Engineering · Electrical engineering

Kirchhoff’s rules for nodes and loops

Lumped network with ideal nodes and no unaccounted charge accumulation. Loop balances must include all sources, including induced voltages; spatially distributed high-frequency systems need more detailed models.

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Formula

Calculate

\sum I_{in} = \sum I_{out}, Masche \sum Δ V = 0

I in A; ΔV in V; sums are signed according to chosen directions

Variables & units

Variables & units
Symbol	Meaning	Unit
$I$	Current in a branch at a junction	A
$ΔV$	Potential change across a component in the chosen traversal direction	V

LaTeX

\sum I_{\mathrm{in}}=\sum I_{\mathrm{out}}, \qquad \sum_{\mathrm{Masche}}\Delta V=0

QUANTA STUDYquanta-study.de

Two sources sharing a return feed an upper node through 4 and 6 ohms. A 3 ohm resistor connects that node to return; arrows mark 5/3, 1/9 and 16/9 amperes. — With 12 V left and 6 V right, the shared node is at 16/3 V. Both sources supply current; the 3 ohm branch carries their sum.

Kirchhoff’s rules calculator

Find the missing current at a junction or the missing voltage in a loop; current flowing in and a rise in potential count as positive.

Calculation model

Known oriented branch current 1

Known oriented branch current 2

Known oriented branch current 3

Filled with example values. Type your own and the result updates straight away.

Working · rearrange for Iₓ, substitute, evaluate

1. Rearrange for I_x
$I_x=-\sum_{j=1}^{3}I_j$
2. Substitute values (in SI units)
$-\left[\left(2\right)+\left(1.5\right)+\left(-1\right)\right]$
3. Result
I_x = -2.5 A

The entered decimal values are summed exactly, and only the result is rounded once, so rounding creates no residual. No tolerance is automatically reported as zero; assess measurement uncertainty and model completeness separately.

Result: Missing current, inward positive I_x

I_x = -2.5 A

With 2 A, 1.5 A and -1 A, I_x = -2.5 A.

Quick tips

Junction rule: what flows in also flows out. Enter currents flowing in as positive and out as negative; if the missing current comes out negative, it flows out.
Loop rule: once round the loop all voltages add up to 0. With an ideal 12 V source and two resistors in series, exactly 12 V drop across them together.

Assumptions and limits

One idealized quasistatic node without relevant net charge storage. Enter all known branch currents at the same instant, inward positive and outward negative. Exactly one additional unknown branch is found from the zero balance. Its result I_x is also counted inward positive; a negative value means the current actually flows out. The restriction to 1 to 100 known rows is only a technical interface limit. Not a general network solver.

Derivation & background

Draw nodes, voltage orientations and initially freely chosen current arrows. Charge conservation gives $\sum I_{\mathrm{in}}-\sum I_{\mathrm{out}}=\frac{\mathrm dQ_{\mathrm{node}}}{\mathrm dt}$ . For an ideal node without modelled charge accumulation, this difference is zero; capacitor currents count as branch currents. Traverse a closed loop in a specified direction and add signed voltage rises and drops. In the lumped model with all sources included, the sum is zero. Unaccounted changing magnetic flux requires an induced voltage. Insert component laws, solve independent equations and check nodes, loops and power balance; negative currents simply reverse the chosen arrow.

Exam blueprint

Validity range

Derivation steps

Charge conservation at nodes and complete loop voltage balances connect component laws.

1Draw nodes, voltage orientations and initially freely chosen current arrows.
2Charge conservation gives $\sum I_{\mathrm{in}}-\sum I_{\mathrm{out}}=\frac{\mathrm dQ_{\mathrm{node}}}{\mathrm dt}$ .
3For an ideal node without modelled charge accumulation, this difference is zero; capacitor currents count as branch currents.
4Traverse a closed loop in a specified direction and add signed voltage rises and drops.
5In the lumped model with all sources included, the sum is zero. Unaccounted changing magnetic flux requires an induced voltage.
6Insert component laws, solve independent equations and check nodes, loops and power balance; negative currents simply reverse the chosen arrow.

Rearrangements

Unknown current

I_{?}=\sum I_{in}-\sum I_{out}

Negative means the opposite direction.

