Intuition The one-sentence ideaA circuit node is like a plumbing junction: water that flows in must flow out — charge cannot pile up or vanish at a point, so the sum of currents entering a node equals the sum leaving .
Definition Kirchhoff's Current LawAt any node (a junction where 2+ wires meet), the algebraic sum of all currents is zero :
∑ k i k = 0 \sum_{k} i_k = 0 ∑ k i k = 0 ∑ i in = ∑ i out \sum i_{\text{in}} = \sum i_{\text{out}} ∑ i in = ∑ i out node = any point connected only by ideal wires (zero resistance). All points on the same wire are the same node.
WHY does it exist? KCL is just conservation of electric charge applied to a point. A node has no capacitance to store charge, so charge in = charge out at every instant.
Charge is conserved. The charge stored inside a tiny region (the node) is Q ( t ) Q(t) Q ( t ) i = d q d t i = \frac{dq}{dt} i = d t d q
The continuity statement: rate of charge accumulating inside = (rate in) − (rate out):
d Q node d t = ∑ i in − ∑ i out \frac{dQ_{\text{node}}}{dt} = \sum i_{\text{in}} - \sum i_{\text{out}} d t d Q node = ∑ i in − ∑ i out
An ideal node cannot store charge (Q node = 0 Q_{\text{node}} = 0 Q node = 0 d Q d t = 0 \frac{dQ}{dt}=0 d t d Q = 0 ∑ i in − ∑ i out = 0 ⟹ ∑ i in = ∑ i out \sum i_{\text{in}} - \sum i_{\text{out}} = 0 \;\Longrightarrow\; \boxed{\sum i_{\text{in}} = \sum i_{\text{out}}} ∑ i in − ∑ i out = 0 ⟹ ∑ i in = ∑ i out
Intuition Why "algebraic sum = 0"?If we adopt a sign convention — say currents leaving are positive , entering are negative — then "in = out" collapses into one clean equation ∑ i k = 0 \sum i_k = 0 ∑ i k = 0
Worked example Example 1 — Simple 3-wire junctionCurrents i 1 = 3 A i_1 = 3\text{ A} i 1 = 3 A i 2 = 2 A i_2 = 2\text{ A} i 2 = 2 A flow into node A. One current i 3 i_3 i 3 flows out . Find i 3 i_3 i 3
Step 1 — Write KCL (in = out).
Why this step? No charge accumulates, so everything in must leave.
i 1 + i 2 = i 3 i_1 + i_2 = i_3 i 1 + i 2 = i 3 Step 2 — Substitute.
3 + 2 = i 3 ⇒ i 3 = 5 A 3 + 2 = i_3 \Rightarrow i_3 = 5\text{ A} 3 + 2 = i 3 ⇒ i 3 = 5 A Why: The single outgoing wire must carry the entire incoming charge flow.
Worked example Example 2 — Four branches with unknownAt node B: i 1 = 4 A i_1 = 4\text{ A} i 1 = 4 A i 2 = 6 A i_2 = 6\text{ A} i 2 = 6 A i 3 = 7 A i_3 = 7\text{ A} i 3 = 7 A i 4 = ? i_4 = ? i 4 = ?
Step 1 — Sum in = sum out.
Why: conservation.
i 1 + i 2 = i 3 + i 4 i_1 + i_2 = i_3 + i_4 i 1 + i 2 = i 3 + i 4 Step 2 — Solve.
4 + 6 = 7 + i 4 ⇒ i 4 = 3 A (out) 4 + 6 = 7 + i_4 \Rightarrow i_4 = 3\text{ A (out)} 4 + 6 = 7 + i 4 ⇒ i 4 = 3 A (out) Why: 10 10 10 7 7 7 i 3 i_3 i 3 3 3 3 i 4 i_4 i 4
Worked example Example 3 — Wrong guess direction (self-correction)At node C, known: 5 A 5\text{ A} 5 A 2 A 2\text{ A} 2 A guessed i x i_x i x in . Use convention "leaving = +", and one branch 8 A 8\text{ A} 8 A
Step 1 — Write ∑ i leave = 0 \sum i_{\text{leave}} = 0 ∑ i leave = 0
Entering currents count as negative:
− 5 − 2 − i x + 8 = 0 -5 -2 -i_x + 8 = 0 − 5 − 2 − i x + 8 = 0 Step 2 — Solve.
i x = 1 A i_x = 1\text{ A} i x = 1 A Why the positive answer is fine: i x = + 1 i_x = +1 i x = + 1 does flow in at 1 A. Had it come out negative, we'd just reverse the arrow — the math self-corrects the guess.
