The Problem Statement: Analyzing an Example of a Direct Current Circuit

When preparing for your fundamentals of electrical engineering exam, you will frequently encounter resistive networks powered by a DC source. A 12V LiFePO4 or lead-acid battery is the most common example of a direct current source used in these textbook problems because it introduces real-world non-ideal characteristics, specifically internal resistance.

Below is a classic exam-style problem. We will solve it using Thevenin's Theorem, which is the most efficient method for finding the current through a specific load in a linear network without solving a system of simultaneous equations.

Practice Problem: LiFePO4 Battery Network

A 12.6V nominal LiFePO4 battery (our example of a direct current source) has an internal resistance (Rint) of 0.05 Ω. It is connected to a voltage divider network where R1 = 10 Ω and R2 = 15 Ω are in series across the battery terminals. A load resistor RL = 20 Ω is connected in parallel with R2.

Objective: Find the exact current flowing through the load resistor RL.

Which method applies and why? We use Thevenin's Theorem. Because we only need to find the current through one specific component (RL), Thevenin allows us to reduce the rest of the complex circuit into a single voltage source and a single series resistor. Using Kirchhoff's Voltage Law (KVL) here would require setting up and solving multiple mesh equations, wasting valuable exam time.

Step-by-Step Solution Using Thevenin's Theorem

Let's break down the algebra. Do not skip steps on your exam; partial credit depends on showing your equivalent circuit transformations.

Step 1: Remove the Load and Find the Thevenin Voltage (Vth)

First, mentally (or on paper) disconnect RL from the circuit. We now have a simple series circuit consisting of the 12.6V source, Rint, R1, and R2.

  • Total Series Resistance: Rtotal = Rint + R1 + R2
  • Rtotal = 0.05 Ω + 10 Ω + 15 Ω = 25.05 Ω
  • Series Current (I): I = V / Rtotal = 12.6V / 25.05 Ω = 0.502994 A
  • Thevenin Voltage (Vth): This is the open-circuit voltage across the terminals where RL used to be (which is exactly the voltage drop across R2).
  • Vth = I × R2 = 0.502994 A × 15 Ω = 7.5449 V

Step 2: Find the Thevenin Resistance (Rth)

Now, we deactivate the independent source. For a DC voltage source, this means replacing it with a short circuit (a wire). Do not short the internal resistance; it remains in the circuit.

  • Looking into the open terminals, Rint and R1 are now in series with each other: Rseries = 0.05 Ω + 10 Ω = 10.05 Ω.
  • This 10.05 Ω combination is in parallel with R2 (15 Ω).
  • Rth = (Rseries × R2) / (Rseries + R2)
  • Rth = (10.05 × 15) / (10.05 + 15) = 150.75 / 25.05 = 6.01796 Ω
The Trap in this Problem: The most common mistake students make is forgetting Rint when calculating Rth. They short the battery and assume R1 and R2 are just in parallel. If you omit the 0.05 Ω internal resistance, your Rth becomes exactly 6 Ω, which throws off your final load current calculation and costs you points.

Step 3: Reconnect the Load and Solve for IL

Draw the Thevenin equivalent circuit: a 7.5449V source in series with a 6.01796 Ω resistor, connected to our 20 Ω load.

  • IL = Vth / (Rth + RL)
  • IL = 7.5449 V / (6.01796 Ω + 20 Ω)
  • IL = 7.5449 / 26.01796 = 0.28998 A (or roughly 290 mA)

Sanity Check and Independent Verification

Never hand in your exam without a sanity check. Here is how you verify the answer independently.

Order of Magnitude and Unit Check

Our Vth is roughly 7.5V. If we ignored Rth entirely, the current through a 20 Ω resistor would be 7.5V / 20 Ω = 0.375 A. Because Rth adds roughly 6 Ω of series resistance, the total resistance is 26 Ω. Therefore, the current must be less than 0.375 A. Our answer of 0.290 A fits this logical boundary perfectly. The units (Volts / Ohms) correctly resolve to Amperes.

Independent Verification via Node Voltage Method

To verify without Thevenin, use the Node Voltage Method at the junction between R1, R2, and RL (let's call this Node A, with the battery negative terminal as ground).

  • KCL at Node A: (VA - 12.6) / (Rint + R1) + VA / R2 + VA / RL = 0
  • (VA - 12.6) / 10.05 + VA / 15 + VA / 20 = 0
  • Multiply through by the common denominator (603) to clear fractions, or use decimal conductances:
  • 0.0995(VA - 12.6) + 0.0666VA + 0.05VA = 0
  • 0.2161VA = 1.2537 → VA = 5.801 V
  • Load Current IL = VA / RL = 5.801 V / 20 Ω = 0.290 A.

The answers match perfectly. For deeper study on DC network analysis, reference the HyperPhysics DC circuit modules hosted by Georgia State University.

Frequently Asked Questions

What is the most common real-world example of a direct current source?

In both practical electronics and exam problems, the most ubiquitous example of a direct current source is the electrochemical cell. This includes 12V lead-acid car batteries, 3.7V lithium-ion cells (like the 18650), and 12.8V LiFePO4 packs. Unlike ideal textbook voltage sources, real-world chemical batteries always possess internal resistance (Equivalent Series Resistance, or ESR), which causes voltage sag under heavy DC loads.

Why do we use Thevenin's theorem instead of Kirchhoff's laws for this DC example?

Kirchhoff's Laws (KVL and KCL) are fundamental and will always work, but they scale poorly. If a circuit has five meshes, KVL requires solving a 5x5 matrix of simultaneous equations. Thevenin's theorem isolates the load from the source network. If the exam question asks you to find the current for three different load resistors, you only have to calculate Vth and Rth once, then use simple Ohm's law for the three different loads.

How does internal resistance affect a direct current circuit in practice?

Internal resistance acts as a hidden voltage divider. In our problem, the battery's 0.05 Ω resistance dropped a small amount of voltage before the current even reached R1. In high-current DC applications—like a 100A draw from a 12V battery to power an inverter—even a tiny 0.02 Ω internal resistance will drop 2 volts (V = I × R = 100 × 0.02), reducing your terminal voltage from 12.6V to 10.6V and potentially triggering a low-voltage brownout shutdown in your electronics.