Task variant

A junction receives 2 A and 0.5 A; 1.2 A leaves. Find the other current.

1.3 A leaves through the other branch.

Common mistakes

Changing signs during the calculation.

Keep chosen directions and interpret negative values as opposite flow.

Exam context

Currents and voltages in multi-loop circuits.

Formula cluster

Electrical networks

Kirchhoff’s rules combine component laws into a solvable network model.

U = R·I

Ohm's Law

Rges = R₁+R₂ ; 1/Rges = 1/R₁+1/R₂

Resistors in Series and Parallel

P = U·I

Electric Power

Worked example

A junction receives 2 A and 0.5 A; if 1.2 A leaves through one branch, another branch carries 1.3 A out. In a loop, 12 V−5 V−7 V=0.

Applications

Solve for unknown currents and voltages in branched DC networks and linear circuits.

How do you calculate with Kirchhoff’s junction and loop rules?

Here is how to work through a typical Kirchhoff’s junction and loop rules (Σ I hinein = Σ I hinaus; Σ ΔV = 0) task step by step:

1
Task
A junction receives 2 A and 0.5 A; 1.2 A leaves. Find the other current.
Solution path
1.3 A leaves through the other branch.

Understand and apply the equation

A single resistance formula is often insufficient in a branched circuit. Kirchhoff’s rules connect the branches: charge is conserved at an ideal node, and voltages balance around a completely modelled loop. Chosen reference arrows turn this into a solvable equation system. A negative result describes direction rather than a reason to rewrite the balance afterwards.

First define nodes, arrows and polarities

A node joins ideal wires at a common potential; a loop is a closed path through the diagram. Choose a reference node and draw current arrows before substituting numbers. Those arrows may initially be guesses. Every component equation must use the same references so that positive and negative results have an unambiguous meaning.[1]

\begin{aligned}\sum I_{\mathrm{in}}-\sum I_{\mathrm{out}}&=0\\{}\sum_{\mathrm{loop}}\Delta V&=0\end{aligned}

Ordinary lumped circuit without unmodelled external flux induction; define current directions and loop traversal first.[1]

Chosen positive current need not match actual flow. Resistor traversal along its current arrow is a drop; crossing an ideal source from minus to plus is a rise. A reference node fixes the otherwise arbitrary potential offset. Branch current, mesh current and direction of loop traversal are distinct concepts.[1]

What is conserved exactly

Conservation statements alone do not determine every current. Add each resistor’s voltage–current relation and each source’s boundary condition. Choose independent equations rather than counting one balance repeatedly. With node potentials, many branch currents follow directly from voltage differences.[1]

\frac{\mathrm dQ_{\mathrm{node}}}{\mathrm dt}=\sum I_{\mathrm{in}}-\sum I_{\mathrm{out}}

The usual algebraic node rule assumes negligible stored net charge at the idealised node.[1]

A capacitor does not violate the node rule: include its branch current. Relevant charge storage at a node belongs in the model as a capacitive/storage contribution. Conservative electrostatic potential differences sum to zero around a closed loop. Changing magnetic flux can add emf that must be modelled as a source or coupling.[1][4]

Explore further: Build an independent equation system

A connected network with N nodes has N−1 independent node balances: the last follows from the others. Add component equations and enough independent loops. Mesh currents offer an alternative for a planar resistor network. A shared resistor carries the difference of adjacent oriented mesh currents, not the sum of their magnitudes. Redundant equations check the solution but add no independent information.[1]

\frac{E_1-V}{R_1}+\frac{E_2-V}{R_2}=\frac{V}{R_3}

A common node V receives two supplied branches and connects through R₃ to the reference node 0 V.[1]

Node potentials, meshes and coupled sources

Two sources supply a common node

Given: Reference node $0\,\mathrm V$ . The left source node at $12\,\mathrm V$ connects through $R_1=4\,\Omega$ to V; the right source node at $6\,\mathrm V$ through $R_2=6\,\Omega$ . $R_3=3\,\Omega$ connects V to the reference node.
Find: Find node voltage and the three branch currents.

