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1.2 — Kirchhoff's Two Laws
Two wires meet at a junction and a third leaves it. Two amps come in on one, three amps on the other. How much leaves on the third?
Five, obviously. You did not need a law for that. But writing down why it has to be five, and then writing down the matching statement about voltages around a loop, gives you the only two rules you will ever need to solve any circuit made of any number of elements. Everything else in circuit theory — mesh analysis, nodal analysis, Thevenin, Norton, superposition, impedance, filters, amplifier bias — is these two statements plus algebra.
Gustav Kirchhoff published them in 1845, while still a student, aged twenty-one.
Both laws below are conservation statements in disguise, and they are conserving the two quantities Chapter 1.1 defined: charge, which cannot pile up or vanish at a junction, and energy per unit charge, which must come back to where it started after a round trip.
1. The vocabulary you need first
Circuit diagrams have three structural words, and using them loosely is where most confusion starts.
A node is a point where two or more components connect, together with every piece of wire connected to it. This is the definition people get wrong. A wire has no resistance in an idealised circuit, so every point along it is at the same voltage, and therefore the whole stretch of connected wire is one single node, no matter how long it is or how many corners it turns. If three components meet at a T-junction and a wire runs from there across the page to a fourth component, that is still one node with four components on it.
A branch is a single path between two nodes containing one component.
A loop is any closed path you can trace through the circuit, returning to where you started without passing through any node twice. A mesh is a loop with nothing inside it — an "empty window" of the circuit when it is drawn flat on the page. Every mesh is a loop; not every loop is a mesh.
Count them in the figure. The top-left wire from the source to R1 to node B is all node A's territory up to R1's left terminal; the junction at the top middle where R1, R2 and R3 meet is node B; and the entire bottom wire, from the source's negative terminal past R2's bottom end to R4's bottom end, is a single node C. Three nodes, four resistors, and two meshes.
2. Kirchhoff's current law
KCL: the sum of currents entering a node equals the sum of currents leaving it.
Written with a sign convention where currents entering are positive and leaving are negative, it is simply
\sum_{k} i_k = 0 \quad \text{at every node}
Read aloud: "the sum over k of i-sub-k equals zero", meaning add up every current at the node, with entering counted positive and leaving counted negative, and you must get zero.
Why it must be true. Charge is conserved (Chapter 1.1) and a node is a point with no volume — it cannot store charge. If more charge arrived than left, charge would pile up at that point, the point would become more and more negatively or positively charged, and the resulting electric field would grow without limit. Nothing like that happens in a circuit at ordinary frequencies. So what goes in comes out, instant by instant.
The honest caveat, because you should know where a law stops working: at very high frequencies, or where two conductors run close together, a junction does store a little charge through stray capacitance, and KCL as stated is an approximation. The fix is not to abandon KCL but to draw the stray capacitance as an explicit component; then KCL holds again at the enlarged node. This is the lumped element assumption, and it is valid as long as the circuit is physically much smaller than the wavelength of the signals in it. Chapter 7.7 is what happens when it is not.
Applying it
At the top junction in the figure, call the current arriving through R1 as i_1, the current going down through R2 as i_2, and the current going right through R3 as i_3. Then
i_1 = i_2 + i_3 \qquad\text{or equivalently}\qquad i_1 - i_2 - i_3 = 0
You do not need to guess the directions correctly. Assume any direction you like, write the equation, and solve. If a current comes out negative, it simply means it flows the other way from the arrow you drew. This is the single most liberating fact in circuit analysis, and it removes any need to stare at a circuit trying to intuit which way things go.
Worked example — a three-branch node
A node has 5 A entering, 2 A leaving through one branch, and an unknown current i_x in a third branch drawn as leaving.
5 = 2 + i_x \quad \Rightarrow \quad i_x = 3\ \text{A}
Positive, so it really does leave. Now change the problem: 5 A entering, 8 A leaving through one branch, i_x drawn as leaving.
5 = 8 + i_x \quad \Rightarrow \quad i_x = -3\ \text{A}
Negative. The arrow was drawn the wrong way; 3 A is actually entering through that branch. The algebra corrected the guess automatically.
3. Kirchhoff's voltage law
KVL: the sum of voltage rises and drops around any closed loop is zero.
\sum_{k} v_k = 0 \quad \text{around every closed loop}
Why it must be true. Voltage is energy per coulomb (Chapter 1.1). Take one coulomb of charge on a walk around a closed loop and bring it back to where it started. It has returned to the same point, so it has the same energy it had when it left — otherwise you would have created or destroyed energy by walking in a circle. So the energy gained in the sources must exactly equal the energy given up in the components.
The gravity analogy is exact here. Walk around a hilly park and return to your front door: the ups and downs cancel to zero, because you are back at the same height. Voltage is height; a loop is a walk that ends where it began.
