

Solution

$$R_{total} = R_1 + R_2 = 330\,\Omega + 470\,\Omega = 800\,\Omega$$

If a voltage of 8 V is applied across this series combination, the current is:

$$I = \frac{V}{R_{total}} = \frac{8}{800} = 10\,\text{mA}$$

This 10 mA current flows through both resistors. The voltage across each resistor is:

$$V_1 = I \cdot R_1 = 10 \times 10^{-3} \cdot 330 = 3.3\,\text{V}$$

$$V_2 = I \cdot R_2 = 10 \times 10^{-3} \cdot 470 = 4.7\,\text{V}$$

Note that $V_1 + V_2 = 3.3\,\text{V} + 4.7\,\text{V} = 8\,\text{V}$, which equals the total applied voltage.

Resistors in Parallel

When resistors are connected in parallel, the reciprocal of the total resistance equals the sum of the reciprocals of the individual resistances:

$$\frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} + \ldots + \frac{1}{R_n}$$

For two resistors in parallel, this simplifies to:

$$R_{total} = \frac{R_1 R_2}{R_1 + R_2}$$

Figure: Resistors in parallel. The total resistance is $R_{total} = \frac{R_1 R_2}{R_1 + R_2}$.

Some key properties of parallel-connected resistors:

• The same voltage appears across each resistor
• The current divides between the resistors inversely proportional to their resistance values
• The total resistance is always less than the smallest individual resistance

Two useful approximations for parallel resistance:

• If $R_1 \gg R_2$ ($R_1$ is much larger than $R_2$), then $R_{total} \approx R_2$
• If $R_1 = R_2$, then $R_{total} = R_1/2$

Example 9: Resistors in Parallel

Calculate the total resistance of a 100 Ω resistor in parallel with a 200 Ω resistor.

Solution

$$R_{total} = \frac{R_1 \cdot R_2}{R_1 + R_2} = \frac{100 \cdot 200}{100 + 200} = \frac{20{,}000}{300} = 66.7\,\Omega$$

If a voltage of 10 V is applied across this parallel combination, the total current is:

$$I_{total} = \frac{V}{R_{total}} = \frac{10\,\text{V}}{66.7\,\Omega} = 0.15\,\text{A} = 150\,\text{mA}$$

This current divides between the two resistors:

$$I_1 = \frac{V}{R_1} = \frac{10}{100} = 100\,\text{mA}, \quad I_2 = \frac{V}{R_2} = \frac{10}{200} = 50\,\text{mA}$$

Verification: $I_1 + I_2 = 100\,\text{mA} + 50\,\text{mA} = 150\,\text{mA} = I_{total}$, confirming KCL.

Example 10: Special Case — Equal Resistors in Parallel

Calculate the equivalent resistance of three identical 120 Ω resistors connected in parallel.

Solution

For $n$ identical resistors of value $R$ connected in parallel, the equivalent resistance is $R_{total} = R/n$.

In this case: $R_{total} = 120\,\Omega / 3 = 40\,\Omega$

Verification:

$$\frac{1}{R_{total}} = \frac{1}{120} + \frac{1}{120} + \frac{1}{120} = \frac{3}{120} = \frac{1}{40} \implies R_{total} = 40\,\Omega$$

Series-Parallel Combinations

Many practical circuits involve combinations of series and parallel connections. To analyze these circuits, simplify them step by step, replacing series or parallel combinations with their equivalent resistances.

Example 11: Series-Parallel Circuit

Calculate the total resistance and current in this circuit.