Approach: With zero reference potential, write one independent balance at the shared node.
$\begin{aligned}I_1+I_2&=I_3\\{}I_1&=\frac{E_1-V}{R_1}\\{}I_2&=\frac{E_2-V}{R_2}\\{}I_3&=\frac{V}{R_3}\end{aligned}$
Define both supplied branch currents toward the node.
$\frac{12-V}{4}+\frac{6-V}{6}=\frac{V}{3}$
Multiply by 12 and solve for V.
$\begin{aligned}36-3V+12-2V&=4V\\{}V&=\frac{16}{3}\,\mathrm V\end{aligned}$
Calculate chosen branch currents from their voltage differences.
$\begin{aligned}I_1&=\frac{5}{3}\,\mathrm A\\{}I_2&=\frac{1}{9}\,\mathrm A\\{}I_3&=\frac{16}{9}\,\mathrm A\end{aligned}$

Result: V≈5.333 V; I₁≈1.667 A and I₂≈0.111 A enter, I₃≈1.778 A leaves.

Check: Node $\frac{5}{3}+\frac{1}{9}=\frac{16}{9}$ . Left loop $12-4(\frac{5}{3})-3(\frac{16}{9})=0$ . Source powers $(20+\frac{2}{3})\,\mathrm W$ equal resistor losses $(20+\frac{2}{3})\,\mathrm W$ .

The right source now absorbs energy

Given: Same network, right source now $3\,\mathrm V$ . Both mesh currents clockwise: left $j_1$ , right $j_2$ .
Find: Solve meshes and interpret negative right branch supply current.

Approach: Orient both meshes clockwise; the shared resistor carries a current difference.
$\begin{aligned}R_1j_1+R_3(j_1-j_2)&=E_1\\{}R_2j_2+R_3(j_2-j_1)&=-E_2\end{aligned}$
Shared R₃ carries j₁−j₂ downward. The right source is traversed from + to −.
$\begin{aligned}7j_1-3j_2&=12\\{}-3j_1+9j_2&=-3\end{aligned}$
Solve the independent pair of equations.
$\begin{aligned}j_1&=\frac{11}{6}\,\mathrm A\\{}j_2&=\frac{5}{18}\,\mathrm A\\{}V&=3(j_1-j_2)=\frac{14}{3}\,\mathrm V\end{aligned}$
The right injection arrow opposes the right mesh current.
$\begin{aligned}I_2&=-j_2=-\frac{5}{18}\,\mathrm A\\{}I_3&=\frac{14}{9}\,\mathrm A\end{aligned}$

Result: The right branch actually flows from the node toward the 3 V source, which absorbs $(\frac{5}{6})\,\mathrm W$ .

Check: Left supply delivers 22 W; right source absorbs $(\frac{5}{6})\,\mathrm W$ ; resistors dissipate $(\frac{127}{6})\,\mathrm W\approx21.1667\,\mathrm W$ . KCL $\frac{11}{6}-\frac{5}{18}=\frac{14}{9}$ holds.

When an ideal voltage source joins two unknown nodes, its current is initially unknown. Balancing both nodes together as a supernode removes that internal current. The source voltage supplies the additional relation needed.[1]

Two nodes linked by an ideal voltage source

Given: Node A connects to reference 0 V through 6 Ω and node B through 3 Ω. An ideal 3 V source connects A (positive) to B (negative). An ideal current source injects 2 A from reference into A. Steady DC; no other branches.
Find: Find node potentials, voltage-source current and power balance.

Approach: balance A and B together. Internal source current cancels, while the voltage source supplies the second equation.
$\begin{aligned}\frac{V_A}{6\,\Omega}+\frac{V_B}{3\,\Omega}&=2\,\mathrm A\\V_A-V_B&=3\,\mathrm V\end{aligned}$
Solve potentials, then use the individual balance at A.
$\begin{aligned}V_B&=3\,\mathrm V\\V_A&=6\,\mathrm V\\I_A&=I_B=1\,\mathrm A\\I_{AB}&=2-1=1\,\mathrm A\end{aligned}$
Current enters the voltage source’s positive terminal, so it absorbs energy.
$\begin{aligned}P_{R,A}+P_{R,B}&=6+3=9\,\mathrm W\\P_{3\mathrm V}&=3\,\mathrm W\\P_{2\mathrm A}&=-6\cdot2=-12\,\mathrm W\end{aligned}$

Result: Potentials are 6 V and 3 V; source current is 1 A from A to B. The 3 V source absorbs 3 W.

