Apply Kirchhoff's Current Law (KCL): Worked examples

Before anything, one reminder of our vocabulary so no symbol is unearned:

The scenario matrix

Every KCL problem you will ever see is one of these case classes. The last column names the example that lands on it. (To avoid confusion, node names below use plain numbers like N1, N2 — never the same letters A–I that label the case classes.)

#	Case class	What makes it tricky	Example
A	All directions known, one unknown	The gentle warm-up	Ex 1
B	Mixed in/out, algebraic-sum form	Signs must be tracked	Ex 2
C	Wrong-guess direction	Answer comes out negative	Ex 3
D	Zero / degenerate input (a branch carries 0 A)	Does the node still balance?	Ex 4
E	Two-node system (KCL twice)	Same unknown links both nodes	Ex 5
F	Time-varying current (capacitor branch)	$i = C d v / d t$ , KCL still holds	Ex 6
G	Real-world word problem	Translate plumbing→current	Ex 7
H	Exam twist: hidden same-node wire	Two "nodes" are secretly one	Ex 8
I	Limiting behaviour (one current → source's full value)	Sanity at extremes	Ex 9

Each example below is tagged with its cell.

Example 1 — Cell A: all known but one

Figure — Apply Kirchhoff's Current Law (KCL)

Figure 1 — Look at how the two mint arrows funnel INTO the purple dot while a single coral arrow drains OUT. The picture makes the "only one exit" fact visible: because there is exactly one outgoing wire, it has no choice but to carry the whole $4 + 3$ . If you covered the number on the coral arrow, the geometry alone would force it to be $7$ . That inevitability is what "in = out" means.

Example 2 — Cell B: algebraic-sum form

Worked example Cell B — mixed directions, one clean equation

At node N2: $i_{1} = 5 A$ in, $i_{2} = 6 A$ in, $i_{3} = 8 A$ out, and unknown $i_{4}$ . Find $i_{4}$ using the algebraic form $\sum i = 0$ with leaving = +.

Forecast: Guess whether $i_{4}$ leaves or enters, and how big.

Step 1 — Assign signs by our convention (algebraic-sum form). Why this step? With currents pointing both ways, "in = out" needs you to sort branches into two piles first. The algebraic form $\sum i = 0$ skips the sorting: every branch gets a sign (leaving = +, entering = −) and drops into one sum. $- i_{1} - i_{2} + i_{3} + i_{4} = 0$ Step 2 — Substitute. Why this step? Replace known values; solve for the lone unknown. $- 5 - 6 + 8 + i_{4} = 0 \Rightarrow - 3 + i_{4} = 0 \Rightarrow i_{4} = 3 A$ Step 3 — Interpret the sign. Why this step? A positive $i_{4}$ under "leaving = +" means $i_{4}$ genuinely leaves the node, magnitude $3$ A.

Verify: In $= 5 + 6 = 11$ A. Out $= 8 + 3 = 11$ A. Balanced. ✓

Figure 2 — The two little sign tags on each arrow (−,−,+,+) are the whole lesson here: notice the tag is set by the arrow's direction, not its size. Trace each arrow into the equation $\sum i = 0$ printed above the node — the picture is literally the equation drawn out, so you can never lose track of which sign a branch carries.

Example 3 — Cell C: wrong guess self-corrects

Worked example Cell C — the arrow you drew was backwards

At node N3, known: $5 A$ in and $2 A$ in, and one branch of $8 A$ out. You guessed the last branch current $i_{x}$ flows in. Use "leaving = +" and find $i_{x}$ .

Forecast: Do you think your "in" guess is right? Predict the sign of $i_{x}$ .

Step 1 — Write $\sum i = 0$ (algebraic-sum form). Why this step? Every arrow, right or wrong, goes into the same balance. Entering currents (including the guessed $i_{x}$ ) are negative; the $8$ A that leaves is positive. $- 5 - 2 - i_{x} + 8 = 0$ Step 2 — Combine the plain numbers, then solve. Why this step? Before isolating $i_{x}$ , gather the three known constants: $- 5 - 2 + 8 = 1$ . That collapses the equation to one number and one unknown. $1 - i_{x} = 0 \Rightarrow i_{x} = 1 A$ Step 3 — Read the sign. Why this step? $i_{x} = + 1$ A means: relative to our "in" arrow, the current is $+ 1$ A → it really does flow in at $1$ A. Our guess happened to be correct.

