Current sources are straightforward in simulation, but they routinely trip up students and junior engineers on paper when placed in shared branches. When an ideal current source sits between two meshes, standard Kirchhoff’s Voltage Law (KVL) breaks down because the voltage across an ideal current source is dictated by the rest of the circuit, not the source itself. In this current source example, we will walk through a classic exam-style supermesh problem, showing every algebraic step, highlighting the traps that cost points, and proving the answer with an independent sanity check.
The Problem Statement & Circuit Parameters
Problem Statement
Find the mesh currents $I_1$ and $I_2$ for a two-window circuit. The left branch contains a 10V DC voltage source (positive terminal facing up). The top-left horizontal branch contains a 2Ω resistor. The top-right horizontal branch contains a 4Ω resistor. The right branch contains a 5Ω resistor. The middle vertical branch (shared by both windows) contains a 3A ideal current source pointing UP. The bottom branch is a continuous ground wire (0Ω). Assume both mesh currents $I_1$ (left window) and $I_2$ (right window) flow clockwise.
Before writing any equations, map the components. A data-dense spec sheet prevents variable mix-ups when the algebra gets heavy.
| Component | Symbol | Value | Role / Branch Location |
|---|---|---|---|
| Voltage Source | $V_s$ | 10V DC | Left outer branch (+ up) |
| Current Source | $I_s$ | 3A DC | Middle shared branch (UP) |
| Resistor 1 | $R_1$ | 2Ω | Top-left horizontal |
| Resistor 2 | $R_2$ | 4Ω | Top-right horizontal |
| Resistor 3 | $R_3$ | 5Ω | Right outer branch |
Method Selection & The 'Shared Branch' Trap
Which method applies and why? We must use Supermesh Analysis. Standard mesh analysis relies on writing KVL around a closed loop. However, an ideal current source has an unknown voltage drop across it that cannot be expressed as $I \times R$. Because the 3A source is shared between Mesh 1 and Mesh 2, neither individual mesh can be solved with standard KVL. By combining the two meshes into a 'supermesh' that bypasses the current source, we eliminate the unknown voltage from our KVL equation.
The most common way students fail this exact circuit isn't the supermesh concept—it's a passive sign convention error on the outer loop KVL, or trying to assign an arbitrary voltage $V_x$ across the current source and brute-forcing the math. Furthermore, attempting a Source Transformation on the 3A source will fail here because it lacks a strictly parallel resistor in its own branch. Stick to the supermesh constraint method.
Step-by-Step Algebraic Solution
According to Electronics Tutorials, a supermesh requires two equations: one KVL for the outer perimeter, and one constraint equation derived from the shared current source.
- Formulate the Constraint Equation (KCL at the middle branch):
The 3A source points UP. Mesh current $I_1$ (clockwise) flows DOWN through the middle branch. Mesh current $I_2$ (clockwise) flows UP through the middle branch. The net upward current must equal the source value:
$I_2 - I_1 = 3$
Rearranging for substitution: $I_2 = I_1 + 3$ (Equation A) - Formulate the Supermesh KVL (Outer Loop):
Trace the outer perimeter clockwise starting from the bottom-left corner.
• Up through 10V source: $+10V$
• Right through $R_1$ (2Ω): We move in the direction of $I_1$, so voltage drops: $-2I_1$
• Right through $R_2$ (4Ω): We move in the direction of $I_2$, so voltage drops: $-4I_2$
• Down through $R_3$ (5Ω): We move in the direction of $I_2$, so voltage drops: $-5I_2$
Summing to zero:
$10 - 2I_1 - 4I_2 - 5I_2 = 0$
$10 - 2I_1 - 9I_2 = 0$ (Equation B) - Solve the System of Equations:
Substitute Equation A into Equation B:
$10 - 2I_1 - 9(I_1 + 3) = 0$
$10 - 2I_1 - 9I_1 - 27 = 0$
$-11I_1 - 17 = 0$
$-11I_1 = 17$
$I_1 = -17/11 \text{ A} \approx -1.545 \text{ A}$ - Solve for $I_2$:
$I_2 = (-17/11) + 3$
$I_2 = -17/11 + 33/11$
$I_2 = 16/11 \text{ A} \approx 1.455 \text{ A}$
Sanity Check & Independent Verification
Never trust a fractional answer on an exam without a quick verification. We will verify using Node Voltage analysis (KCL) at the top-middle node, a technique heavily emphasized in MIT OpenCourseWare's Circuits and Electronics curriculum.
1. Calculate Node Voltages:
Let the bottom wire be Ground (0V).
• Top-Left Node ($V_A$): Directly connected to the 10V source. $V_A = 10V$.
• Top-Middle Node ($V_B$): $V_B = V_A - (I_1 \times R_1) = 10 - (-17/11 \times 2) = 10 + 34/11 = 144/11 \text{ V} \approx 13.09V$.
• Top-Right Node ($V_C$): $V_C = I_2 \times R_3 = (16/11) \times 5 = 80/11 \text{ V} \approx 7.27V$.
2. Verify KCL at Top-Middle Node ($V_B$):
Current entering from left = $(V_A - V_B) / 2 = (10 - 144/11) / 2 = (-34/11) / 2 = -17/11 \text{ A}$.
Current leaving to right = $(V_B - V_C) / 4 = (144/11 - 80/11) / 4 = (64/11) / 4 = 16/11 \text{ A}$.
Current leaving downwards (through source) = -3A (since source points UP).
Sum of currents leaving = $(16/11) + (-3) = 16/11 - 33/11 = -17/11 \text{ A}$.
Current entering equals current leaving. The math is flawless.
Frequently Asked Questions
Q: Can I just use Nodal Analysis instead?
A: Yes. Nodal analysis is often faster when current sources are present. In this circuit, you only have one unknown node voltage ($V_B$) if you define the left and right nodes via the sources. However, mastering the supermesh is mandatory for exams that specifically test KVL-based methods.
Q: What does the negative sign on $I_1$ mean physically?
A: It means the actual current flows counter-clockwise in the left window. The 3A current source pushing up into the middle node forces more current to the right than the 10V source can naturally supply through the 2Ω resistor, causing a reverse flow in the left loop.