Check: At A, $2=1+1$ , and at B, $1=1$ , in amperes. The passive power sum is $9+3-12=0\,\mathrm W$ . Being called a voltage source therefore does not automatically mean delivering power.

Translate negative solutions correctly

A negative current solution means actual flow opposes the defined arrow. Keep that signed value in all voltage and power balances. Only when describing the result replace it by magnitude and the reversed arrow. Current direction can make an ideal source absorb energy without changing its specified voltage. The example explicitly uses ideal sources, not a complete real-battery operating model.[1][3]

Source current can reverse direction

QUANTA STUDYquanta-study.de

For E1=12 V, R1=4 Ω, R2=6 Ω and R3=3 Ω,

I_2=\frac{7E_2-36\,\mathrm V}{54\,\Omega}

, positive from the right source to the node. The zero means equal node and source voltage.[1]

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Ohm for components, conservation for the network

\begin{aligned}[RI]&=\Omega\,\mathrm A=\mathrm V\\{}\sum_k p_k&=0\end{aligned}

The power check includes sources and all modelled components using one passive sign convention.[1][3]

The introductory upper case remains: a node receives 2 A and 0.5 A; after 1.2 A in one outgoing branch, 1.3 A remains for the other. The separate 12 V loop with 5 V and 7 V drops gives 12−5−7=0. These checks are not yet a general multi-loop solution. In larger networks each equation term should explicitly have amps or volts. Currents are not added to voltages; node and loop equations have different roles.[1]

Five sign-aware exercises

Calculate · Foundation

0.60 A enters a node; 0.20 A leaves in one branch. Find the second outgoing current.

Show hint

Conserve charge at the ideal node.

Learning goal: Find a missing branch current from the node balance.

Show solution and calculation

Approach: the ideal node stores no charge, so signed currents must balance.
I=0.40 A outward.
Check 0.60−0.20−0.40=0.

I=0.60-0.20=0.40\,\mathrm A

Calculate · Foundation

Ideal 10 V source, 2 Ω and 3 Ω in series. Find current and both drops.

Show hint

One loop, one current.

Learning goal: Find series current and individual voltage drops.

Show solution and calculation

Approach: one loop relates source voltage to both ohmic voltage drops.
I=2 A; drops 4 V and 6 V.
Loop 10−4−6=0; resistor power 8+12=20 W=10×2.

10-(2+3)I=0\Rightarrow I=2\,\mathrm A

Transfer · Challenge

An externally imposed flux increases in the positive loop orientation at 0.12 Wb/s. Loop resistance 8 Ω, self-inductance neglected. Find signed steady model current.

Show hint

Include Faraday emf as a source.

Learning goal: Integrate induced emf with loop orientation into a current calculation.

Show solution and calculation

Approach: increasing positive flux generates negative induced emf along the selected positive loop direction.
Induced emf=−0.12 V; I=−0.015 A.
The current opposes the flux increase.
This added emf must not be omitted from the loop balance.

\begin{aligned}\mathcal E&=-\frac{\mathrm d\Phi}{\mathrm dt}=-0.12\,\mathrm V\\{}I&=\frac{\mathcal E}{8}=-0.015\,\mathrm A\end{aligned}

Interpret · Foundation

At an ideal node, one 0.40 A arrow points inward and another 0.70 A arrow outward. Unknown current I also has an outward reference arrow. The node has no relevant charge storage. Find I and describe its actual direction. Should a negative result be replaced by zero?

Show hint

Write the node balance using the chosen arrows. A negative value reverses actual direction relative to its reference arrow.

Learning goal: Interpret a negative branch current as direction information.

Show solution and calculation

Approach: count inward current positively and both outward-defined currents negatively.
This gives I=−0.30 A.
Actual flow is therefore 0.30 A inward.
Do not replace it by zero or insert it as positive outward current.

Check: actual inflows 0.40 A and 0.30 A together equal outflow 0.70 A.

\begin{aligned}0&=0.40\,\mathrm A-0.70\,\mathrm A-I\\I&=-0.30\,\mathrm A\\I_{\mathrm{in,actual}}&=0.40\,\mathrm A+0.30\,\mathrm A=0.70\,\mathrm A\end{aligned}

[1]

Error analysis · Application

An ideal unbranched loop contains two sources and a 4 Ω resistor. In the chosen traversal, the first source is crossed from minus to plus (12 V), the second from plus to minus (4 V), then the resistor along defined current I. A solution writes $12\,\mathrm V+4\,\mathrm V-(4\,\Omega)I=0$ and obtains 4 A. Correct it and check power balance.