Now flip the guess to see self-correction: suppose instead we'd drawn $i_{x}$ as out (so it enters the $+$ side). The three constants now combine the same way ( $- 5 - 2 + 8 = 1$ ): $- 5 - 2 + i_{x} + 8 = 0 \Rightarrow 1 + i_{x} = 0 \Rightarrow i_{x} = - 1 A$ The negative answer says: "your out-arrow is backwards; the current actually flows in, magnitude $1$ A." Same physics, opposite bookkeeping.

Verify: In $= 5 + 2 + 1 = 8$ A, out $= 8$ A. Balanced either way. ✓

Figure 3 — Compare the two panels side by side. Same node, same physical current — only the arrow we drew differs. In the left panel the dashed "in" arrow yields $+ 1$ ; in the right the "out" arrow yields $- 1$ , and the solid arrow shows the true flow snapping back to "in". The visual proves the sign is a property of our drawing, not of the electrons.

Example 4 — Cell D: zero / degenerate branch

Figure 4 — The grey dashed arrow is the point: it is drawn (the wire physically exists) yet labelled $0$ A. Notice the mint and coral arrows already balance each other perfectly, which is exactly why the third branch is forced to zero — there is no leftover charge for it to carry. The dashed style is a visual code for "present but idle".

Example 5 — Cell E: two nodes, KCL twice

Worked example Cell E — the unknown links two junctions

A chain: source pushes $9 A$ into node N5. From N5, a branch $i_{a}$ goes down to ground and a branch $i_{l ink}$ carries the rest to node N6. At N6, $i_{l ink}$ splits into $i_{b} = 2 A$ and $i_{c} = 4 A$ , both to ground. Find $i_{l ink}$ and $i_{a}$ .

Forecast: Guess $i_{l ink}$ first (what does node N6 demand?), then $i_{a}$ .

Step 1 — KCL at node N6 first (balance form). Why this step? N6 has only the incoming $i_{l ink}$ and two known exits — one unknown, so start where the algebra is easiest. $i_{l ink} = i_{b} + i_{c} = 2 + 4 = 6 A$ Step 2 — KCL at node N5 (balance form). Why this step? Now $i_{l ink}$ is known, N5 has one remaining unknown $i_{a}$ . $9 = i_{a} + i_{l ink} \Rightarrow i_{a} = 9 - 6 = 3 A$

Verify: N5: in $= 9$ , out $= i_{a} + i_{l ink} = 3 + 6 = 9$ . ✓ N6: in $= i_{l ink} = 6$ , out $= 2 + 4 = 6$ . ✓ Whole circuit: source $9$ A leaves via $i_{a} + i_{b} + i_{c} = 3 + 2 + 4 = 9$ A to ground. ✓ This chaining of KCL node-by-node is exactly what Nodal Analysis automates.

Figure 5 — Follow the $i_{l ink}$ wire (coral) connecting the two lavender dots: it is the shared branch that couples the two equations. See how N6 hangs off the end with only known exits — that is the visual cue telling you to solve N6 first, then feed its answer leftward into N5. The picture ranks the equations for you by counting arrows at each dot.

Example 6 — Cell F: time-varying (capacitor branch)

Worked example Cell F — a capacitor branch does NOT break KCL

At node N7 there are exactly three branches (we list them all up front, because KCL only balances if every physical wire is accounted for): a steady $i_{1} = 3 A$ in, a resistor branch $i_{R}$ out, and a capacitor branch $i_{C}$ out with $C = 2 F$ whose voltage rises at $\frac{d v}{d t} = 1 V/s$ . At this instant $i_{R} = 1 A$ . Find $i_{C}$ and confirm the node balances.

Forecast: Does the capacitor "storing charge" let the node imbalance? Guess $i_{C}$ .

