Advanced DC Circuit Examples: Solving Unbalanced Bridges via Thevenin

When studying dc circuit examples for exams or real-world bench troubleshooting, the unbalanced Wheatstone bridge is a notorious hurdle. Unlike balanced bridges where the center current is zero, an unbalanced bridge requires rigorous network reduction. This walkthrough dissects a complex bridge problem, applies the optimal theorem, shows every algebraic step, and terminates in a concrete physical component selection for your breadboard.

The Problem: Unbalanced Bridge with a Sensor Load

Method Selection: Why Thevenin's Theorem Wins Here

When evaluating dc circuit examples with a single load variable, you have three primary choices: Mesh Current Analysis, Nodal Voltage Analysis, or Thevenin's Theorem.

Mesh analysis would require setting up and solving a 3x3 matrix of simultaneous equations. Nodal analysis requires a 2x2 matrix. While both are mathematically valid, they are prone to arithmetic errors under exam time limits and obscure the physical relationship between the source and the load.

Thevenin's Theorem is the definitive choice here. By removing the load and reducing the remaining bridge to a single equivalent voltage source ($V_{th}$) and a single series resistance ($R_{th}$), we transform a complex multi-loop network into a trivial single-loop series circuit. As detailed in Khan Academy's circuit analysis modules, Thevenin's approach isolates the load, making it vastly superior when you need to evaluate how different load values affect the circuit.

Step-by-Step Algebraic Solution

Step 1: Calculate Thevenin Voltage ($V_{th}$)
Remove $R_L$ from the circuit. We now have two independent voltage dividers connected in parallel across the 12.0V source.

• Voltage at Node A ($V_A$):
  $V_A = V_s \times [R_2 / (R_1 + R_2)]$
  $V_A = 12.0 \times [200 / (100 + 200)]$
  $V_A = 12.0 \times [200 / 300]$
  $V_A = 12.0 \times 0.6667 = 8.00V$
• Voltage at Node B ($V_B$):
  $V_B = V_s \times [R_4 / (R_3 + R_4)]$
  $V_B = 12.0 \times [400 / (150 + 400)]$
  $V_B = 12.0 \times [400 / 550]$
  $V_B = 12.0 \times 0.72727 = 8.727V$
• Thevenin Voltage ($V_{th}$):
  $V_{th} = V_B - V_A$ (Defining B as positive relative to A based on the higher potential)
  $V_{th} = 8.727V - 8.00V = 0.727V$

Step 2: Calculate Thevenin Resistance ($R_{th}$)
Deactivate the independent source by replacing the 12.0V voltage source with a short circuit (a wire). Looking back into the open terminals A and B, the circuit topology changes: $R_1$ is now in parallel with $R_2$, and $R_3$ is in parallel with $R_4$. These two parallel combinations are in series with each other.

• Left parallel pair:
  $R_{left} = (R_1 \times R_2) / (R_1 + R_2)$
  $R_{left} = (100 \times 200) / (100 + 200) = 20,000 / 300 = 66.67\Omega$
• Right parallel pair:
  $R_{right} = (R_3 \times R_4) / (R_3 + R_4)$
  $R_{right} = (150 \times 400) / (150 + 400) = 60,000 / 550 = 109.09\Omega$
• Total Thevenin Resistance:
  $R_{th} = R_{left} + R_{right}$
  $R_{th} = 66.67\Omega + 109.09\Omega = 175.76\Omega$

Step 3: Reattach Load and Solve for Current and Power
Reconnect $R_L$ to the Thevenin equivalent circuit (a 0.727V source in series with 175.76Ω and the 300Ω load).

• Load Current ($I_L$):
  $I_L = V_{th} / (R_{th} + R_L)$
  $I_L = 0.727V / (175.76\Omega + 300\Omega)$
  $I_L = 0.727 / 475.76 = 0.001528A$
  $I_L = 1.53mA$
• Load Power ($P_L$):
  $P_L = (I_L)^2 \times R_L$
  $P_L = (0.001528)^2 \times 300$
  $P_L = 0.000002334 \times 300 = 0.000700W$
  $P_L = 0.70mW$

The Trap, Sanity Check, and Independent Verification

The Trap: The most common mistake in these dc circuit examples is assuming the bridge is balanced without verifying the ratios. A bridge is only balanced if $R_1/R_2 = R_3/R_4$. Here, $100/200 = 0.5$, but $150/400 = 0.375$. Because they are unequal, current must flow through the load. A second trap is forgetting to short the voltage source when calculating $R_{th}$, which leads to wildly incorrect series-parallel reductions.

Sanity Check: Let's verify the order of magnitude and units. The driving voltage ($V_{th}$) is less than 1V. The total resistance is roughly 500Ω. Using Ohm's law mentally: $1V / 500\Omega = 2mA$. Our calculated 1.53mA aligns perfectly with this estimate. Unit analysis confirms: Volts divided by Ohms yields Amperes; Amperes squared multiplied by Ohms yields Watts. The math holds.

Independent Verification: Never trust hand-calculated bridge math on the first pass. Verify this independently using Analog Devices LTspice. Build the schematic, run a '.op' (DC operating point) simulation, and probe the current through $R_L$. Alternatively, build it on a breadboard using 1% tolerance metal film resistors and measure the voltage drop across $R_L$ with a 4.5-digit bench multimeter, then calculate $I = V/R$.

Component Selection Decision Tree

Calculating the theoretical current is only half the engineering task. You must select a physical resistor for $R_L$ that handles the 0.70mW dissipation while surviving real-world thermal and surge conditions. Use this decision path to select your exact part: