Kirchhoff's current law (KCL), Kirchhoff's voltage law (KVL): Visual walkthrough

We only need two everyday truths: charge cannot appear or vanish and a round trip on a hill returns to the same height. Everything else follows.

Step 1 — What is a "current"? Count the charges going past

WHAT we did: turned the vague idea "electricity flowing" into a precise count — charge per second through a chosen gate, with a chosen positive direction.

WHY: before we can say "current in = current out" we must be able to count current and know its sign. The arrow is the whole point — a current of $- 3$ A through a rightward arrow just means $3$ A actually flowing leftward.

PICTURE (s01): the blue arrow labelled "chosen + direction" is our reference; little red charges drift through the yellow gate, and the caption under the wire shows $I = Δ Q /Δ t$ — the arrow is what fixes the sign of that $I$ .

Figure — Kirchhoff's current law (KCL), Kirchhoff's voltage law (KVL)

Step 2 — A node is just a meeting point that can't store charge

WHAT we did: named the junction and stated its one special property — zero storage.

WHY: this single fact ("can't hoard charge") is going to become KCL in two steps. Everything hinges on it, so we make it explicit and picture it.

PICTURE (s02): three wires meet at the red node dot; the two blue arrows ( $I_{1}, I_{2}$ ) point in, the yellow arrow ( $I_{3}$ ) points out. The dashed circle is a tiny "accounting box" drawn around the node — we do the charge bookkeeping inside it in the next step.

Step 3 — Bookkeeping the charge inside the box

Watch the amount of charge $Q_{node}$ trapped inside the dashed box. It can only change if more flows in than out:

$how fast stored charge grows \frac{d Q _{node}}{d t} = coulombs/sec entering \sum I_{in} - coulombs/sec leaving \sum I_{out}$

Reading it term by term:

$\frac{d Q _{node}}{d t}$ — the rate of change of stored charge. " $d / d t$ " is just "how fast, per second". If it's positive, charge is piling up; negative, draining.
$\sum I_{in}$ — add up every current whose arrow points into the box.
$\sum I_{out}$ — add up every current whose arrow points out.

WHAT we did: wrote the exact charge balance for the box.

WHY: it is nothing but honest accounting — the only way the pile inside can grow is if inflow beats outflow.

PICTURE (s03): the box with a blue " $\sum I_{in}$ " pipe entering on the left and a red " $\sum I_{out}$ " pipe leaving on the right; the yellow " $Q$ inside" meter in the middle is what the equation below the box, $d Q / d t = \sum I_{in} - \sum I_{out}$ , is tracking.

Step 4 — Set the storage to zero → KCL is born

The ideal node can't store charge (Step 2). So the pile inside never grows or shrinks:

$\frac{d Q _{node}}{d t} = 0.$

Drop that into the balance of Step 3:

$0 = \sum I_{in} - \sum I_{out} ⟹ \sum I_{in} = \sum I_{out}$

Now the promised concrete sign rule that turns this into the single compact sum. Once you have drawn an arrow on every wire (arrows may point any way you like — the maths sorts out the truth), assign each current a sign at this node:

So the two sides of the boxed equation above merge into one clean sum:

$k \sum I_{k} = 0$

$\sum_{k}$ — "add over every wire $k$ touching the node".
$I_{k}$ — the current in wire $k$ , carrying the $\pm$ from the rule above (positive if its arrow enters, negative if it leaves).

Worked sign example. Suppose three arrows are drawn: two into the node ( $I_{1}, I_{2}$ ) and one out ( $I_{3}$ ). The rule gives $+ I_{1} + I_{2} - I_{3} = 0$ . If instead you had drawn $I_{3}$ 's arrow into the node too, the rule would give $+ I_{1} + I_{2} + I_{3} = 0$ — and solving would hand you a negative $I_{3}$ , correctly telling you that current really leaves.

WHAT we did: forced $d Q / d t = 0$ , then gave an explicit arrow→sign rule so any set of drawn arrows collapses into $\sum_{k} I_{k} = 0$ .

WHY: because "no storage" is the physical truth about an ideal junction — KCL contains no new physics, it is charge conservation wearing circuit clothes; the sign rule is just consistent bookkeeping of the arrows.

PICTURE (s04): the water-junction with the sign tags shown — $+ 3 A$ and $+ 2 A$ enter (blue, " $+$ "), so exactly $5 A$ must leave (green, " $-$ "); the caption $\sum I_{in} = \sum I_{out} \Rightarrow 3 + 2 = 5$ reads straight off the arrows.

Step 5 — Switch to energy: potential is the height of a hill

Now the second law. First we need what "voltage" means at a point.

The key property: each point has exactly one height. Point $A$ cannot be at two altitudes at once.

WHAT we did: reinterpreted voltage as a height, one number per location.

WHY: KVL is about round trips, and round trips are easiest to reason about on a landscape — walk out, walk back, and you must land at the altitude you started from. That "single height per point" fact is the seed of KVL.

