Textbook circuit problems love to use clean, round numbers like 10Ω and 20V. But when you sit down at the exam bench—or your actual workbench—you are dealing with E12 series resistors, nominal voltage drops, and parasitic tolerances. Mastering Kirchhoff's theorem (specifically Kirchhoff's Voltage Law and Kirchhoff's Current Law) isn't just about memorizing formulas; it is about systematically tracking energy and charge through a messy, real-world network without dropping a negative sign.
This walkthrough dissects a classic 2-loop DC circuit exam problem. We will define the parameters using real-world component values, expose the most common algebraic trap, solve the system step-by-step, and verify the result using Tellegen's power balance theorem.
The Problem Statement and Real-World Component Data
A 2-loop DC circuit consists of a left branch with a 12V source (V1, positive terminal up) and resistor R1. The right branch contains a 5V source (V2, positive terminal up) and resistor R2. The middle shared branch contains resistor R3. All sources are ideal. Calculate the branch currents I1, I2, and I3, and determine the total power dissipated by the resistive network.
Before writing a single equation, a competent engineer maps theoretical values to physical components. In a lab setting, you cannot buy a '330.000Ω' resistor; you buy an E12 series component with a tolerance. Below is the spec sheet for the physical build of this exam problem.
| Component | Theoretical Value | Physical E12 Value | Tolerance | Power Rating |
|---|---|---|---|---|
| R1 (Left Branch) | 100 Ω | 100 Ω (Brown-Black-Brown) | ±5% (Gold) | 1/4W (250mW) |
| R2 (Right Branch) | 220 Ω | 220 Ω (Red-Red-Brown) | ±5% (Gold) | 1/4W (250mW) |
| R3 (Shared Branch) | 330 Ω | 330 Ω (Orange-Orange-Brown) | ±5% (Gold) | 1/4W (250mW) |
| V1 (Left Source) | 12.0 V | 12.1 V (Measured) | N/A | N/A |
| V2 (Right Source) | 5.0 V | 4.95 V (Measured) | N/A | N/A |
Note: For the algebraic solution below, we will use the exact theoretical values (12V, 5V, 100Ω, 220Ω, 330Ω) to match standard exam grading rubrics. Always use measured values when debugging physical hardware.
Method Selection and The 'Shared Branch' Trap
Which theorem applies and why? We use Kirchhoff's Voltage Law (KVL) via Mesh Analysis to write the loop equations, backed by Kirchhoff's Current Law (KCL) to define the shared branch current. Mesh analysis is chosen over Nodal analysis here because the circuit has two distinct voltage sources in separate outer branches, making current tracking more intuitive than solving for a single node voltage.
Step-by-Step Algebraic Solution
Let's define two clockwise mesh currents: I1 for the left loop, and I2 for the right loop. The physical branch currents will be: Left = I1, Right = I2, Middle (downward) = I1 - I2.
- Write KVL for Loop 1 (Left):
Starting at the bottom left node and moving clockwise:
+12V (rise across V1) - 100·I1 (drop across R1) - 330·(I1 - I2) (drop across R3) = 0
12 - 100·I1 - 330·I1 + 330·I2 = 0
430·I1 - 330·I2 = 12 (Equation A) - Write KVL for Loop 2 (Right):
Starting at the bottom right node and moving clockwise:
+330·(I1 - I2) (rise across R3, since we are moving against the net downward current) - 220·I2 (drop across R2) - 5V (drop across V2, moving + to -) = 0
330·I1 - 330·I2 - 220·I2 - 5 = 0
330·I1 - 550·I2 = 5 (Equation B) - Solve the System of Equations:
Isolate I1 in Equation A: I1 = (12 + 330·I2) / 430
Substitute into Equation B:
330 · [(12 + 330·I2) / 430] - 550·I2 = 5
Multiply entire equation by 430 to clear the denominator:
330 · (12 + 330·I2) - 236,500·I2 = 2,150
3,960 + 108,900·I2 - 236,500·I2 = 2,150
-127,600·I2 = -1,810
I2 = 0.01418 A (14.18 mA) - Back-Substitute to find I1:
I1 = (12 + 330 · 0.01418) / 430
I1 = (12 + 4.6794) / 430 = 16.6794 / 430
I1 = 0.03879 A (38.79 mA) - Calculate Shared Branch Current (I3):
I3 (downward) = I1 - I2 = 38.79 mA - 14.18 mA
I3 = 24.61 mA
Sanity Checks and Independent Verification
Never hand in an exam or finalize a PCB design without a sanity check. We verify our answers using Tellegen's Theorem, which states that the sum of power supplied must exactly equal the sum of power dissipated (conservation of energy).
1. Power Supplied vs. Absorbed by Sources:
- V1 (12V): Current I1 (38.79 mA) exits the positive terminal. P = 12V × 0.03879A = 465.48 mW (Supplied).
- V2 (5V): Current I2 (14.18 mA) flows down through V2, entering the positive terminal. This means V2 is acting as a load (absorbing power, like a battery on charge). P = 5V × 0.01418A = 70.90 mW (Absorbed).
- Net Power Supplied to Network: 465.48 mW - 70.90 mW = 394.58 mW.
2. Power Dissipated by Resistors (I²R):
- R1: (0.03879)² × 100 = 150.46 mW
- R2: (0.01418)² × 220 = 44.15 mW
- R3: (0.02461)² × 330 = 199.92 mW
- Total Dissipated: 150.46 + 44.15 + 199.92 = 394.53 mW.
The 0.05 mW difference is purely rounding error from our 4-decimal truncation. The power balance holds perfectly. Furthermore, an order-of-magnitude check confirms that none of the resistors exceed their 250mW (1/4W) physical rating, meaning the physical E12 components will not overheat on the breadboard.
Frequently Asked Questions
Q: What if I assumed I2 was counter-clockwise?
A: Kirchhoff's theorem doesn't care about your initial assumption. If you assume counter-clockwise, your KVL equation for Loop 2 changes, but your final calculated value for I2 will simply be -14.18 mA. The negative sign indicates the physical current flows opposite to your assumed arrow.
Q: How do I measure the shared branch current on a real breadboard?
A: You cannot measure current in parallel. You must either break the circuit and insert your multimeter in series with R3, or measure the voltage drop across R3 and use Ohm's law (I = V/R). For a 24.61 mA current across a 330Ω resistor, you should read exactly 8.12V across the shared branch.
