Apply Kirchhoff's Voltage Law (KVL): Visual walkthrough

We assume nothing but arithmetic and the idea of "how high up am I." Every new word gets a picture the moment it appears.

Step 1 — What "voltage" even is: a height on a hill

WHAT. Before any law, we need one honest picture of the word voltage. Voltage at a point is like an altitude — a number telling you how "high up" you are on an electrical hill. We write the altitude at a point $A$ as $V_{A}$ .

WHY this picture and not another? Because the whole law is about a quantity that only cares where you are, not how you got there. Altitude is the everyday thing that behaves exactly like that: two hikers standing on the same rock are at the same height, no matter which trail each walked. We will lean on that fact hard, so we plant it first.

PICTURE. Look at the hill. Two points $A$ and $B$ sit at different heights. The voltage difference $V_{A B}$ is just the vertical gap between them:

$V_{A B} = height at A V_{A} - height at B V_{B}$

Each symbol is doing one job: $V_{A}$ is where you start, $V_{B}$ is where you end, and the minus sign measures the change from start to finish.

Figure — Apply Kirchhoff's Voltage Law (KVL)

Step 2 — A loop is a walk that ends where it began

WHAT. A closed loop is any path through the circuit that starts at some point and returns to that very same point. Draw a wire, wander through batteries and resistors, and come home.

WHY. KVL is not about circuits in general — it is specifically about loops. The magic ingredient is "same start and same finish," so we must nail down what that means before we use it.

PICTURE. The dashed loop leaves point $A$ , climbs, drops, climbs, and lands back on $A$ . Because $A$ is one single rock on the hill, it has one altitude. So the height where you finish equals the height where you started — always.

$V_{start} = V_{finish} because both are V_{A}$

Step 3 — Net change of a round trip is zero

WHAT. Add up every little height-change along the loop. The total must equal (final height) minus (starting height).

WHY. This is just bookkeeping: if you climb $+ 3$ , drop $- 1$ , climb $+ 2$ , drop $- 4$ , your net move is $3 - 1 + 2 - 4 = 0$ , and you'd better be back where you started. We turn that arithmetic into the shape of the law.

PICTURE. Each colored segment is one height-change. Call the signed height-change across the $k$ -th element $V_{k}$ : it is $+$ when that element raised you and $-$ when it dropped you. Stack them tip-to-tail on the right of the figure; the stack returns to the baseline — the sum of the arrows is zero.

$all the signed ups and downs k \sum V_{k} = = 0 from Step 2 V_{finish} - V_{start} = 0$

Here $V_{k}$ is the ==signed height-change you experience as you cross the $k$ -th element== (a battery, a resistor) — a rise counts $+$ , a drop counts $-$ . It is not a raw terminal-to-terminal magnitude; the sign is baked in by the direction you walk (that's exactly what Step 5 pins down). The Greek $\sum$ ("sigma") just means "add them all up," and $k$ counts them $1, 2, 3, \dots$

Step 4 — Who pushes up, who pulls down: sources vs resistors

WHAT. Two kinds of elements change your height. A battery (source) is an escalator that lifts you up. A resistor is a slide that drops you down when current runs through it.

WHY. The law only counts if we know the sign of each height-change $V_{k}$ . To get signs, we need to know which elements raise altitude and which lower it. That's a physical fact about the parts.

PICTURE. On the left, the battery: its long line is the $+$ terminal (top of the escalator). On the right, the resistor with current $I$ flowing through it — Ohm's Law says the height you lose going from its $+$ terminal to its $-$ terminal, in the direction of the current, is exactly

$V_{R} = current (amps) I \times resistance (ohms) R$

The current $I$ is how much charge flows per second; $R$ measures how "steep" the slide is. Bigger $I$ or bigger $R$ ⇒ bigger drop. Note the sign convention: $V_{R} = I R$ is a positive magnitude of the drop; whether it enters the loop sum as $+ V_{R}$ or $- V_{R}$ is decided by the exit-terminal rule of Step 5.

Step 5 — The sign rule: read it off the terminal you exit

WHAT. As you walk the loop, each element gets a $+$ or a $-$ in the sum. The rule: look at the terminal you leave through.

Exit through the $+$ terminal → you climbed → write $+ V$ (a rise).
Exit through the $-$ terminal → you descended → write $- V$ (a drop).

WHY. Step 3 needs signed changes $V_{k}$ . "Exit-terminal" is the one rule that never lies, because the terminal you come out of literally tells you whether you ended up higher or lower.