Step 1 — Compute the capacitor current. Why this step? The capacitor's branch current is governed by $i_{C} = C \frac{d v}{d t}$ (the capacitance $C$ from our vocabulary; see Capacitor i-v Relationship). This is why the derivative enters: current through a capacitor answers the question "how fast is its voltage changing?", not "what is its voltage?". $i_{C} = C \frac{d v}{d t} = 2 \times 1 = 2 A$ Step 2 — Apply KCL "in = out" (balance form) at node N7. Why this step? Charge stores on the plates inside the capacitor, not at the node. Whatever current enters the capacitor terminal equals what leaves the other terminal; at node N7 it's just an ordinary branch current. Since all three branches were listed at the start, we can balance directly: $i_{1} = i_{R} + i_{C} \Rightarrow 3 = ? 1 + 2$ That gives $3 = 3$ — the node balances with no missing branch. (Had we forgotten to list a branch, KCL would have flagged the imbalance for us; that is why listing every physical wire before balancing is step zero.)

Verify: In $= 3$ A, out $= i_{R} + i_{C} = 1 + 2 = 3$ A. Balanced at this instant. ✓ KCL holds moment-by-moment even with a time-varying capacitor current — the capacitor is just one more branch.

Figure 6 — Spot the two-plate capacitor symbol on the lower-right branch: that is the only clue this branch is "special", yet its arrow enters the balance exactly like the plain resistor arrow above it. The figure deliberately draws the capacitor current $i_{C} = 2$ A the same weight as the others, hammering home that a time-varying branch is still just one arrow in "in = out".

Example 7 — Cell G: real-world word problem

Worked example Cell G — the smart power strip

A wall outlet feeds a power strip drawing $12 A$ total. Plugged in: a laptop charger ( $2 A$ ), a monitor ( $3 A$ ), and a space heater ( $i_{h e a t}$ ). All three draw current out of the strip's internal node. What does the heater draw?

Forecast: The strip is a node — what current must the heater eat so the books balance?

Step 1 — Identify the node and directions (balance form). Why this step? The strip's internal junction is one node: $12$ A in from the wall, three loads out. $i_{in} = i_{l a pt o p} + i_{m o ni t or} + i_{h e a t}$ Step 2 — Solve. Why this step? Everything known but $i_{h e a t}$ . $12 = 2 + 3 + i_{h e a t} \Rightarrow i_{h e a t} = 12 - 5 = 7 A$

Verify: Out $= 2 + 3 + 7 = 12$ A $=$ in. ✓ Real meaning: the wall must supply exactly the sum of all loads — which is why overloading a strip trips a breaker. If you added a $4$ A kettle, the strip would demand $16$ A and exceed a $15$ A rating. KCL is your household safety arithmetic.

Figure 7 — The power strip is drawn as one lavender box, not a dot, to make the "everything inside is one node" idea concrete. Watch the single mint arrow from the wall fan out into three coral load arrows: the width of the incoming supply must exactly account for all three exits — that visual conservation is what sizes a breaker.

Example 8 — Cell H: the hidden same-node twist

Worked example Cell H — two "nodes" that are secretly one

A circuit looks like it has node N8a (currents $6 A$ in, $x$ out) joined by a plain wire to node N8b (currents $x$ in, $4 A$ out and $y$ out). A student writes two separate KCLs. Find $y$ — and spot the trap.

Forecast: Are N8a and N8b really different nodes?

Step 1 — Collapse the pure wire. Why this step? N8a and N8b are connected by an ideal wire (zero resistance, no voltage drop). By the definition of a node, all points on one wire are the same node. So N8a and N8b are ONE node — call it N8. The current $x$ was never a real branch between them; it was an internal fiction. Step 2 — Write KCL once at the merged node N8 (balance form). Why this step? On the single node, list only the external branches: $6 A$ in, $4 A$ out, $y$ out. $6 = 4 + y \Rightarrow y = 12 - 10 (rearranged) = 2 A$

More plainly: $6 = 4 + y \Rightarrow y = 6 - 4 = 2 A$ .

Verify: In $= 6$ , out $= 4 + 2 = 6$ . ✓

The trap: treating a single node as two invents a phantom current $x$ and doubles your equations. Always merge wire-connected points first (see Ideal Wires and Nodes).

Figure 8 — The two panels are the whole trap. On the left, the grey dashed "phantom $x$ " sits on the wire between two dots that only look separate. On the right, that wire has shrunk to nothing and the dots have merged into one — and crucially, $x$ has vanished. Watching $x$ disappear as the dots fuse is the visual proof that it was never a real branch.