Show hint

Traversal contains one source voltage rise and one source voltage drop. The second ideal source can absorb power.

Learning goal: Form a voltage balance for an opposing second source with consistent loop orientation.

Show solution and calculation

Approach: assign each element its sign from the stated traversal.
Resultant source voltage is 8 V, giving 2 A.
The 12 V source supplies 24 W; the 4 V source absorbs 8 W and the resistor 16 W.

Check: 24 W exactly equals total absorbed power. Incorrect addition ignores the second source’s polarity.

\begin{aligned}0&=12\,\mathrm V-4\,\mathrm V-(4\,\Omega)I\\I&=\frac{12-4}{4}\,\mathrm A=2\,\mathrm A\\P_1&=12\cdot2\,\mathrm W=24\,\mathrm W\\P_2+P_R&=4\cdot2\,\mathrm W+2^2\cdot4\,\mathrm W=24\,\mathrm W\end{aligned}

[1][3]

Lumped circuits and field limitations

\oint\vec E\cdot\mathrm d\vec\ell=-\frac{\mathrm d\Phi_B}{\mathrm dt}

For a spatially fixed loop, time-varying linked magnetic flux makes the electric field nonconservative. For a moving loop, total emf must additionally include the motional contribution.[4]

Kirchhoff equations with suitable inductances and sources remain useful circuit models. Unmodelled coupling, propagation delay, radiation and relevant distributed storage make a simple purely resistive loop balance incomplete. Compare excitation time scale with propagation time. An apparent conflict with a zero-sum rule then identifies missing model contributions, not a breakdown of charge conservation.[4][1]

In brief

Choose arrows and a reference first, then combine conservation with component laws. Solve independent equations and substitute into node, loop and power balances. A source may deliver or absorb energy depending on current direction.

Sources for this explanation

Quanta · Updated October 2, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

This explanation, 3 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

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Change history

October 2, 2026Added relevant links to related formulas and named the supernode at the combined node balance; retained existing explanations, calculations, exercises, figures and sources.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Added an accessible learning sequence, an additional model example, a quantitative diagram and a closing summary; retained existing examples, exercises and sources.

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Quanta. Kirchhoff’s junction and loop rules. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/kirchhoff-knotenmaschen

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Frequently asked questions about Kirchhoff’s junction and loop rules

What does the junction rule state?+

The junction rule follows from charge conservation: at an idealised node with negligible net charge storage, total incoming current equals total outgoing current. Choose arrows first and retain their signs throughout all equations. Include a capacitor branch with its time-dependent current; it does not violate the rule. The retained case with 2 A and 0.5 A incoming and 1.2 A leaving through one branch gives 1.3 A in the second outgoing branch. If relevant charge storage at the node itself is modelled, include its time derivative. Another redundant node balance provides no new independent information.

What does the loop rule state?+

In an ordinary lumped circuit, signed voltage changes around a closed traversal sum to zero when every source and component contribution is modelled. A resistor traversed along its defined current arrow gives a drop; an ideal source from negative to positive gives a rise. Choose either loop direction but remain consistent. In the retained 12 V case, 5 V and 7 V drops balance it. Changing linked magnetic flux may introduce additional emf. That contribution must not be ignored as a missing zero-sum remainder: include it as a source or inductive coupling in the appropriate circuit model.

What does a negative current result mean?+

A negative solution means actual current opposes the positive arrow chosen beforehand. It is not automatically a calculation error. Initially keep the signed value unchanged in every voltage and power balance. Afterwards state its magnitude and describe the reversed physical arrow. In the original two-source network with a right 3 V source, the right branch defined toward the common node carries minus five eighteenths ampere. Actual flow is toward the right source, which absorbs energy. Sources may supply or absorb; the specified ideal voltage alone does not decide this. Independent node and power checks confirm the interpretation.