Solution

1. Identify that $R_2$ and $R_3$ are in parallel:

$$R_{parallel} = \frac{R_2 \cdot R_3}{R_2 + R_3} = \frac{12 \cdot 4}{12 + 4} = \frac{48}{16} = 3\,\Omega$$

1. The parallel combination is in series with $R_1$:

$$R_{total} = R_1 + R_{parallel} = 6\,\Omega + 3\,\Omega = 9\,\Omega$$

1. Calculate the total current:

$$I_1 = \frac{V}{R_{total}} = \frac{24}{9} = 2.67\,\text{A}$$

1. The voltage across the parallel combination is:

$$V_{parallel} = I_1 \cdot R_{parallel} = 2.67 \cdot 3 = 8\,\text{V}$$

1. Currents through each parallel resistor:

$$I_2 = \frac{V_{parallel}}{R_2} = \frac{8}{12} = 0.67\,\text{A}, \quad I_3 = \frac{V_{parallel}}{R_3} = \frac{8}{4} = 2\,\text{A}$$

1. Verification using KCL: $I_1 = I_2 + I_3 = 0.67\,\text{A} + 2\,\text{A} = 2.67\,\text{A}$ ✓

Additional Resources

• Series and parallel circuits (Wikipedia)
• Resistors in Series (Khan Academy)
• Resistors in Parallel (Khan Academy)

Voltage Divider

Podcast: Voltage Divider

A voltage divider is a simple yet fundamental circuit configuration that produces a fraction of a source voltage. It consists of two or more resistors connected in series across a voltage source, with an output taken across one of the resistors.

Basic Voltage Divider Principle

Consider the basic voltage divider circuit below, consisting of two resistors $R_1$ and $R_2$ connected in series across a voltage source $V$.

Figure: A basic voltage divider circuit. The output voltage $V_B$ is taken across resistor $R_2$.

In this circuit:

• $V$ is the input voltage (from the source)
• $V_A$ is the voltage at the top node
• $V_B$ is the voltage at the middle node (the output)
• $V_C$ is the voltage at the bottom node (ground, 0 V)

Deriving the Voltage Divider Formula

To find the output voltage $V_B$, apply Ohm's Law and the principles of series circuits:

• The same current $I$ flows through both resistors.
• Using Ohm's Law, the voltage across $R_2$ is $V_B = I \times R_2$.
• The total voltage $V$ equals the sum of voltages across both resistors: $V = I \times (R_1 + R_2)$, so $I = \frac{V}{R_1 + R_2}$.
• Substituting:

$$V_B = I \times R_2 = \frac{V}{R_1 + R_2} \times R_2 = V \times \frac{R_2}{R_1 + R_2}$$

This gives the voltage divider formula:

$$V_{out} = V_{in} \times \frac{R_2}{R_1 + R_2}$$

Example 12: Basic Voltage Divider

Calculate the output voltage $V_B$ in a voltage divider with $V = 12\,\text{V}$, $R_1 = 4.7\,\text{k}\Omega$, and $R_2 = 3.3\,\text{k}\Omega$.

Solution

$$\begin{aligned} > V_B &= V \times \frac{R_2}{R_1 + R_2} \\ > &= 12 \times \frac{3.3 \times 10^{3}}{4.7 \times 10^{3} + 3.3 \times 10^{3}} \\ > &= 12\,\text{V} \times \frac{3.3}{8} \\ > &= 12\,\text{V} \times 0.4125 \\ > &= 4.95\,\text{V} > \end{aligned}$$

The output voltage $V_B$ is 4.95 V.

Applications of Voltage Dividers

Voltage dividers have numerous practical applications:

• Level shifting: Converting between different voltage levels in a circuit
• Biasing: Setting a specific operating point for transistors and other components
• Reference voltages: Creating stable reference voltages for comparators and operational amplifiers
• Sensor interfaces: Converting sensor resistance changes to voltage changes (e.g., with thermistors or photoresistors)
• Potentiometers: Variable voltage dividers used for volume controls, dimmer switches, and similar applications
• Measurement: Reducing high voltages to measurable levels for instrumentation

Limitations and Loading Effects

The voltage divider formula assumes that no current is drawn from the output node. In practice, if a load is connected to the output, it will draw current and change the voltage division ratio. This is known as loading.

Figure: A voltage divider with a load resistor $R_L$.

