Kirchhoff's Laws: what to do when Ohm's law isn't enough
Two statements about conservation that let you solve any network, however ugly.
Start here: what problem are we even solving?
Ohm's law works when there is one battery and one path. Multiply, divide, done.
But look at a circuit with two batteries pushing against each other, and wires that split and rejoin. Now there is no single current to find — different wires carry different currents, and you cannot tell which way some of them flow just by looking.
Kirchhoff's two laws are the tools for exactly that situation. They are not new physics. They are two things you already believe, written down carefully enough to do arithmetic with.
The first law: water in pipes
Imagine water flowing through a pipe that splits into two smaller pipes.
If 6 litres a second flows in, and one branch takes 4 litres a second, how much goes down the other branch?
2 litres a second. You did not need a formula. Water cannot vanish at the join, and it cannot pile up there either — whatever arrives has to leave.
Electricity behaves exactly the same way. Current that arrives at a join must leave it. That is Kirchhoff's Current Law, and you already understood it before you read this sentence.
💡 The place where wires meet is called a node. That is the only new word so far, and it just means "a join". A road junction, a pipe junction, a wire junction — same idea.
In: 6 A
Out: 4 A + ?
? = 6 − 4 = 2 A
Current in = current out. That is the whole law.
The second law: walking up and down a hill
Now picture walking from your front door, up a hill, around the block, and back to your front door.
You went up. You went down. When you get back to where you started, how much higher are you than when you left?
Zero. Obviously — you are standing in the same place. Every metre you climbed, you came back down.
Voltage works the same way. Voltage is electrical "height". Go all the way around a circuit and return to your starting point, and the ups must exactly cancel the downs. That is Kirchhoff's Voltage Law.
💡 A path that goes all the way round and comes back to where it started is called a loop. Second new word, and again it means exactly what it sounds like.
Where the "ups" and "downs" come from
In a circuit there are only two things that change your electrical height:
🔋 Going up and coming down
A batteryPushes you UP. It is the thing doing the lifting — like a staircase or an escalator.
A resistorDrops you DOWN. Current squeezing through it loses height, and the amount you drop is I × R — Ohm's law again, doing the same job it always did.
So walking a loop is just bookkeeping: add up all the ups, add up all the downs, and the two must be equal because you finished where you began.
A simple loop: a 10 V battery and a 4 Ω resistor.
Going round:
up 10 V (through the battery)
down I × 4 (through the resistor)
They must cancel: 10 = I × 4
I = 2.5 A
That is Ohm's law — but now you can see it is really
KVL applied to the simplest possible loop.
Two batteries: why you need this at all
Here is where Ohm's law alone gives up. Two batteries, and a resistor in the middle that both of them push current through.
You cannot say "the current is V over R", because there are two Vs and they are fighting. Some current comes from the left, some from the right, and the middle wire carries whatever the combination works out to be.
The method: guess the directions, write down the two laws, and let the algebra correct you. If you guess a direction wrongly, the answer comes out negative — which tells you it flows the other way and the size was right anyway. You are allowed to guess. That is the whole point.
What the two methods are, in one line each
There are two standard ways to organise the bookkeeping, and both give the same answer.
🔧 Pick whichever gives you less work
Mesh analysisWalk around each loop and use the hill rule (KVL). Good when the circuit has few loops.
Nodal analysisStand at each join and use the water rule (KCL). Good when the circuit has few joins.
Count the loops, count the joins, and use whichever number is smaller. That is the entire basis of the choice — fewer equations to solve.
Mesh analysis, step by step
You can follow this before you fully understand it. Understanding tends to arrive around the second or third attempt.
1. Find the loops. A loop is any closed path you can
walk round. In a circuit drawn as two squares side
by side, there are two obvious loops — the left
square and the right square.
2. Draw a circular arrow inside each loop.
Make them all CLOCKWISE. Always clockwise.
You are not predicting anything, you are choosing
a direction to measure in.
3. Name them I₁, I₂, …
4. Walk each loop in the direction of its arrow, and
write down every up and every down. Set the total
to zero.
5. Solve the equations.
6. Check: put your answers back into an equation you
did not use to solve, and see that it comes to zero.
💡 A negative answer is not a mistake. It means the current really flows opposite to the arrow you drew. The number is still correct. Nobody knows the real directions before solving — that is exactly why we guess and let the maths sort it out.
The one part that genuinely confuses people
When two loops sit side by side, they share the wire between them. That shared resistor has both loop currents flowing through it — one going down, one going up.
So when you walk loop 1 through the shared resistor, the current in it is not just I₁. It is I₁ − I₂: your own current, minus the one coming the other way through the same component.
That single term is the only thing that makes mesh analysis harder than solving one loop on its own. If you understand why it is a subtraction, you understand the method.
💡 Why subtract? Because they flow through that shared wire in opposite directions. Two people pushing a door from opposite sides — the net push is the difference, not the sum.
The sign rules, written out
Walking a loop, at each component ask two questions: what is it, and which way am I going through it?
➕➖ The four cases
Resistor, going WITH the currentYou are going downhill. Write +IR.
Resistor, going AGAINST the currentYou are going uphill. Write −IR.
Battery, entering − and leaving +It lifted you. Write −V.
Battery, entering + and leaving −You went down through it. Write +V.