PICTURE (s05): a hilly altitude profile. The green ramp labelled "battery = lift $(+ ε)$ " raises the height; each red ramp ("R1 slide", "R2 slide") lowers it. The vertical axis is literally "height = potential $V$ ", and the yellow dot marks the start point $A$ .

Step 6 — Meet the battery: what EMF actually is

Before signing steps, we must formally name the lift.

Now the sign of each step. As we walk the loop in a chosen direction, each element changes our height by some $Δ V$ :

Through a battery from − to +: height rises → $Δ V = + ε$ . ( $ε$ = the EMF / lift height, just defined.)
Through a resistor along the current (with Ohm's law $V = I R$ ): height drops → $Δ V = - I R$ .
Through the same element against your walking direction: flip the sign — a drop becomes a rise and vice-versa.

Term by term for the resistor step $Δ V = - I R$ :

$I$ — the current through the resistor.
$R$ — its resistance.
$I R$ — the size of the potential drop (this is Ohm's law under the passive sign convention of Step 1).
the minus — because moving with the current we go downhill.

WHAT we did: defined EMF properly, then assigned a signed height-change to every kind of element.

WHY: without a firm sign rule the sum is meaningless — half the "voltage errors" in circuits are just a dropped minus sign.

PICTURE (s06): one small loop drawn flat; the green node tags " $+ ε$ (EMF, up)" and the two red tags " $- I R_{1}$ (down)", " $- I R_{2}$ (down)" mark each element's signed $Δ V$ , with a blue "walk dir" arrow showing the direction those signs were read in.

Step 7 — Choose a loop orientation, take a round trip → KVL is born

Before summing, we must fix a loop orientation — and commit to it.

Now start at point $A$ at height $V_{A}$ . Walk the whole loop in the chosen orientation, adding every signed step, and return to $A$ :

$V_{A} + Δ V_{1} + Δ V_{2} + \dots + Δ V_{n} = V_{A}$

The left $V_{A}$ — your starting height.
$Δ V_{1}, \dots, Δ V_{n}$ — the signed rises and drops of each element you pass, read in the chosen orientation.
The right $V_{A}$ — your ending height, which is the same point, so the same height (Step 5!).

Subtract $V_{A}$ from both sides — it cancels because it's identical on both sides:

$k \sum Δ V_{k} = 0$

$\sum_{k} V_{k} = 0, V_{k} \equiv Δ V_{k} (branch change, not a node potential) .$

What if you traverse against your orientation? You cannot — for one lap you commit to a single direction. But if you re-solve the same loop with the opposite orientation, every $Δ V_{k}$ flips sign at once (every rise becomes a drop and vice-versa), so the whole equation is just multiplied by $- 1$ : $\sum_{k} (- Δ V_{k}) = 0$ is the same equation. The physics — and the current you compute — is identical; only the paperwork sign convention flipped. The illustrated example (s07) shows both orientations landing on the same current.

WHAT we did: fixed a loop orientation, summed every signed height-change around it, used "same point ⇒ same height", then carefully renamed $Δ V_{k}$ to $V_{k}$ while flagging that these are branch changes, not node potentials.

WHY: the round trip is the trick — the messy $V_{A}$ vanishes precisely because you came home, leaving only the ups and downs, which must cancel.

PICTURE (s07): the loop as an altitude profile — pump up by $+ ε$ (green), slide down through $- I R_{1}$ and $- I R_{2}$ (red), landing exactly back on the starting altitude at $A$ (blue dots); the inset shows the reversed orientation giving the identical current.

Step 8 — When KVL needs a fix: changing magnetic fields

The whole round-trip argument secretly assumed potential is single-valued — true only when the electric field is conservative, i.e. $\oint E \cdot d l = 0$ .

Common mistake KVL is

not universally valid Faraday's law says a time-varying magnetic flux $Φ_{B}$ through the loop induces a non-conservative electric field: $\oint E \cdot d l = - \frac{d Φ _{B}}{d t} \neq = 0 (when flux changes) .$ When flux through your loop changes (an inductor, a nearby AC field, a transformer), the round trip does not return to the same "height" — $\sum_{k} V_{k} \neq = 0$ . Two honest fixes:

Model the source of flux as an element. Put the inductor's voltage $V_{L} = L d I / d t$ into the loop as its own $Δ V$ ; then $\sum_{k} V_{k} = 0$ works again because you moved the flux term to the element side.
Or keep the raw field form $\oint E \cdot d l = - d Φ_{B} / d t$ and use that instead of $\sum V_{k} = 0$ .

Rule of thumb: plain KVL ( $\sum V_{k} = 0$ ) is exact for DC and for lumped circuits where all magnetic flux is confined inside labelled inductor elements. It fails for a bare loop threaded by changing external flux.

WHAT we did: named the one situation where Step 7's premise breaks.

WHY: honesty — a reader who later meets inductors or transformers must know that " $\sum V = 0$ " is the conservative-field special case, not a law of the universe.