When a load resistor $R_L$ is connected to the output, it forms a parallel combination with $R_2$. The effective resistance becomes:

$$R_{2,eff} = \frac{R_2 \times R_L}{R_2 + R_L}$$

The modified output voltage becomes:

$$V_{out} = V_{in} \times \frac{R_{2,eff}}{R_1 + R_{2,eff}}$$

which can be simplified to:

$$V_{out} = V_{in} \times \frac{R_2 \times R_L}{R_1(R_2 + R_L) + R_2 \times R_L}$$

Example 13: Voltage Divider with Load

For a voltage divider with $V = 12\,\text{V}$, $R_1 = 10\,\text{k}\Omega$, and $R_2 = 10\,\text{k}\Omega$, calculate the output voltage when:
a) No load is connected ($R_L = \infty$)
b) $R_L = 10\,\text{k}\Omega$
c) $R_L = 1\,\text{k}\Omega$

Solution

a) Without load:

$$V_{out} = V_{in} \times \frac{R_2}{R_1 + R_2} = 12 \times \frac{10}{10 + 10} = 6\,\text{V}$$

b) With $R_L = 10\,\text{k}\Omega$:

$$\begin{aligned} > R_{2,eff} &= \frac{10 \times 10}{10 + 10} = 5\,\text{k}\Omega \\ > V_{out} &= 12 \times \frac{5}{10 + 5} = 4\,\text{V} > \end{aligned}$$

c) With $R_L = 1\,\text{k}\Omega$:

$$\begin{aligned} > R_{2,eff} &= \frac{10 \times 1}{10 + 1} \approx 0.909\,\text{k}\Omega \\ > V_{out} &= 12 \times \frac{0.909}{10 + 0.909} \approx 1\,\text{V} > \end{aligned}$$

A load reduces the output voltage; the lower the load resistance, the greater the reduction.

Practical Design Considerations

When designing voltage dividers for practical applications, consider the following:

• Loading effects: Choose $R_1$ and $R_2$ values small enough that loading effects are minimized, but large enough to minimize power consumption.
• Power rating: Ensure resistors can handle the power dissipation: $P_1 = \frac{V^2 \times R_1}{(R_1 + R_2)^2}$ and $P_2 = \frac{V^2 \times R_2}{(R_1 + R_2)^2}$. See Example 7 for a worked verification.
• Tolerance: Consider the effect of resistor tolerance on the accuracy of the output voltage.
• Buffer circuits: For sensitive applications, use a buffer amplifier (such as a voltage follower, covered later) to minimize loading effects.

Additional Resources

• Voltage Divider (Wikipedia)
• Voltage Divider (Khan Academy)

Simplification of Resistor Circuits

Podcast: Circuit Simplification Techniques

While Kirchhoff's Laws provide a systematic approach to solving any resistive circuit, they often lead to multiple simultaneous equations that can be tedious to solve. For many circuits, simplification techniques based on series and parallel connections and voltage divider principles yield solutions more efficiently.

The Simplification Approach

The general approach to circuit simplification involves:

1. Identifying series and parallel combinations of resistors
2. Replacing these combinations with their equivalent resistances
3. Applying voltage divider or current divider principles
4. Working backward to find the values in the original circuit

Example 14: Simplifying a Parallel Current Source Circuit

Revisiting Example 1, find currents $I_1$ and $I_2$ using circuit simplification techniques.

Solution

1. Calculate the equivalent parallel resistance of $R_1$ and $R_2$:

$$R_{parallel} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{3 \times 4}{3 + 4} = \frac{12}{7} \approx 1.714\,\Omega$$

1. Find the voltage across the parallel combination:

$$V_0 = I_0 \times R_{parallel} = 2 \times 1.714 \approx 3.429\,\text{V}$$

1. Calculate the current through each resistor using Ohm's Law:

$$\begin{aligned} > I_1 &= \frac{V_0}{R_1} = \frac{3.429}{3} \approx 1.143\,\text{A} \\ > I_2 &= \frac{V_0}{R_2} = \frac{3.429}{4} \approx 0.857\,\text{A} > \end{aligned}$$

1. Verification using KCL: $I_0 = I_1 + I_2 = 1.143 + 0.857 = 2\,\text{A}$ ✓

This approach is more direct than setting up simultaneous equations, particularly for circuits with clear series or parallel structures.