How do I solve a multi-loop network?+

First choose a reference node and unambiguous branch-current arrows. Write independent node balances and component relations. Alternatively, a planar network permits oriented mesh currents; a shared resistor then carries the difference of adjacent mesh currents. In the original 12 V/3 V network, the equations are seven times left mesh current minus three times right equals 12, and minus three times left plus nine times right equals minus 3. Recover physical branch currents afterwards. Check at least one node, one loop and total power including sources. Additional equations help only when independent; duplicated balance conditions do not determine another unknown.

When are Kirchhoff’s rules insufficient as simple circuit equations?+

Simple purely resistive equations require an appropriate lumped model. Relevant time-varying magnetic flux, propagation delay, line capacitance, self-inductance or radiation may require added contributions. Kirchhoff equations with suitable sources and storage remain useful; conservation laws are not repealed. In the original flux example, an increase of 0.12 Wb/s creates emf of minus 0.12 V in the chosen loop orientation. Without this source the purely resistive zero-sum calculation is incomplete. Charge storage is accounted for as storage current. The key question is whether the model captures essential processes on the excitation’s time and length scales.

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Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]OpenStax, University Physics Volume 2, §10.3 “Kirchhoff’s Rules”.

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Kirchhoff’s rules for nodes and loops

AdvancedUpper secondaryUniversity/collegeBachelor foundations

Free to browse · flashcards or a multiple-choice quiz · import from Essential

Symbol

Meaning

Unit

I

Current in a branch at a junction

ΔV

Potential change across a component in the chosen traversal direction

Derivation & background

Exam blueprint

Validity range

Derivation steps

Charge conservation at nodes and complete loop voltage balances connect component laws.

1Draw nodes, voltage orientations and initially freely chosen current arrows.
2Charge conservation gives $\sum I_{\mathrm{in}}-\sum I_{\mathrm{out}}=\frac{\mathrm dQ_{\mathrm{node}}}{\mathrm dt}$ .
3For an ideal node without modelled charge accumulation, this difference is zero; capacitor currents count as branch currents.
4Traverse a closed loop in a specified direction and add signed voltage rises and drops.
5In the lumped model with all sources included, the sum is zero. Unaccounted changing magnetic flux requires an induced voltage.
6Insert component laws, solve independent equations and check nodes, loops and power balance; negative currents simply reverse the chosen arrow.

Understand and apply the equation

First define nodes, arrows and polarities

\begin{aligned}\sum I_{\mathrm{in}}-\sum I_{\mathrm{out}}&=0\\{}\sum_{\mathrm{loop}}\Delta V&=0\end{aligned}

Ordinary lumped circuit without unmodelled external flux induction; define current directions and loop traversal first.[1]

What is conserved exactly

\frac{\mathrm dQ_{\mathrm{node}}}{\mathrm dt}=\sum I_{\mathrm{in}}-\sum I_{\mathrm{out}}

The usual algebraic node rule assumes negligible stored net charge at the idealised node.[1]

Explore further: Build an independent equation system

\frac{E_1-V}{R_1}+\frac{E_2-V}{R_2}=\frac{V}{R_3}

A common node V receives two supplied branches and connects through R₃ to the reference node 0 V.[1]

Node potentials, meshes and coupled sources

Two sources supply a common node

Given: Reference node $0\,\mathrm V$ . The left source node at $12\,\mathrm V$ connects through $R_1=4\,\Omega$ to V; the right source node at $6\,\mathrm V$ through $R_2=6\,\Omega$ . $R_3=3\,\Omega$ connects V to the reference node.
Find: Find node voltage and the three branch currents.

Approach: With zero reference potential, write one independent balance at the shared node.
$\begin{aligned}I_1+I_2&=I_3\\{}I_1&=\frac{E_1-V}{R_1}\\{}I_2&=\frac{E_2-V}{R_2}\\{}I_3&=\frac{V}{R_3}\end{aligned}$
Define both supplied branch currents toward the node.
$\frac{12-V}{4}+\frac{6-V}{6}=\frac{V}{3}$
Multiply by 12 and solve for V.
$\begin{aligned}36-3V+12-2V&=4V\\{}V&=\frac{16}{3}\,\mathrm V\end{aligned}$
Calculate chosen branch currents from their voltage differences.
$\begin{aligned}I_1&=\frac{5}{3}\,\mathrm A\\{}I_2&=\frac{1}{9}\,\mathrm A\\{}I_3&=\frac{16}{9}\,\mathrm A\end{aligned}$

Result: V≈5.333 V; I₁≈1.667 A and I₂≈0.111 A enter, I₃≈1.778 A leaves.