💡 If the signs feel arbitrary, use the hill picture instead of memorising the table. Ask "did I just go up, or down?" — and be consistent all the way round the loop. Consistency matters far more than which convention you picked.
Nodal analysis, step by step
Same idea, organised around joins instead of loops.
1. Pick the join with the most wires and call it
GROUND. Its voltage is 0 — not because it is
special, but because heights have to be measured
from SOMETHING, and this is the sea level you chose.
2. Label the other joins V_A, V_B, …
3. At each labelled join, write:
"all the current leaving here adds up to zero"
Current leaving through a resistor R towards
somewhere at voltage V_other is:
(this voltage − other voltage) / R
Always "mine minus theirs". Keep that order at
every single one and the signs look after themselves.
4. Solve. Multiply everything by the lowest common
denominator first — it clears the fractions and
removes most of the arithmetic mistakes.
💡 Why always "leaving"? Because if you write some currents as entering and some as leaving, you will drop a minus sign somewhere. Pick one direction, use it everywhere, and a current that is really entering simply comes out negative — which is correct, and which the algebra handles for you.
KCL — Kirchhoff's Current Law
The algebraic sum of currents at any node is zero.
Σ I_in = Σ I_out or Σ I = 0 at a node
Physical basis: charge cannot pile up at a junction.
Whatever flows in must flow out.
KVL — Kirchhoff's Voltage Law
The algebraic sum of voltages around any closed loop is zero.
Σ V = 0 around a loop
Physical basis: energy conservation. Walk a loop and return
to where you started, and you must be at the same potential.
Sign convention is where nearly all the marks are lost. Fix one rule and never change it mid-problem: going −ve → +ve through a source is a rise (+); crossing a resistor along the assumed current direction is a drop (−). If you guess a current's direction wrongly, the algebra returns a negative value — which is the correct answer, just flowing the other way. Don't "fix" it.
Mesh analysis — the KVL method
Assign a loop current to each independent mesh, write KVL for each, solve simultaneously. Best when the circuit has few loops.
Worked numerical 1 — mesh analysis, two loops
A 10 V source and a 4 V source share a middle branch. Left loop: 10 V source, 2 Ω, and the shared 3 Ω. Right loop: 4 V source, 1 Ω, and the shared 3 Ω. Find both mesh currents.
Let I₁ = left mesh (clockwise), I₂ = right mesh (clockwise).
The shared 3 Ω carries (I₁ − I₂) downward.
Loop 1 (KVL): −10 + 2I₁ + 3(I₁ − I₂) = 0
5I₁ − 3I₂ = 10 … (i)
Loop 2 (KVL): 3(I₂ − I₁) + 1·I₂ + 4 = 0
−3I₁ + 4I₂ = −4 … (ii)
Solve: from (i) I₁ = (10 + 3I₂)/5
Substitute into (ii):
−3(10 + 3I₂)/5 + 4I₂ = −4
−30 − 9I₂ + 20I₂ = −20
11I₂ = 10 → I₂ = 0.909 A
I₁ = (10 + 2.727)/5 = 2.545 A
Current in shared 3 Ω = I₁ − I₂ = 1.636 A (downward)
Verify with loop 1: −10 + 2(2.545) + 3(1.636)
= −10 + 5.09 + 4.91 = 0 ✔
Nodal analysis — the KCL method
Pick a reference node (ground), write KCL at each remaining node in terms of node voltages, solve. Best when the circuit has few nodes but many loops.
Worked numerical 2 — nodal analysis
Node A connects: a 12 V source through 4 Ω, a 6 Ω to ground, and an 8 V source through 2 Ω. Find V_A.
Take ground as reference. KCL at A — sum of currents
LEAVING A equals zero:
(V_A − 12)/4 + V_A/6 + (V_A − 8)/2 = 0
Multiply through by 12 (the LCM):
3(V_A − 12) + 2V_A + 6(V_A − 8) = 0
3V_A − 36 + 2V_A + 6V_A − 48 = 0
11V_A = 84
V_A = 7.64 V
Branch currents:
through 4 Ω: (7.64 − 12)/4 = −1.09 A (flows INTO A)
through 6 Ω: 7.64/6 = 1.27 A (out of A)
through 2 Ω: (7.64 − 8)/2 = −0.18 A (flows INTO A)
Check KCL: −1.09 + 1.27 − 0.18 = 0 ✔
The negative signs mean those currents flow opposite to the assumed direction — into node A rather than out. The algebra handled it automatically.
🔀 Mesh or nodal — which to pick?
Use meshFewer loops than nodes. Circuit is planar and drawn as neat windows. Sources are mostly voltage sources.
Use nodalFewer nodes than loops. Sources are mostly current sources. You need node voltages as the answer anyway.
Count firstMeshes needed = B − N + 1 (B branches, N nodes). Node equations needed = N − 1. Pick whichever number is smaller — that's fewer simultaneous equations to solve under time pressure.
💡 Exam angle: this is a guaranteed high-mark question, usually 8–10 marks with two sources. Always draw the circuit and mark your assumed current directions before writing a single equation — examiners award marks for correct KVL/KCL equations even if the simultaneous solution goes wrong. And finish by substituting your answer back into one original equation; it catches sign errors in seconds.
Syllabus points
KCL and KVL
Mesh & nodal analysis (numerical)
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