Example 15: Simplifying a Circuit with Two Voltage Sources

Revisiting Example 5, find the current $I$ in this circuit with two voltage sources and two resistors. The circuit has already been shown to simplify to:

Solution

1. The resistors are in series, so the total resistance is:

$$R_{series} = R_1 + R_2 = 2\,\Omega + 6\,\Omega = 8\,\Omega$$

1. The voltage sources oppose each other. Moving clockwise, the 11 V source increases potential while the 3 V source decreases it. The effective voltage is:

$$V_{series} = 11 - 3 = 8\,\text{V}$$

1. Applying Ohm's Law:

$$I = \frac{V_{series}}{R_{series}} = \frac{8}{8} = 1\,\text{A}$$

The two voltage sources combine into one equivalent source, and the two resistors combine into one equivalent resistance.

Example 16: Series-Parallel Simplification

Revisiting Example 11, consider a circuit with a 24 V battery, a series resistor $R_1 = 6\,\Omega$, and two parallel resistors $R_2 = 12\,\Omega$ and $R_3 = 4\,\Omega$. Calculate all currents in the circuit.

Solution

1. Calculate the equivalent resistance of the parallel combination:

$$\begin{aligned} > \frac{1}{R_{parallel}} &= \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{12} + \frac{1}{4} = \frac{1+3}{12} = \frac{4}{12} = \frac{1}{3} \\ > R_{parallel} &= 3\,\Omega > \end{aligned}$$

1. Total resistance: $R_{total} = R_1 + R_{parallel} = 6\,\Omega + 3\,\Omega = 9\,\Omega$

2. Calculate the main current: $I_1 = \frac{V}{R_{total}} = \frac{12}{9} = 1.33\,\text{A}$

3. Find the voltage across the parallel combination:

$$V_{parallel} = I_1 \times R_{parallel} = 1.33 \times 3 = 4\,\text{V}$$

(Alternatively, the voltage divider formula gives: $V_{parallel} = 12\,\text{V} \times \frac{3}{9} = 4\,\text{V}$)

1. Calculate the individual currents through the parallel resistors:

$$I_2 = \frac{V_{parallel}}{R_2} = \frac{4}{12} = 0.33\,\text{A}, \quad I_3 = \frac{V_{parallel}}{R_3} = \frac{4}{4} = 1\,\text{A}$$

1. Verification using KCL: $I_1 = I_2 + I_3 = 0.33\,\text{A} + 1\,\text{A} = 1.33\,\text{A}$ ✓

Example 17: Analyzing a More Complex Circuit

Consider this circuit with multiple resistors. Find the current through each resistor.

Solution

1. Calculate series combinations:
$R_{left} = 2 + 3 = 5\,\text{k}\Omega$ and $R_{right} = 1.5 + 6 = 7.5\,\text{k}\Omega$

2. Calculate equivalent parallel resistance:

$$R_{equiv} = \frac{R_{left} \times R_{right}}{R_{left} + R_{right}} = \frac{5 \times 7.5}{12.5} = 3\,\text{k}\Omega$$

3. Calculate currents:
- Total current: $I_{total} = \frac{V}{R_{equiv}} = \frac{24}{3 \times 10^3} = 8\,\text{mA}$
- Left branch: $I_{left} = I_{total} \times \frac{R_{right}}{R_{left} + R_{right}} = 8\,\text{mA} \times \frac{7.5}{12.5} = 4.8\,\text{mA}$
- Right branch: $I_{right} = I_{total} \times \frac{R_{left}}{R_{left} + R_{right}} = 8\,\text{mA} \times \frac{5}{12.5} = 3.2\,\text{mA}$

