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Engineering Reference · Electrical
Track current at a junction. Track voltage around a loop. Use both to turn a circuit diagram into equations you can solve.
Scope: ideal lumped circuits, with DC resistor examples · Updated 1 November 2026
Σ I = 0
Σ I entering = Σ I leaving
Charge is conserved. In the ideal circuit model, a node does not accumulate charge.
Σ ΔV = 0
Add signed voltage changes along a closed path. Voltage rises and drops balance when you return to the starting point.
Use Ohm’s law, V = IR, to relate a resistor’s voltage drop to its current. KCL gives current relationships; KVL gives voltage relationships.
Explore the equations
Change a value or direction and watch the diagram and equation update. Values stay in this browser session and are not saved.
Add the branches at a junction. Choose a reference direction for each current and one unknown to solve.
Changing units relabels all values; it does not convert the entered numbers. Enter magnitudes ≥ 0; the direction or polarity sets the sign.
Arrows show reference directions. A negative result means the actual current flows against its arrow.
Equation → substitution → result
+ 5 − 2 − x = 0 (A)
3 − x = 0
x = 3 A
3 A leaves the node.
Check: known signed sum + signed unknown = 3 + (-3) = 0 A (before display rounding).
Walk clockwise around the loop. Add voltage changes as rises or drops, then choose one to solve.
Changing units relabels all values; it does not convert the entered numbers. Enter magnitudes ≥ 0; the direction or polarity sets the sign.
Each symbol shows entry/exit polarity when travelling clockwise. A negative result reverses the unknown’s chosen polarity.
Equation → substitution → result
+ 5 − 2 − x = 0 (V)
3 − x = 0
x = 3 V
3 V drop in the clockwise traversal.
Check: known signed sum + signed unknown = 3 + (-3) = 0 V (before display rounding).
One ideal DC source and two resistors in series. Current is referenced from the source’s positive terminal through R₁ and R₂.
12 − I × 2 − I × 4 = 0
I = 12 / (2 + 4)
I = 2 A
Total resistance: 6 Ω
R₁ drop: 4 V · R₂ drop: 8 V
Loop check: +12 − 4 − 8 = 0 V (before display rounding).
Node and loop tools solve one unknown among 2–6 entries. Use consistent units. These are ideal lumped-circuit tools; the loop tool balances specified voltage changes, rather than solving an arbitrary connected circuit.
Take entering currents as positive and leaving currents as negative. Either choice works if you keep it consistent.
+5 − 2 − I = 0
I = 3 A
The downward branch carries 3 A away from A. Check: 5 A enters and 2 + 3 = 5 A leaves.
If the answer is negative: keep your reference arrow and interpret the result. For example, 2 A entering and 5 A leaving through another branch gives I = −3 A for an assumed outward third current. That branch actually carries 3 A into the node.
Pick a starting point and a traversal direction. Record the voltage change across each element until you return to the start.
| Crossing | Voltage change |
|---|---|
| Source: − to + | +Vₛ (rise) |
| Source: + to − | −Vₛ (drop) |
| Resistor: along its reference current | −IR |
| Resistor: against its reference current | +IR |
| Ideal connecting wire | 0 V |
Start below the source and travel clockwise, crossing the source from − to +, then both resistors along I.
12 − 2I − 4I = 0
6I = 12 → I = 2 A
V₁ = 2 × 2 = 4 V
V₂ = 2 × 4 = 8 V
Check: +12 − 4 − 8 = 0 V. Travelling the other way gives −12 + 8 + 4 = 0 V: the same physical result.
A 10 V supply connects through a 1 kΩ resistor to node A. From A, a 2 kΩ resistor and a 3 kΩ resistor each connect to the 0 V reference. Call the voltage at A Vₐ.
10 V ── 1 kΩ ── A ── 2 kΩ ── 0 V
│
└──── 3 kΩ ── 0 VEach branch voltage is the difference between its endpoint potentials, consistent with KVL. Apply V = IR, then balance the currents at A:
(10 − Vₐ)/1 = Vₐ/2 + Vₐ/3
60 − 6Vₐ = 3Vₐ + 2Vₐ
Vₐ = 60/11 ≈ 5.455 V
Here volts divided by kΩ gives mA. Supply current = 50/11 ≈ 4.545 mA; the branch currents are 30/11 ≈ 2.727 mA and 20/11 ≈ 1.818 mA. Their sum equals the supply current. Use unrounded values when checking.
Write your equation before entering an answer. These activities use the reference directions stated in each question.
These examples assume ideal wires and lumped components. With changing magnetic flux through a loop, induced emf must be included; at high frequencies, distributed effects can require a different circuit model. Capacitor and inductor circuits also need their component equations.
Further reading: OpenStax, University Physics Volume 2, §10.3: Kirchhoff’s Rules. The numerical examples, diagrams, and activities on this page were created for this reference.
AI-assisted reference material. Recheck assumptions against your circuit and course conventions when applying it.