The same caveat applies: this holds when there is no changing magnetic field passing through the loop. If there is one — a transformer, an inductor, a loop of wire near a motor — then a voltage is induced around the loop and the naive sum is not zero. Again the fix is not to abandon the law but to draw the induced voltage as an explicit element, which is exactly what an inductor symbol is. Then KVL holds again.
The sign rule that stops the mistakes
Pick a direction to walk the loop — clockwise, say — and stick to it. Then:
- When you walk through an element and enter at the terminal marked - and leave at +, you have gained voltage. Write +V.
- When you enter at + and leave at -, you have dropped voltage. Write -V.
- For a resistor, the terminal the current enters is the + terminal. Current flows from high potential to low through a resistance, always.
That last line is the one people forget. Mark the resistor's polarity from your assumed current direction, then walk the loop, and the signs take care of themselves.
Walking clockwise from the bottom-left corner: up through the battery from - to + is a rise of +12. Along the top through R1 entering the + side is a drop of -2I. Through R2 the same, -4I. Down through R3, -6I. Back to where we started. Total:
12 - 2I - 4I - 6I = 0 \quad\Rightarrow\quad 12 = 12I \quad\Rightarrow\quad I = 1\ \text{A}
Check the result by conservation of power. The battery supplies P = VI = 12 \times 1 = 12 W. The resistors absorb I^2R each: 1^2\times2 = 2 W, 1^2\times4 = 4 W, 1^2\times6 = 6 W, totalling 12 W. They match, so the answer is right. Get into the habit of this check — it catches sign errors that nothing else will.
4. Worked problem set
These are the standard forms. Work each one before reading its solution; the point is to build the reflex of choose directions, write equations, solve, check power.
Problem 1 — Two loops, one shared branch
A 24 V source drives R_1 = 4\ \Omega into a node. From that node, R_2 = 12\ \Omega goes to ground and R_3 = 6\ \Omega also goes to ground. Find every current.
Solution. Call the middle node voltage V_B, with the bottom rail as 0 V. That single choice makes this a one-unknown problem.
KCL at node B: current in through R_1 = current out through R_2 + current out through R_3.
\frac{24 - V_B}{4} = \frac{V_B}{12} + \frac{V_B}{6}
Multiply everything by 12 to clear the fractions:
3(24 - V_B) = V_B + 2V_B
72 - 3V_B = 3V_B
72 = 6V_B \quad\Rightarrow\quad V_B = 12\ \text{V}
Now every current follows from Ohm's law:
i_1 = \frac{24-12}{4} = 3\ \text{A}, \qquad i_2 = \frac{12}{12} = 1\ \text{A}, \qquad i_3 = \frac{12}{6} = 2\ \text{A}
Check KCL: 3 = 1 + 2 ✓. Check power: source supplies 24\times3 = 72 W; resistors absorb 3^2\times4 = 36, 1^2\times12 = 12, 2^2\times6 = 24, total 72 W ✓.
Problem 2 — Two sources fighting each other
A 10 V source in series with 2\ \Omega connects to a node; a 6 V source in series with 3\ \Omega connects to the same node from the other side; a 6\ \Omega resistor runs from that node to ground. Both sources have their negative terminals on the ground rail. Find the node voltage.
Solution. Let the node be V. Currents into the node from each source branch, and out through the resistor:
\frac{10 - V}{2} + \frac{6 - V}{3} = \frac{V}{6}
Multiply by 6:
3(10-V) + 2(6-V) = V
30 - 3V + 12 - 2V = V
42 = 6V \quad\Rightarrow\quad V = 7\ \text{V}
Branch currents: (10-7)/2 = 1.5 A in from the 10 V source, (6-7)/3 = -0.333 A in from the 6 V source — negative, so that source is actually absorbing current, being charged by the stronger one — and 7/6 = 1.167 A out through the resistor.
Check KCL: 1.5 - 0.333 = 1.167 ✓.
The negative result is the interesting part. The 6 V source is being driven backwards. If it were a real battery it would be charging; if it were a bench power supply without a sink capability, its voltage would rise instead and the real circuit would not match this model. This is the kind of thing the algebra tells you and intuition does not.
Problem 3 — A loop containing two sources
A single loop contains a 15 V source, a 5 V source connected in opposition (its plus facing the first source's plus), and resistances of 3\ \Omega and 2\ \Omega. Find the current.
Solution. Walk clockwise, taking current I clockwise. The 15 V source is a rise, the 5 V source opposes so it is a drop, and both resistors drop:
15 - 5 - 3I - 2I = 0 \quad\Rightarrow\quad 10 = 5I \quad\Rightarrow\quad I = 2\ \text{A}
Power check with a twist. The 15 V source supplies 15\times2 = 30 W. The 5 V source absorbs 5\times2 = 10 W, because current is being pushed into its positive terminal. The resistors absorb 2^2\times3 = 12 and 2^2\times2 = 8, which is 20 W. Total absorbed = 10+20 = 30 W ✓.