4. Individual resistor currents:
- 2 kΩ and 3 kΩ resistors: $I_{left} = 4.8\,\text{mA}$
- 1.5 kΩ and 6 kΩ resistors: $I_{right} = 3.2\,\text{mA}$

5. Verification using KCL: $I_{total} = I_{left} + I_{right} = 4.8 + 3.2 = 8\,\text{mA}$ ✓

General Strategy for Circuit Analysis

Based on the examples above, here is a general strategy for analyzing DC resistive circuits:

• Identify the circuit structure: look for series and parallel connections.
• Simplify the circuit: replace series and parallel combinations with their equivalent resistances until the circuit reduces to a simple form.
• Calculate key parameters: find the total current or voltage in the simplified circuit.
• Work backward: use voltage divider or current divider principles to find voltages and currents in the original components.
• Verify the results: use Kirchhoff's Laws to confirm that the solution satisfies both KCL and KVL.

This approach is often faster and less error-prone than setting up and solving multiple simultaneous equations, particularly for circuits with clear series and parallel structures.

Limitations of the Simplification Approach

While circuit simplification is powerful, it has some limitations:

• Not all circuits can be reduced to simple series-parallel combinations
• Circuits with multiple voltage or current sources may require more sophisticated techniques
• For circuits with many components arranged in complex networks, techniques like node analysis or mesh analysis may be more appropriate

For most practical circuits encountered in introductory courses, however, the simplification approach provides a direct and intuitive method for analysis.

Measurement Instruments and Practical Considerations

Podcast: Measurement and Practical Considerations

When analyzing circuits on paper, we work with ideal components and perfect measurements. In real-world applications, however, measuring instruments interact with the circuits they measure, and components like batteries have limitations. Understanding these practical considerations is essential for successful circuit design and testing.

Oscilloscopes and Voltmeters

An oscilloscope is an electronic instrument that visually displays how voltage changes over time, while a voltmeter measures the potential difference (voltage) between two points in a circuit. Both instruments are essential tools for circuit analysis and troubleshooting.

Internal Resistance

A key characteristic of any voltage measuring instrument is its internal resistance. Ideally, this resistance should be infinitely large to avoid disturbing the circuit being measured.

Figure: A voltmeter with internal resistance $R_m$ measuring the voltage across resistor $R_1$ in a voltage divider circuit.

When a voltmeter or oscilloscope is connected to measure voltage, it is placed in parallel with the component being measured. The internal resistance of the instrument forms a parallel combination with the component, altering the effective resistance and potentially changing the voltage being measured.

For example, the ADALM2000's oscilloscope and voltmeter have an internal resistance of approximately 1 MΩ. This becomes significant when measuring across high-value resistors.

Example 18: Loading Effect of a Voltmeter

Consider a voltage divider with $R_1 = 90\,\text{k}\Omega$ and $R_2 = 10\,\text{k}\Omega$ connected to a 10 V source. Calculate the voltage across $R_2$ both with and without a voltmeter (with 1 MΩ internal resistance) connected for measurement.

Solution

Without the voltmeter:

$$V_{R_2} = V_{source} \times \frac{R_2}{R_1 + R_2} = 10\,\text{V} \times \frac{10}{10 + 90} = 1\,\text{V}$$

With the voltmeter, the effective resistance across $R_2$ becomes the parallel combination of $R_2$ and $R_m$:

$$R_{parallel} = \frac{R_2 \times R_m}{R_2 + R_m} = \frac{(90 \times 10^3)(1 \times 10^6)}{90 \times 10^3 + 1 \times 10^6} \approx 89.6\,\text{k}\Omega$$

The voltage divider equation now gives:

$$V_{measured} = 10\,\text{V} \times \frac{R_{2}}{R_2 + R_{parallel}} = 10\,\text{V} \times \frac{10}{10+ 89.569} \approx 1.08\,\text{V}$$

The voltmeter reads approximately 1.08V, about 0.08V (8% error) more than the actual voltage.