The right source now absorbs energy

Given: Same network, right source now $3\,\mathrm V$ . Both mesh currents clockwise: left $j_1$ , right $j_2$ .
Find: Solve meshes and interpret negative right branch supply current.

Approach: Orient both meshes clockwise; the shared resistor carries a current difference.
$\begin{aligned}R_1j_1+R_3(j_1-j_2)&=E_1\\{}R_2j_2+R_3(j_2-j_1)&=-E_2\end{aligned}$
Shared R₃ carries j₁−j₂ downward. The right source is traversed from + to −.
$\begin{aligned}7j_1-3j_2&=12\\{}-3j_1+9j_2&=-3\end{aligned}$
Solve the independent pair of equations.
$\begin{aligned}j_1&=\frac{11}{6}\,\mathrm A\\{}j_2&=\frac{5}{18}\,\mathrm A\\{}V&=3(j_1-j_2)=\frac{14}{3}\,\mathrm V\end{aligned}$
The right injection arrow opposes the right mesh current.
$\begin{aligned}I_2&=-j_2=-\frac{5}{18}\,\mathrm A\\{}I_3&=\frac{14}{9}\,\mathrm A\end{aligned}$

Result: The right branch actually flows from the node toward the 3 V source, which absorbs $(\frac{5}{6})\,\mathrm W$ .

Two nodes linked by an ideal voltage source

Given: Node A connects to reference 0 V through 6 Ω and node B through 3 Ω. An ideal 3 V source connects A (positive) to B (negative). An ideal current source injects 2 A from reference into A. Steady DC; no other branches.
Find: Find node potentials, voltage-source current and power balance.

Approach: balance A and B together. Internal source current cancels, while the voltage source supplies the second equation.
$\begin{aligned}\frac{V_A}{6\,\Omega}+\frac{V_B}{3\,\Omega}&=2\,\mathrm A\\V_A-V_B&=3\,\mathrm V\end{aligned}$
Solve potentials, then use the individual balance at A.
$\begin{aligned}V_B&=3\,\mathrm V\\V_A&=6\,\mathrm V\\I_A&=I_B=1\,\mathrm A\\I_{AB}&=2-1=1\,\mathrm A\end{aligned}$
Current enters the voltage source’s positive terminal, so it absorbs energy.
$\begin{aligned}P_{R,A}+P_{R,B}&=6+3=9\,\mathrm W\\P_{3\mathrm V}&=3\,\mathrm W\\P_{2\mathrm A}&=-6\cdot2=-12\,\mathrm W\end{aligned}$

Result: Potentials are 6 V and 3 V; source current is 1 A from A to B. The 3 V source absorbs 3 W.

Check: At A, $2=1+1$ , and at B, $1=1$ , in amperes. The passive power sum is $9+3-12=0\,\mathrm W$ . Being called a voltage source therefore does not automatically mean delivering power.

Translate negative solutions correctly

Source current can reverse direction

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For E1=12 V, R1=4 Ω, R2=6 Ω and R3=3 Ω,

I_2=\frac{7E_2-36\,\mathrm V}{54\,\Omega}

, positive from the right source to the node. The zero means equal node and source voltage.[1]

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Ohm for components, conservation for the network

\begin{aligned}[RI]&=\Omega\,\mathrm A=\mathrm V\\{}\sum_k p_k&=0\end{aligned}

The power check includes sources and all modelled components using one passive sign convention.[1][3]

Five sign-aware exercises

Calculate · Foundation

0.60 A enters a node; 0.20 A leaves in one branch. Find the second outgoing current.

Show hint

Conserve charge at the ideal node.

Learning goal: Find a missing branch current from the node balance.

Show solution and calculation

Approach: the ideal node stores no charge, so signed currents must balance.
I=0.40 A outward.
Check 0.60−0.20−0.40=0.

I=0.60-0.20=0.40\,\mathrm A

Calculate · Foundation

Ideal 10 V source, 2 Ω and 3 Ω in series. Find current and both drops.