This is exactly what a phone charger does: a higher voltage source pushes current backwards into a lower voltage battery, and the battery absorbs power.
Problem 4 — Three unknown currents, three equations
A bridge-shaped network: node A at the top connects through R_1 = 10\ \Omega to node B and through R_2 = 20\ \Omega to node C. Node B connects to ground through R_3 = 30\ \Omega; node C connects to ground through R_4 = 40\ \Omega; and a R_5 = 50\ \Omega bridges B to C. A 12 V source drives node A relative to ground.
Solution. Two unknown node voltages, V_B and V_C, so two KCL equations.
At node B, current in from A equals current out through R_3 plus current out through R_5 towards C:
\frac{12 - V_B}{10} = \frac{V_B}{30} + \frac{V_B - V_C}{50}
At node C, current in from A plus current in from B equals current out through R_4:
\frac{12 - V_C}{20} + \frac{V_B - V_C}{50} = \frac{V_C}{40}
Clear the first equation by multiplying by 150:
15(12 - V_B) = 5V_B + 3(V_B - V_C)
180 - 15V_B = 5V_B + 3V_B - 3V_C
180 = 23V_B - 3V_C \qquad (1)
Clear the second by multiplying by 200:
10(12 - V_C) + 4(V_B - V_C) = 5V_C
120 - 10V_C + 4V_B - 4V_C = 5V_C
120 + 4V_B = 19V_C \qquad (2)
From (2), V_C = (120 + 4V_B)/19. Substitute into (1):
180 = 23V_B - \frac{3(120+4V_B)}{19}
Multiply through by 19:
3420 = 437V_B - 360 - 12V_B
3780 = 425V_B \quad\Rightarrow\quad V_B = 8.894\ \text{V}
Then V_C = (120 + 35.576)/19 = 8.188\ \text{V}.
What the numbers mean. V_B is higher than V_C, so current flows through the bridging resistor from B to C, specifically (8.894-8.188)/50 = 14.1 mA. If the four resistors had been chosen so that R_1/R_3 = R_2/R_4, then V_B would equal V_C and no current at all would flow through the bridge. That is the balanced bridge condition, and it is the operating principle of the Wheatstone bridge used in every strain gauge, load cell and precision resistance measurement — you adjust one resistor until the bridge current reads exactly zero, and zero is a thing you can detect far more precisely than any absolute value.
Problem 5 — When the answer comes out negative
A loop has a 9 V source, and resistances 6\ \Omega and 3\ \Omega, but you accidentally assume the current flows anticlockwise while the source pushes it clockwise.
Walking anticlockwise with assumed current I: you now pass through the source from + to -, so it is a drop:
-9 - 6I - 3I = 0 \quad\Rightarrow\quad I = -1\ \text{A}
The magnitude is right, the sign says the real current is clockwise. Nothing was lost by guessing wrong. Confirm this on your own paper once, and you will never again waste time worrying about direction before writing equations.
5. How many equations do you need?
For a circuit with n nodes and b branches:
- KCL gives n-1 independent equations. Not n. Write KCL at every node and the last one is the sum of all the others rearranged, carrying no new information. So you pick one node as the reference (ground) and write KCL at the rest.
- KVL gives b - n + 1 independent equations, which for a flat drawing is exactly the number of meshes.
Add them: (n-1) + (b-n+1) = b, which is exactly the number of unknown branch currents. The two laws always supply precisely enough equations, never too few and never a surplus of useful ones. That is not luck — it is a theorem of graph theory, the same subject Volume II, Chapter 8.6 develops, applied to the circuit's connection diagram.
Knowing the count tells you which method to reach for. A circuit with few nodes and many meshes wants nodal analysis; one with few meshes and many nodes wants mesh analysis. That choice, and the systematic way to write both without thinking, is Chapter 1.3.
Kirchhoff's laws are complete but tedious — writing b equations for a twenty-branch circuit is not a life. The next chapter turns them into two mechanical procedures that a person can run without thinking, and adds the series and parallel shortcuts that collapse most circuits before you write a single equation.
Every formula above, built from scratch
None of the results in this chapter are worth memorising, because each one can be rebuilt in under a minute from something simpler. What follows is that rebuilding, one result at a time, so the formula and the reason for it sit on the same page as the explanation that needed them.
Kirchhoff's laws
\sum i = 0 \ \text{at a node} \qquad\qquad \sum v = 0 \ \text{around a loop}
KCL follows from conservation of charge: a node has no volume and cannot store charge, so what arrives must leave.
KVL follows from conservation of energy: a coulomb taken around a closed loop returns with the energy it started with, so the rises must equal the drops.
Equation count: n-1 independent KCL equations for n nodes, and b-n+1 independent KVL equations for b branches, totalling b — exactly the number of unknown branch currents.