PICTURE. Two walkers cross the same battery in opposite directions. The one exiting $+$ writes $+ V$ ; the one exiting $-$ writes $- V$ . Same element, opposite signs — because they ended at opposite heights.

$exit + \Rightarrow + V exit - \Rightarrow - V$

Step 6 — Walk a real loop and watch it sum to zero

WHAT. Put it together on the single-loop circuit from the parent: a $12 V$ battery, then $R_{1} = 4 Ω$ , then $R_{2} = 2 Ω$ , all in series. We walk clockwise starting at the battery's $-$ terminal, and we guess the current clockwise too (so here walker and current point the same way).

WHY. This is where the abstract " $\sum V_{k} = 0$ " becomes a number you can solve. Each crossing gives one signed term; setting the sum to zero hands us the current.

PICTURE. Follow the walker. The battery lifts $+ 12$ (exit $+$ ). Because the current runs clockwise, each resistor's $+$ terminal faces the walker first, so the walker exits through $-$ → a drop → we write $- I R_{1}$ and $- I R_{2}$ :

$climb the battery + 12 slide down R_{1} - I R_{1} slide down R_{2} - I R_{2} = 0$

Substitute $R_{1} = 4$ , $R_{2} = 2$ :

$12 = I (4 + 2) = 6 I \Rightarrow I = 2 A$

Step 7 — Edge case: what if you guessed the current backwards?

WHAT. Keep walking clockwise (walker unchanged), but draw the current arrow counter-clockwise — the wrong-way guess — and solve anyway. What happens?

WHY. Beginners freeze here, afraid a wrong guess ruins everything. This is precisely the case where the walker direction and the current arrow disagree, and the picture shows KVL is self-correcting — so you never fear the arrow.

PICTURE. Same walker (clockwise), but the reversed current now makes each resistor's $-$ terminal face the walker first, so the walker exits through $+$ → a rise → the resistor signs flip to $+ I R$ . The algebra spits out a negative current:

$12 + I R_{1} + I R_{2} = 0 \Rightarrow I = \frac{- 12}{6} = - 2 A$

The minus is the circuit talking back: "the current actually runs opposite to the arrow you drew" — i.e. clockwise, agreeing with Step 6. The magnitude, $2 A$ , is still perfectly correct.

Step 8 — Degenerate case: the loop with no source

WHAT. What if there's no battery at all — just resistors in a closed loop, and nobody pushing?

WHY. The law must survive the boring cases too, or it isn't a law. This is the "zero input" check every derivation owes you.

PICTURE. A loop of resistors with no escalator. KVL says the signed sum is still zero:

$- I R_{1} - I R_{2} - I R_{3} = 0 \Rightarrow I (R_{1} + R_{2} + R_{3}) = 0$

Since the resistances are positive, the only way the sum vanishes is $I = 0$ . No push, no flow — the hill is flat, everyone stays put. The law didn't break; it correctly predicted nothing happens.

The one-picture summary

Everything above compressed into a single frame: the hill (Step 1), the round trip returning to one height (Step 2), the ups and downs stacking back to zero (Step 3), and the escalator-vs-slide signs (Steps 4–5) feeding the single equation $\sum V_{k} = 0$ .

Recall Retell the whole walkthrough (Feynman style)

Voltage is height on a hill. A point on the hill has only one height, so if I take any walk and come back to the same point, my up-and-down changes have to cancel to exactly zero. Batteries are escalators that push me up; resistors are slides that drop me down by $I R$ . When I walk a loop I give each element a sign — plus if I climbed out its top, minus if I slid out its bottom — and I add them all up. That sum is zero. If I set it to zero and solve, I get the current. Guess the current backwards? The math hands me a minus sign that just says "walk the other way." No battery at all? It correctly says nothing flows. The only way to break the rule is a changing magnetic field secretly acting like a hidden escalator — Faraday's territory — but plain DC circuits never have that.

Recall Why is the loop sum exactly zero and not "small"?

Question ::: Because a single point has exactly one electric potential; ending where you started means finish-height minus start-height is precisely zero, so the signed sum of all changes is precisely zero.

Connections

1.2.04 Apply Kirchhoff's Voltage Law (KVL) (Hinglish) — the parent topic this page derives visually.
Ohm's Law — turns each resistor's drop into $I R$ inside the loop sum.
Voltage Divider — the $8 V /4 V$ split in Step 6 is this in action.
Conservation of Energy — the "one height per point" fact is energy conservation per unit charge.
Kirchhoff's Current Law (KCL) — the companion law for nodes.
Mesh Analysis — applies this loop-walk systematically to many loops.
Faraday's Law — the hidden-escalator caveat from Step 8.