The Problem Statement: Analyzing Our Schematic Diagram Example

Exam Problem Statement:
Analyze the provided example of a schematic diagram. The circuit consists of a 24V DC ideal source ($V_s$) connected in series with $R_1$ (100Ω) to Node A. A 200Ω resistor ($R_2$) connects Node A to ground. A 150Ω resistor ($R_3$) connects Node A to Node B. Terminals at Node B and ground are left open for a load.

Tasks:
  1. Determine the Thevenin equivalent voltage ($V_{th}$) and resistance ($R_{th}$) at Node B.
  2. Select a standard E24 physical resistor for the load ($R_L$) to maximize power transfer, specifying the exact wattage rating and a real-world part series.

When you are handed an example of a schematic diagram in an exam or on the bench, the immediate goal is to strip away the visual noise and find the core electrical behavior at your terminals of interest. Here, we care exclusively about what the circuit looks like from the perspective of Node B.

Method Selection: Why Thevenin's Theorem Applies Here

Which theorem applies and why? We use Thevenin's Theorem because we are evaluating a varying load at a specific terminal pair (Node B to Ground). While Nodal Analysis would let us find the voltage at Node A, it forces us to rebuild and re-solve the entire KCL matrix every time the load resistance changes. Thevenin reduces the entire linear network upstream of Node B into a single voltage source ($V_{th}$) and a single series resistor ($R_{th}$).

Circuit Goal Best Method Why?
Find all node voltages in a complex DC grid Nodal Analysis Solves KCL systematically via matrix math.
Find current through a single specific branch Mesh Analysis Reduces KVL loops to minimal equations.
Evaluate varying loads at a terminal pair Thevenin's Theorem Reduces network to one $V_{th}$ and $R_{th}$.

Step-by-Step Algebraic Solution (No Skipped Steps)

Let's solve for the Thevenin equivalent. The most common trap in this problem is forgetting that an open terminal means zero current flows through the series resistor attached to it, or mistakenly including $R_3$ in the parallel resistance calculation.

Step 1: Find the Thevenin Voltage ($V_{th}$)

  1. Remove the load at Node B (it is already open in our example of a schematic diagram).
  2. Because the terminal is open, no current flows through $R_3$. Therefore, there is no voltage drop across $R_3$ ($V = I \times R = 0 \times 150 = 0V$).
  3. This means the voltage at Node B is exactly equal to the voltage at Node A ($V_B = V_A$).
  4. Node A is a simple voltage divider formed by $V_s$, $R_1$, and $R_2$.
  5. Algebra: $V_A = V_s \times \frac{R_2}{R_1 + R_2}$
  6. Substitution: $V_A = 24V \times \frac{200\Omega}{100\Omega + 200\Omega}$
  7. Calculation: $V_A = 24 \times \frac{200}{300} = 24 \times 0.6667 = 16V$.
  8. Result: $V_{th} = 16V$.

Step 2: Find the Thevenin Resistance ($R_{th}$)

  1. Turn off the independent source: Short the 24V DC source to ground.
  2. Look into the circuit from Node B towards the source.
  3. With $V_s$ shorted, $R_1$ is now connected to ground. This puts $R_1$ in parallel with $R_2$.
  4. Algebra for parallel combo: $R_{1||2} = \frac{R_1 \times R_2}{R_1 + R_2}$
  5. Substitution: $R_{1||2} = \frac{100 \times 200}{100 + 200} = \frac{20000}{300} = 66.67\Omega$.
  6. Looking from Node B, $R_3$ is in series with this parallel combination.
  7. Algebra: $R_{th} = R_3 + R_{1||2}$
  8. Substitution: $R_{th} = 150\Omega + 66.67\Omega = 216.67\Omega$.
  9. Result: $R_{th} = 216.67\Omega$.

Sanity Check and Independent Verification

Before moving to component selection, we must verify our math. An answer without a sanity check is just a guess.

  • Order of Magnitude & Units: $V_{th}$ is 16V. This is correctly bounded between 0V and the 24V source. $R_{th}$ is 216.67Ω. It must be strictly greater than $R_3$ (150Ω) but less than $R_3 + R_1$ (250Ω). The value 216.67Ω fits perfectly in this window.
  • How to verify independently: Build the source network on a breadboard using a bench power supply dialed to exactly 24.0V. Measure Node B to ground with a high-impedance DMM (like a Fluke 87V); it should read 16.0V. To verify $R_{th}$, power down the supply, place a jumper wire across the supply terminals to short them, and use the DMM's ohmmeter function to probe Node B to ground. It should read ~217Ω (accounting for 5% resistor tolerances and lead resistance).

Decision Path: Selecting the Physical Load Resistor

According to the Maximum Power Transfer Theorem, maximum power is delivered to the load when $R_L = R_{th}$. Therefore, our target resistance is 216.67Ω. We must now map this theoretical number to a physical component.

Decision Tree for Component Selection:
  • IF target $R_L$ falls exactly on an E24 standard value THEN pick that value.
  • IF target $R_L$ falls between E24 values THEN pick the closest standard value (200Ω or 220Ω). 220Ω is closer to 216.67Ω, so we select 220Ω.
  • Calculate Power Dissipation:
    • $I_{load} = \frac{V_{th}}{R_{th} + R_L} = \frac{16V}{216.67\Omega + 220\Omega} = \frac{16}{436.67} = 0.03664A$ (36.6 mA).
    • $P_{load} = I^2 \times R_L = (0.03664)^2 \times 220 = 0.001342 \times 220 = 0.295W$.
  • IF $P_{load} < 0.20W$ THEN select a standard 1/4W (0.25W) resistor.
  • IF $P_{load} > 0.25W$ THEN a 1/4W resistor will overheat and drift or fail. You must step up to a 1/2W (0.5W) or 1W package.

The Concrete Pick: Since our calculated dissipation is 0.295W, a 1/4W resistor is a fire hazard. We default to a 1/2W metal film resistor. Specifically, specify the Vishay PR02 series, 220Ω, 0.5W, 5% tolerance (or an equivalent Yageo/Stackpole 0.5W metal film). For standard E-series resistor values, always verify the physical wattage rating against your calculated dissipation plus a 20% safety margin.

Frequently Asked Questions

What if the 24V source isn't ideal?

If the schematic diagram example included an internal source resistance ($R_s$), you would simply add $R_s$ in series with $R_1$ before starting the voltage divider and Thevenin resistance calculations. The method remains identical.

Does maximum power transfer mean maximum efficiency?

No. When $R_L = R_{th}$, the circuit is exactly 50% efficient—half the total power is wasted as heat inside the Thevenin equivalent resistance of the source network. If your goal is efficiency (like in a battery-powered sensor), you want $R_L \gg R_{th}$, not $R_L = R_{th}$.