When you sit down for a circuits exam or debug a multi-source PCB on the bench, abstract definitions of Kirchhoff's laws won't save you. You need a reliable, repeatable algebraic process. This kirchhoff law example walks through a classic two-loop DC network, highlighting the exact sign-convention traps that cause students and junior engineers to lose points or misdiagnose a board.
According to Electronics Tutorials, Kirchhoff's Voltage Law (KVL) states that the directed sum of the potential differences around any closed loop is zero. We will apply this via Mesh Analysis, which is the most efficient method for planar circuits where branch currents are the target variables.
The Problem Statement: Two-Loop DC Network
Given Circuit:
- Left Branch: A 10V DC voltage source (positive terminal facing up) in series with a 2Ω resistor.
- Right Branch: A 5V DC voltage source (positive terminal facing up) in series with a 3Ω resistor.
- Middle Branch: A single 4Ω resistor shared between the two loops.
Objective: Find the mesh currents I₁ (left loop) and I₂ (right loop), and calculate the exact current flowing downward through the shared 4Ω resistor.
Step-by-Step Solution: Applying Kirchhoff's Voltage Law
We use Mesh Analysis (a direct application of KVL) because the circuit is planar and we are solving for branch currents. This method naturally exposes the shared-resistor mutual coupling that trips up most beginners.
Step 1: Assign Mesh Currents and Polarities
Assign a clockwise mesh current I₁ to the left loop and a clockwise mesh current I₂ to the right loop. By convention, we assume all currents flow clockwise. If our final answer is negative, it simply means the physical current flows counter-clockwise. Mark the voltage polarities across each resistor based on these assumed current directions: the terminal where current enters is positive.
Step 2: Write the KVL Equation for Loop 1
Trace Loop 1 clockwise, starting from the bottom-left corner:
- Go up through the 10V source: We move from negative to positive, which is a voltage rise. (+10)
- Go right through the 2Ω resistor: We move in the direction of I₁, which is a voltage drop. (-2 * I₁)
- Go down through the shared 4Ω resistor: The net current flowing downward is (I₁ - I₂). Moving in the direction of this net current is a drop. (-4 * (I₁ - I₂))
Set the sum to zero:
10 - 2*I₁ - 4*(I₁ - I₂) = 0
Distribute and simplify:
10 - 2*I₁ - 4*I₁ + 4*I₂ = 0
10 - 6*I₁ + 4*I₂ = 0
6*I₁ - 4*I₂ = 10
Equation 1: 3*I₁ - 2*I₂ = 5 (Divided by 2 for simplicity)
Step 3: Write the KVL Equation for Loop 2 (The Trap)
Trace Loop 2 clockwise, starting from the bottom-right corner:
- Go up through the 5V source: Moving from negative to positive is a rise. (+5)
- Go left through the 3Ω resistor: Moving in the direction of I₂ is a drop. (-3 * I₂)
- Go up through the shared 4Ω resistor: Here is where exams are won or lost. We are moving against the assumed downward net current (I₁ - I₂). Moving against current is a voltage rise. Alternatively, define the upward net current as (I₂ - I₁) and treat it as a drop. Let's use the latter: (-4 * (I₂ - I₁))
⚠️ The Common Trap: Many students write the shared resistor drop as just -4*I₂, completely ignoring the mutual coupling from Loop 1. Others write -4*(I₁ - I₂) without realizing they are tracing upward, flipping the sign of the entire term. Always explicitly define the net current in the direction of your trace.
Set the sum to zero:
5 - 3*I₂ - 4*(I₂ - I₁) = 0
Distribute and simplify:
5 - 3*I₂ - 4*I₂ + 4*I₁ = 0
5 - 7*I₂ + 4*I₁ = 0
Equation 2: 4*I₁ - 7*I₂ = -5
Step 4: Solve the Linear System
We now have a system of two equations:
1) 3*I₁ - 2*I₂ = 5
2) 4*I₁ - 7*I₂ = -5
Multiply Equation 1 by 4, and Equation 2 by 3 to eliminate I₁:
12*I₁ - 8*I₂ = 20
12*I₁ - 21*I₂ = -15
Subtract the second equation from the first:
(-8*I₂) - (-21*I₂) = 20 - (-15)
13*I₂ = 35
I₂ = 35/13 A ≈ 2.692 A
Substitute I₂ back into Equation 1:
3*I₁ - 2*(35/13) = 5
3*I₁ = 5 + 70/13
3*I₁ = 65/13 + 70/13 = 135/13
I₁ = 45/13 A ≈ 3.462 A
The current flowing downward through the shared 4Ω resistor is the difference between the two mesh currents:
I_shared = I₁ - I₂ = 45/13 - 35/13 = 10/13 A ≈ 0.769 A
Sanity Check: Verifying the Answer Independently
Never hand in an exam or finalize a simulation without a sanity check. In circuit theory, MIT OpenCourseWare lectures heavily emphasize Tellegen's Theorem: the total power supplied by sources must exactly equal the total power dissipated by the resistors. Let's build a power balance table to verify our algebra.
| Component | Formula | Exact Fraction | Decimal (W) |
|---|---|---|---|
| 10V Source (Supplied) | 10 * I₁ | 450/13 | 34.615 |
| 5V Source (Supplied) | 5 * I₂ | 175/13 | 13.462 |
| Total Power Supplied | 625/13 | 48.077 | |
| 2Ω Resistor (Dissipated) | I₁² * 2 | 4050/169 | 23.964 |
| 3Ω Resistor (Dissipated) | I₂² * 3 | 3675/169 | 21.746 |
| 4Ω Resistor (Dissipated) | (I₁ - I₂)² * 4 | 400/169 | 2.367 |
| Total Power Dissipated | 8125/169 | 48.077 |
Notice that 625/13 is exactly equal to 8125/169 (since 625 * 13 = 8125). The power balance is perfect. The order of magnitude makes physical sense (a ~48W dissipation in a low-voltage, low-resistance bench circuit), and the units are strictly in Watts. The algebra is verified.
Frequently Asked Questions
How do I choose between Kirchhoff's Current Law and Voltage Law for an example?
The choice between KCL (Nodal Analysis) and KVL (Mesh Analysis) depends on the circuit topology and what you need to find. Use Mesh Analysis (KVL) when the circuit is planar (no crossing wires) and you primarily need to find branch currents, especially in circuits with many series voltage sources. Use Nodal Analysis (KCL) when the circuit has many parallel current sources, or when the circuit is non-planar. As a rule of thumb, count the essential nodes versus the essential meshes; choose the method that yields the fewest simultaneous equations.
What is the most common mistake in a Kirchhoff law example with dependent sources?
The most frequent error is failing to express the controlling variable of the dependent source in terms of the assigned mesh or node variables. For instance, if a voltage-controlled voltage source (VCVS) depends on the voltage across a 5Ω resistor, you must explicitly write that control voltage as 5 * (I_a - I_b) before substituting it into your KVL equations. If you leave the control variable as an isolated symbol (like V_x), your system of equations will be unsolvable because you will have more unknowns than equations.
Can I apply a Kirchhoff law example to AC circuits with complex impedance?
Yes. Kirchhoff's laws apply universally to lumped-parameter circuits, regardless of whether the sources are DC or AC. However, in AC analysis, you must transition from scalar algebra to complex phasor algebra. Resistances become impedances (e.g., Z_R = R, Z_L = jωL, Z_C = 1/(jωC)), and voltages/currents become complex numbers representing magnitude and phase. The exact same mesh and nodal frameworks apply, but you must be meticulous with complex arithmetic, particularly keeping track of the j operator when distributing terms across shared branches.