Show hint

One loop, one current.

Learning goal: Find series current and individual voltage drops.

Show solution and calculation

Approach: one loop relates source voltage to both ohmic voltage drops.
I=2 A; drops 4 V and 6 V.
Loop 10−4−6=0; resistor power 8+12=20 W=10×2.

10-(2+3)I=0\Rightarrow I=2\,\mathrm A

Transfer · Challenge

An externally imposed flux increases in the positive loop orientation at 0.12 Wb/s. Loop resistance 8 Ω, self-inductance neglected. Find signed steady model current.

Show hint

Include Faraday emf as a source.

Learning goal: Integrate induced emf with loop orientation into a current calculation.

Show solution and calculation

Approach: increasing positive flux generates negative induced emf along the selected positive loop direction.
Induced emf=−0.12 V; I=−0.015 A.
The current opposes the flux increase.
This added emf must not be omitted from the loop balance.

\begin{aligned}\mathcal E&=-\frac{\mathrm d\Phi}{\mathrm dt}=-0.12\,\mathrm V\\{}I&=\frac{\mathcal E}{8}=-0.015\,\mathrm A\end{aligned}

Interpret · Foundation

At an ideal node, one 0.40 A arrow points inward and another 0.70 A arrow outward. Unknown current I also has an outward reference arrow. The node has no relevant charge storage. Find I and describe its actual direction. Should a negative result be replaced by zero?

Show hint

Write the node balance using the chosen arrows. A negative value reverses actual direction relative to its reference arrow.

Learning goal: Interpret a negative branch current as direction information.

Show solution and calculation

Approach: count inward current positively and both outward-defined currents negatively.
This gives I=−0.30 A.
Actual flow is therefore 0.30 A inward.
Do not replace it by zero or insert it as positive outward current.

Check: actual inflows 0.40 A and 0.30 A together equal outflow 0.70 A.

\begin{aligned}0&=0.40\,\mathrm A-0.70\,\mathrm A-I\\I&=-0.30\,\mathrm A\\I_{\mathrm{in,actual}}&=0.40\,\mathrm A+0.30\,\mathrm A=0.70\,\mathrm A\end{aligned}

[1]

Error analysis · Application

An ideal unbranched loop contains two sources and a 4 Ω resistor. In the chosen traversal, the first source is crossed from minus to plus (12 V), the second from plus to minus (4 V), then the resistor along defined current I. A solution writes $12\,\mathrm V+4\,\mathrm V-(4\,\Omega)I=0$ and obtains 4 A. Correct it and check power balance.

Show hint

Traversal contains one source voltage rise and one source voltage drop. The second ideal source can absorb power.

Learning goal: Form a voltage balance for an opposing second source with consistent loop orientation.

Show solution and calculation

Approach: assign each element its sign from the stated traversal.
Resultant source voltage is 8 V, giving 2 A.
The 12 V source supplies 24 W; the 4 V source absorbs 8 W and the resistor 16 W.

Check: 24 W exactly equals total absorbed power. Incorrect addition ignores the second source’s polarity.

\begin{aligned}0&=12\,\mathrm V-4\,\mathrm V-(4\,\Omega)I\\I&=\frac{12-4}{4}\,\mathrm A=2\,\mathrm A\\P_1&=12\cdot2\,\mathrm W=24\,\mathrm W\\P_2+P_R&=4\cdot2\,\mathrm W+2^2\cdot4\,\mathrm W=24\,\mathrm W\end{aligned}

[1][3]

Lumped circuits and field limitations

\oint\vec E\cdot\mathrm d\vec\ell=-\frac{\mathrm d\Phi_B}{\mathrm dt}

For a spatially fixed loop, time-varying linked magnetic flux makes the electric field nonconservative. For a moving loop, total emf must additionally include the motional contribution.[4]

In brief

Sources for this explanation

Quanta · Updated October 2, 2026

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An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

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Change history

October 2, 2026Added relevant links to related formulas and named the supernode at the combined node balance; retained existing explanations, calculations, exercises, figures and sources.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Added an accessible learning sequence, an additional model example, a quantitative diagram and a closing summary; retained existing examples, exercises and sources.

Cite this reference page

Quanta. Kirchhoff’s junction and loop rules. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/kirchhoff-knotenmaschen

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