Example 9 — Cell I: limiting behaviour

Worked example Cell I — pushing a split to its extreme

A $10 A$ source enters node N9 and splits into two resistor branches $i_{1}$ and $i_{2}$ . As one resistor's value grows without bound (an open circuit), $i_{1} \to 0$ . What does $i_{2}$ approach? Confirm the limit is physical.

Forecast: If one branch shuts off, where does all the current go?

Step 1 — Write KCL, keep it general (balance form). Why this step? We do not yet fix numbers for $i_{1}$ and $i_{2}$ , because we want a formula that stays true while the circuit changes. KCL is not a one-off fact — it holds at every instant for any branch values, so we can treat $i_{1}$ as a knob we will turn and read $i_{2}$ off the equation. $10 = i_{1} + i_{2} \Rightarrow i_{2} = 10 - i_{1}$ Step 2 — Take the limit as $i_{1} \to 0$ . Why this step? "Open circuit" is not a single number — it's a process: the resistor's value climbs toward infinity, squeezing $i_{1}$ smaller and smaller toward $0$ but never letting us just write $i_{1} = 0$ outright. The limit is the precise tool for "what value does $i_{2}$ head toward as $i_{1}$ is driven to $0$ ?" — it reads off the destination of a moving quantity. Substituting the trend into our formula: $i_{2} = 10 - i_{1} i_{1} \to 0 10 A$ Step 3 — Check the opposite extreme. Why this step? A limit argument must cover both ends so no scenario is left unshown. If instead $i_{1} \to 10$ A (branch 2 opens), the same formula gives $i_{2} \to 0$ . Total is always $10$ .

Verify: At $i_{1} = 0$ : out $= 0 + 10 = 10 =$ in. ✓ At $i_{1} = 10$ : out $= 10 + 0 = 10 =$ in. ✓ KCL forces the entire source current down whichever path stays open — exactly what happens when a bulb burns out and its series partner still glows. No current is ever lost at the limit.

Figure 9 — Read the two sloping lines as a see-saw: as $i_{1}$ (lavender) slides down to $0$ , $i_{2}$ (coral) rises to fill in. The flat mint line at $10$ A on top is their sum — it never budges. That unmoving flat line is KCL: no matter how the split shifts, the total leaving equals the $10$ A entering. The endpoints of the see-saw are the two limiting cases of Step 3.

Coverage check

Recall Did we hit every cell of the matrix?

A→Ex1, B→Ex2, C→Ex3, D→Ex4, E→Ex5, F→Ex6, G→Ex7, H→Ex8, I→Ex9. Every sign case, the zero/degenerate branch, a two-node chain, a time-varying capacitor, a word problem, the hidden-node exam twist, and both limiting extremes. Nothing left unshown.

Active Recall

3 A and 2 A enter a node, one wire leaves — what leaves?

5 A (Cell A logic: in = out).

Under "leaving = +", an entering current gets what sign?

Negative — entering currents subtract in the algebraic sum

\sum i = 0

Your solved current is

- 1

A — what does that mean?

Correct magnitude, but the real direction is opposite to your drawn arrow; just flip the arrow.

A branch reads exactly 0 A — does the node still balance?

Yes; zero is a valid current, KCL holds.

Two components joined by a plain wire — how many nodes?

One node (ideal wire, no voltage drop).

A capacitor branch carries

i_{C} = C d v / d t

— does KCL fail?

No; it's an ordinary branch current, KCL holds every instant.

A 10 A source splits; one branch opens (

i_{1} \to 0

) — where does the current go?

All 10 A flows through the remaining branch,

i_{2} \to 10

Connections

Apply Kirchhoff's Current Law (KCL) — the parent law these examples exercise.
Nodal Analysis — automates the node-by-node KCL of Example 5.
Conservation of Charge — why "in = out" must hold.
Current and Current Density — defines $i = d q / d t$ .
Ideal Wires and Nodes — the merge rule behind Example 8.
Capacitor i-v Relationship — the $i = C d v / d t$ used in Example 6.
Kirchhoff's Voltage Law (KVL) — the loop counterpart.

Worked examples — Apply Kirchhoff's Current Law (KCL)