Worked example Example 4 — Node with a current sourceA 10 A 10\text{ A} 10 A i 1 i_1 i 1 i 2 i_2 i 2 i 1 = 6 A i_1 = 6\text{ A} i 1 = 6 A i 2 i_2 i 2
Step 1 — KCL. 10 = i 1 + i 2 10 = i_1 + i_2 10 = i 1 + i 2 Step 2. i 2 = 10 − 6 = 4 A i_2 = 10 - 6 = 4\text{ A} i 2 = 10 − 6 = 4 A Why: the source current is fully distributed among exits; none is lost.
Common mistake "I'll add magnitudes without caring about direction."Why it feels right: currents are "amounts," and adding amounts seems natural.
Why it's wrong: direction is physical — an incoming and outgoing current subtract, they don't add.
Fix: always assign arrows + signs; entering and leaving have opposite signs.
Common mistake "Two resistors on the same wire = two nodes."Why it feels right: they look like separate components.
Why it's wrong: an ideal wire has no voltage drop; every point on it is electrically identical — one node .
Fix: collapse all points joined by pure wire into a single node before writing KCL.
Common mistake "A capacitor breaks KCL because charge stores on its plates."Why it feels right: capacitors literally store charge.
Why it's wrong: charge stores on the plates inside the component , not at the node . The current i = C d v / d t i = C\,dv/dt i = C d v / d t Fix: treat capacitor terminal current like any branch current.
Recall Feynman: explain to a 12-year-oldImagine a road roundabout where cars enter and leave. Cars can't magically disappear or spawn on the roundabout — so the number of cars driving in per minute must equal the number driving out per minute. Electric current is just charges (tiny cars) moving through wires. At any junction of wires, whatever charge flows in must flow out. That's KCL!
"In = Out, Charge Can't Sprout." Nothing is created or destroyed at a node.
Cover the answers. State KCL two ways. Why can't charge accumulate at an ideal node? What does a negative computed current mean?
State Kirchhoff's Current Law (algebraic form). The algebraic sum of currents at any node is zero:
∑ i k = 0 \sum i_k = 0 ∑ i k = 0 .
KCL is a direct consequence of which conservation law? Conservation of electric charge.
Why does an ideal node force ∑ i i n = ∑ i o u t \sum i_{in} = \sum i_{out} ∑ i in = ∑ i o u t An ideal node stores no charge, so
d Q / d t = 0 dQ/dt = 0 d Q / d t = 0 , meaning current in must equal current out.
What is a "node" in circuit analysis? A junction of two or more branches; all points connected by ideal (zero-resistance) wire form the SAME node.
If you guess a current's direction wrong, what happens when you solve? You get a negative value; its magnitude is correct and you simply reverse the assumed arrow.
Do two components on the same continuous wire form one node or two? One node — an ideal wire has no voltage drop, so it's all one electrical point.
Does a capacitor violate KCL? No — the terminal current
i = C d v / d t i = C\,dv/dt i = C d v / d t enters one terminal and leaves the other; KCL holds at every node.
3 A and 2 A enter a node; one wire leaves. What current leaves? 5 A.
Kirchhoff's Voltage Law (KVL) — the voltage counterpart (energy conservation around a loop).Nodal Analysis — systematically applies KCL at every node to solve whole circuits.Conservation of Charge — the physical foundation of KCL.Current and Current Density — defines i = d q / d t i = dq/dt i = d q / d t Ideal Wires and Nodes — why same-wire points are one node.Capacitor i-v Relationship — clarifies the capacitor "storage" misconception.
Ideal node stores no charge
Intuition Hinglish mein samjhoDekho, KCL ka core idea bahut simple hai: kisi bhi node (jahan do ya zyada wires milte hain) par jitna current andar aa raha hai, utna hi bahar bhi jaana chahiye. Kyun? Kyunki charge na to gayab ho sakta hai, na paida — yeh conservation of charge ka seedha result hai. Ideal node koi charge store nahi karta, isliye ∑ i i n = ∑ i o u t \sum i_{in} = \sum i_{out} ∑ i in = ∑ i o u t ∑ i k = 0 \sum i_k = 0 ∑ i k = 0
Ek roundabout socho: jitni cars andar aati hain, utni hi bahar nikalti hain, warna traffic jam ho jayega (charge pile-up). Wire ki har junction bhi aisi hi hai. Ek important baat — jitne bhi points ek hi continuous wire se jude hain, woh sab ek hi node count hote hain, chahe do resistor beech mein dikhte hon.
Practical tip: current ki direction ka arrow apni marzi se guess kar lo. Agar galat guess kiya, to answer negative aayega — bas arrow ulta kar do, magnitude sahi rahega. Yeh self-correcting nature exam mein tension nahi lene deta.
Yeh law itna important isliye hai kyunki Nodal Analysis puri iski hi buniyaad par khadi hai — bade circuits solve karne ka standard method. KVL (voltage law) ke saath milkar KCL har circuit ko poori tarah solve kar deta hai.