Note: As a rule of thumb, voltmeter measurements are accurate to within 1% when the resistance being measured is at least 100 times smaller than the internal resistance of the voltmeter. For the ADALM2000 with 1 MΩ internal resistance, measurements across resistors up to about 10 kΩ will be reasonably accurate.

Signal Generators

A signal generator produces electrical signals with controllable frequency and amplitude for testing circuit behavior. Like any real source, it has an internal resistance, known as its output impedance, which forms a voltage divider with any load connected to it. The ADALM2000 signal generator has an output impedance of approximately 50 Ω. When the load resistance is large compared to 50 Ω, the loading effect is negligible and the output voltage is close to the set value. When the load resistance is comparable to 50 Ω, a significant fraction of the set voltage drops across the internal impedance and the load receives less than expected. The full treatment of signal generators, including periodic waveforms, frequency, and amplitude, is covered in the Signals chapter.

Real Batteries

Real batteries have an internal resistance that affects their performance, particularly under load.

Figure: Comparison of an ideal battery (left) and a real battery modeled with internal resistance (right).

The internal resistance of a battery causes its terminal voltage to drop when current is drawn:

$$V_{terminal} = V_{nominal} - I \times R_{internal}$$

Different batteries have different internal resistances:

• Car batteries: very low internal resistance (less than 10 mΩ), capable of delivering high currents
• 9 V batteries: higher internal resistance (approximately 100 Ω), limited current capability
• Rechargeable batteries: internal resistance increases with age and use

Example 19: Battery Performance Under Load

A 12 V car battery with an internal resistance of 5 mΩ is used to start a car, drawing 200 A. Calculate the terminal voltage during starting.

Solution

$$\begin{aligned} > V_{terminal} &= V_{nominal} - I \times R_{internal} \\ > &= 12 - 200 \times 0.005 \\ > &= 11\,\text{V} > \end{aligned}$$

Despite the high current draw, the voltage drops by only 1 V due to the battery's low internal resistance.

In contrast, a 9 V battery with 100 Ω internal resistance cannot deliver anywhere near this current. The maximum current it can supply is:

$$I_{max} = \frac{V_{nominal}}{R_{internal}} = \frac{9}{100} = 0.09\,\text{A} = 90\,\text{mA}$$

A 9 V battery cannot start a car because it can only supply approximately 90 mA.

Practical Considerations for Circuit Design

When designing and analyzing real circuits, keep these practical considerations in mind:

• Measurement loading: Choose measuring instruments with sufficiently high input impedance relative to the circuit being measured.
• Source loading: Account for the output impedance of signal sources when precise amplitudes are needed.
• Battery limitations: Consider both the voltage and internal resistance when selecting power sources for a given application.
• Component tolerances: Real components have variations from their nominal values, typically ±1%, ±5%, or ±10% for resistors.
• Temperature effects: Component values can change with temperature, particularly for resistors and semiconductors.
• Power ratings: Ensure components can handle the power they will dissipate ($P = I^2R = V^2/R$).

Understanding these practical aspects helps bridge the gap between theoretical circuit analysis and real-world implementation, leading to more successful designs and more accurate measurements.

Chapter Summary

This chapter introduced the fundamental principles for analyzing direct current (DC) circuits. We explored Kirchhoff's Current Law (KCL) and Kirchhoff's Voltage Law (KVL), which are based on the conservation of charge and energy, respectively. We calculated power in DC circuits and verified that total power supplied equals total power absorbed. We examined how resistors behave when connected in series and parallel, and how to use voltage divider principles to determine voltages at different points in a circuit. We also discussed circuit simplification techniques that combine these concepts to solve complex problems more efficiently. Finally, we considered how measurement instruments like oscilloscopes and voltmeters interact with circuits, and how batteries behave in real-world applications. These fundamental concepts and techniques provide the foundation for analyzing and designing more complex electrical systems in future chapters.