Learn › Topic 2

Kirchhoff's Laws: Currents, Voltages and a Two-Mesh Example

8 min read · bitGateLab Team

Ohm's law is enough when the current has only one path to follow. As soon as there are several branches and more than one source, you need two more rules: Kirchhoff's laws. One deals with the currents at nodes and the other with the voltages around a closed loop. Together with Ohm's law, they let you solve any resistive circuit. Here you'll see both laws, the sign convention that prevents most mistakes, and a complete example with two meshes.

Nodes, branches and meshes

Before applying the laws, it helps to name the parts of a circuit:

Node: a point where two or more elements meet. All the wires joined together with no elements between them form a single node.
Branch: a path between two nodes that contains one element (or several in series). The same current flows through the whole branch.
Loop: any closed path that does not pass through the same node twice.
Mesh: a loop with no other loops inside it, like a single pane of a window.

Kirchhoff's current law (KCL)

Charge does not pile up at a node: everything that flows in has to flow out. That is why the sum of the currents entering a node equals the sum of the currents leaving it. Put another way, if you count the entering currents as positive and the leaving currents as negative, the algebraic sum is zero:

Σ I_in = Σ I_out        or        Σ I = 0

Example 1. A current of 5 mA flows into a node. Two branches carry away 2 mA and 1.5 mA. How much leaves through the third branch?

KCL: 5 mA = 2 mA + 1.5 mA + I_x.
I_x = 5 − 3.5 = 1.5 mA, leaving the node.

Kirchhoff's voltage law (KVL)

If you travel around a closed loop and come back to your starting point, you are back at the same potential. That is why the algebraic sum of the voltages around any loop is zero: the voltage rises (in the sources) are exactly balanced by the drops (in the resistors).

Σ V = 0   around a closed loop

Example 2. A 9 V battery feeds three resistors in series. Across two of them you measure 3 V and 4.5 V. How much drops across the third?

KVL: 9 V − 3 V − 4.5 V − V₃ = 0.
V₃ = 1.5 V.

Sign convention

Most mistakes with Kirchhoff's laws are sign errors, not conceptual ones. Always follow the same procedure:

Assume a direction for each current and draw it as an arrow. It doesn't matter if you guess wrong: if the result comes out negative, the actual current flows the other way and its magnitude is correct.
Mark the polarity of each resistor according to the assumed current: the current enters at the + end and leaves at the − end. That way, the drop across a resistor is I · R in the direction of the current.
Go around the loop in one fixed direction. When you cross an element from − to +, you add its voltage (a rise); when you cross it from + to −, you subtract it (a drop).

With this convention, the sign for a source depends only on its physical polarity, and the sign for a resistor depends on whether you cross it with or against the assumed current.

Complete example: a two-mesh circuit

Look at the circuit in the figure. It has two sources, V₁ = 12 V and V₂ = 6 V, both with the positive terminal at the top, and three resistors: R₁ = 1 kΩ, R₂ = 2 kΩ (the middle branch, shared by the two meshes) and R₃ = 1 kΩ.

Two sources feed node A. The arrows show the assumed directions of I₁, I₂ and I₃.

Example 3. Find the three branch currents and the voltage at node A with respect to node B.

Assumed directions: I₁ leaves V₁ and flows toward A through R₁; I₃ leaves V₂ and flows toward A through R₃; I₂ flows down from A to B through R₂.
KCL at node A: I₁ + I₃ = I₂.
KVL around the left mesh (clockwise from B): you rise 12 V in V₁, drop 1·I₁ across R₁ and drop 2·I₂ across R₂. Working in kΩ and mA: 12 − I₁ − 2I₂ = 0.
KVL around the right mesh (counterclockwise from B): you rise 6 V in V₂, drop 1·I₃ across R₃ and drop 2·I₂ across R₂: 6 − I₃ − 2I₂ = 0.
Solve for the currents: I₁ = 12 − 2I₂ and I₃ = 6 − 2I₂.
Substitute into KCL: (12 − 2I₂) + (6 − 2I₂) = I₂, that is, 18 = 5I₂, so I₂ = 3.6 mA.
Then I₁ = 12 − 7.2 = 4.8 mA and I₃ = 6 − 7.2 = −1.2 mA.
The negative sign tells you that I₃ actually flows from A toward V₂: the 12 V source is "charging" the 6 V one.
Voltage at node A: V_AB = 2 kΩ × 3.6 mA = 7.2 V.

Checks. KCL: 4.8 + (−1.2) = 3.6 mA. Power balance: V₁ delivers 12 × 4.8 = 57.6 mW; V₂ absorbs 6 × 1.2 = 7.2 mW; the resistors dissipate 4.8² × 1 + 3.6² × 2 + 1.2² × 1 = 23.04 + 25.92 + 1.44 = 50.4 mW. And 57.6 − 7.2 = 50.4 mW. Everything adds up.

Key point: in a circuit with n nodes there are only n − 1 independent KCL equations, and the number of independent KVL equations equals the number of meshes. Here: 2 nodes give 1 KCL equation, plus 2 meshes, for a total of 3 equations in 3 unknowns.

Common mistakes

Changing the direction of travel halfway around a loop, or mixing conventions between one mesh and another.
Choosing the sign of the drop across a resistor without looking at the assumed direction of the current.
"Fixing" a negative current by flipping its sign in the equations that follow. Use it exactly as it came out; the sign already carries the information.
Writing KCL at every node: one of the equations always turns out to be redundant, so the system looks solvable but is not independent.
Forgetting that a source with current flowing into its positive terminal absorbs power instead of delivering it.

Try it in the simulator

The two-mesh example is ideal for checking with measurements, especially the negative current in the right-hand branch. Build the circuit and compare each reading with your calculations.

Place two DC sources, one of 12 V and one of 6 V, with the positive terminal facing up.
Place R₁ = 1 kΩ, R₂ = 2 kΩ and R₃ = 1 kΩ and connect them as in the figure.
Run the DC analysis and look at the voltage at node A; it should read 7.2 V.
Measure the current in each branch with the multimeter. Notice that the current in R₃ flows opposite to the assumed direction.
Change V₂ to 9 V and measure again: is I₃ still negative?

Exercises

Currents of 3 mA and 7 mA flow into a node, and 4 mA and an unknown current I flow out. Find I.

Show answer

3 + 7 = 4 + I, so I = 6 mA, leaving the node.
A 15 V source feeds three resistors in series. The first drops 4 V and the second 6 V. How much drops across the third?

Show answer

15 − 4 − 6 − V₃ = 0, so V₃ = 5 V.
In the circuit in the figure, replace V₂ with a wire (0 V). Find I₁, I₂, I₃ and V_AB.

Show answer

Equations: I₁ = 12 − 2I₂, I₃ = 0 − 2I₂ = −2I₂. KCL: 12 − 2I₂ − 2I₂ = I₂, so I₂ = 2.4 mA. Then I₁ = 7.2 mA, I₃ = −4.8 mA and V_AB = 2 × 2.4 = 4.8 V. Check: 7.2 − 4.8 = 2.4 mA.
You solve a circuit and get I = −2 mA in a branch where you assumed the current flowed downward. What does that mean?

Show answer

That 2 mA actually flows upward. The calculation is correct; only the assumed direction was the wrong way around.

Review the basics in Ohm's Law and Power and continue with series, parallel and dividers.