PICTURE (s08): the blue circuit loop is pierced by the growing red magnetic flux $Φ_{B}$ (dots = field into page); the yellow arrow marks the induced non-conservative field, and the caption $\oint E \cdot d l = - d Φ_{B} / d t \neq = 0$ is exactly why the trip fails to close in height.

Step 9 — Degenerate & edge cases (never get surprised)

Here are four cases that "look weird" — each obeys the very same two lines, and the figure (s09) shows all four as labelled panels.

Worked example The four edge cases, spelled out

(a) Two-wire node. Only one current in and one out ⇒ $I_{in} = I_{out}$ . This is why current is the same through series resistors — a resistor is not a node with storage, so nothing splits off. (Panel a: one blue arrow in, one green arrow out of a red dot.)

(b) A loop with no source. Remove the EMF ( $ε = 0$ ) but keep a real resistance $R > 0$ . KVL gives $\sum (- I R) = 0$ , i.e. $- I (R_{1} + R_{2}) = 0$ , which forces $I = 0$ — no pump, no persistent current. (And a nonzero EMF across zero resistance is unphysical — infinite current — so a real loop always carries some $R > 0$ .) (Panel b: a closed loop labelled "I = 0".)

(c) Negative answer. If solving the KCL/KVL equations gives $I = - 2 A$ , your assumed arrow was drawn backward. The magnitude $2 A$ is correct; just flip the arrow and the sign becomes positive. Nothing went wrong — the algebra told you the true direction. (Panel c: an arrow assumed one way, annotation "got −2 A ⇒ reverse".)

(d) Ideal wire / zero-ohm branch ( $R = 0$ ). Then its drop is $Δ V = - I R = 0$ : both ends sit at the same height, no matter how much current flows. All the current happily pours through this zero-resistance path (a "short"). (Panel d: a thick green wire, "R = 0, same height".)

PICTURE (s09): four mini-panels (a)–(d), one per case above, each carrying its own verdict so you can match panel → paragraph directly.

The one-picture summary

Both laws on a single diagram: at the red node the signed arrows balance ( $\sum I_{k} = 0$ ); around the blue loop the yellow height profile returns home ( $\sum V_{k} = 0$ , conservative case). Read the node arrows with the Step-4 sign rule and the loop with a fixed orientation from Step 7.

Recall Feynman: the whole walkthrough in plain words

Two ideas, nothing more. Charge is like water and never disappears. So at any junction, the water pouring in must equal the water pouring out — it can't pile up because the junction has no bucket. Draw an arrow on each pipe: count in-flows as $+$ , out-flows as $-$ , and the total is exactly zero. That's KCL, and if a number comes out negative you just drew that arrow backward. Voltage is like height on a hill. The battery is a lift that carries you up; each resistor is a slide that takes you down. Pick one direction to walk the loop and stick to it. Walk a full lap and come back to where you started — you must be at the same height, so all the ups and downs cancel to zero. That's KVL — as long as no changing magnetic field is threading your loop. If it is, the hill secretly shifts under your feet (Faraday), and you must add the inductor's voltage as its own slide to make the trip close again. Everything else — the minus signs, the negative currents, the shorts — is just carefully counting which way the water flows and which way you're walking on the hill.

Recall

Why can we set

d Q / d t = 0

at a node?

An ideal junction has no capacity to store charge (stray capacitance negligible), so the charge inside never changes.

Once arrows are drawn, how do you sign each

I_{k}

\sum_{k} I_{k} = 0

Arrow pointing into the node →

+ I_{k}

; arrow pointing out →

- I_{k}

. The

\pm

comes purely from the drawn arrow.

In the round-trip derivation of KVL, why does

V_{A}

cancel?

You return to the same point, which has a single height, so start and end

V_{A}

are identical and subtract away.

In KVL, what does

V_{k}

stand for — and what is it NOT?

The signed potential change

Δ V_{k}

across element $k$ as you traverse the loop (a branch voltage change) — it is not the potential at a node.

Why can you never "traverse against your orientation" mid-loop, and what if you reverse it entirely?

You commit to one direction for a lap; reversing the whole orientation flips every

Δ V_{k}

sign at once, multiplying the equation by

- 1

— same physics, same current.

When does plain KVL (

\sum V_{k} = 0

) fail?

When a changing magnetic flux threads the loop (

\oint E \cdot d l = - d Φ_{B} / d t \neq = 0

); fix it by modelling the inductor's voltage as an element.

For a zero-ohm branch, what is the potential difference across it?

Zero, since

Δ V = - I R = 0

— both ends are at the same potential.

If solving a network gives a negative current, what does it mean?

The assumed arrow was drawn backward; the magnitude is correct, just reverse the arrow.

Connections

Ohm's Law — gives the resistor drop $Δ V = - I R$ used in Step 6.
Conservation of Charge — the "no storage" fact that becomes KCL.
Conservation of Energy — the round-trip-returns-home fact that becomes KVL (and Faraday's law, its limit).
Electric Potential — the "one height per point" property in Step 5.
Series and Parallel Resistors — Step 9(a) explains same series current.
Wheatstone Bridge · Mesh and Nodal Analysis — where both laws are combined systematically.