DC Circuits

Overview

DC (Direct Current) circuits involve the analysis of circuits with constant voltage sources. Understanding how resistors combine and how to analyze circuit behavior is essential.

Resistors in Series

$$R_{\text{eq}} = R_1 + R_2 + R_3 + \cdots$$

Properties

• Same current through each resistor
• Voltage divides proportionally to resistance
• Total resistance increases

Voltage Divider

$$V_1 = V_{\text{total}} \times \frac{R_1}{R_1 + R_2}$$

Resistors in Parallel

$$\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \cdots$$

For two resistors:

$$R_{\text{eq}} = \frac{R_1 R_2}{R_1 + R_2}$$

Properties

• Same voltage across each resistor
• Current divides inversely to resistance
• Total resistance decreases

Current Divider

$$I_1 = I_{\text{total}} \times \frac{R_2}{R_1 + R_2}$$

Circuit Analysis Methods

Node Analysis

1. Identify nodes (connection points)
2. Choose a reference node (ground)
3. Write KCL equations at each node
4. Solve for node voltages

Mesh Analysis

1. Identify independent loops (meshes)
2. Assign mesh currents
3. Write KVL equations for each mesh
4. Solve for mesh currents

Power in Circuits

Power Delivered by Source

$$P = \varepsilon I$$

Power Dissipated in Resistor

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

Efficiency

$$\eta = \frac{P_{\text{load}}}{P_{\text{total}}}$$

Maximum Power Transfer

Maximum power delivered to load when:

$$R_{\text{load}} = R_{\text{internal}}$$

Maximum power:

$$P_{\max} = \frac{\varepsilon^2}{4R_{\text{internal}}}$$

Internal Resistance

Real voltage source with internal resistance $r$:

Terminal Voltage

$$V = \varepsilon - Ir$$

Short Circuit Current

$$I_{\text{short}} = \frac{\varepsilon}{r}$$

Open Circuit Voltage

$$V_{\text{open}} = \varepsilon$$

RC Circuits

Charging

$$Q(t) = C\varepsilon(1 - e^{-t/\tau})$$ $$I(t) = \frac{\varepsilon}{R}e^{-t/\tau}$$ $$V_C(t) = \varepsilon(1 - e^{-t/\tau})$$

Discharging

$$Q(t) = Q_0 e^{-t/\tau}$$ $$I(t) = -\frac{Q_0}{RC}e^{-t/\tau}$$ $$V_C(t) = V_0 e^{-t/\tau}$$

Time Constant

$$\tau = RC$$

Examples

Example 1: Series Circuit

Three resistors (4 Ω, 6 Ω, 2 Ω) in series with 24 V battery.

$$R_{\text{eq}} = 4 + 6 + 2 = 12 \text{ Ω}$$ $$I = \frac{V}{R} = \frac{24}{12} = 2 \text{ A}$$ $$V_1 = IR_1 = 2 \times 4 = 8 \text{ V}$$ $$V_2 = IR_2 = 2 \times 6 = 12 \text{ V}$$ $$V_3 = IR_3 = 2 \times 2 = 4 \text{ V}$$

Example 2: Parallel Circuit

Three resistors (6 Ω, 3 Ω, 2 Ω) in parallel with 12 V battery.

$$\frac{1}{R_{\text{eq}}} = \frac{1}{6} + \frac{1}{3} + \frac{1}{2} = \frac{1}{6} + \frac{2}{6} + \frac{3}{6} = 1$$ $$R_{\text{eq}} = 1 \text{ Ω}$$ $$I_{\text{total}} = \frac{V}{R_{\text{eq}}} = \frac{12}{1} = 12 \text{ A}$$ $$I_1 = \frac{12}{6} = 2 \text{ A}, \quad I_2 = \frac{12}{3} = 4 \text{ A}, \quad I_3 = \frac{12}{2} = 6 \text{ A}$$

Example 3: Series-Parallel Combination

$R_1 = 4$ Ω in series with parallel combination of $R_2 = 6$ Ω and $R_3 = 3$ Ω. Battery = 18 V.

$$R_{\text{parallel}} = \frac{6 \times 3}{6 + 3} = 2 \text{ Ω}$$ $$R_{\text{total}} = 4 + 2 = 6 \text{ Ω}$$ $$I = \frac{18}{6} = 3 \text{ A}$$ $$V_1 = 3 \times 4 = 12 \text{ V}$$ $$V_{23} = 3 \times 2 = 6 \text{ V}$$ $$I_2 = \frac{6}{6} = 1 \text{ A}, \quad I_3 = \frac{6}{3} = 2 \text{ A}$$

Example 4: Voltage Divider

Design a voltage divider to get 5 V from a 15 V source using $R_1$ and $R_2 = 10$ kΩ.

$$\frac{V_2}{V} = \frac{R_2}{R_1 + R_2}$$ $$\frac{5}{15} = \frac{10}{R_1 + 10}$$ $$R_1 + 10 = 30$$ $$R_1 = 20 \text{ kΩ}$$

Example 5: Internal Resistance

A battery produces 12 V with no load and 11 V when 0.5 A is drawn.

$$V = \varepsilon - Ir$$ $$11 = 12 - 0.5r$$ $$r = 2 \text{ Ω}$$

Example 6: Maximum Power

A battery ($\varepsilon = 9$ V, $r = 3$ Ω) powers a variable load. Find $R$ for maximum power.

$$R = r = 3 \text{ Ω}$$ $$P_{\max} = \frac{\varepsilon^2}{4r} = \frac{81}{12} = 6.75 \text{ W}$$

Example 7: RC Circuit

A 10 μF capacitor charges through 100 kΩ resistor from 20 V source.

$$\tau = RC = 100 \times 10^3 \times 10 \times 10^{-6} = 1 \text{ s}$$

At $t = 1$ s:

$$V_C = 20(1 - e^{-1}) = 20 \times 0.632 = 12.64 \text{ V}$$

At $t = 5\tau = 5$ s:

$$V_C = 20(1 - e^{-5}) = 20 \times 0.993 = 19.86 \text{ V (essentially fully charged)}$$

Circuit Theorems

Thévenin's Theorem

Any linear circuit can be replaced by:

• Voltage source $V_{Th}$ (open-circuit voltage)
• Series resistance $R_{Th}$ (equivalent resistance)

Norton's Theorem

Any linear circuit can be replaced by:

• Current source $I_N$ (short-circuit current)
• Parallel resistance $R_N$ (equivalent resistance)

Relationship:

$$V_{Th} = I_N \times R_N$$ $$R_{Th} = R_N$$