Intuition What this page is forThe parent note taught you the rule : walk a closed loop, add up the voltage changes with signs, set the total to zero. But a rule is only as good as your ability to fire it at any circuit that shows up. This page enumerates every kind of case KVL can throw at you and works each one from zero — signs, opposing sources, a resistor at zero ohms, a source with no load, two loops sharing a wire, a real-world battery-and-cable problem, and an exam trick. Nothing is skipped.
Before we start, one reminder of the two tools we will lean on the whole way:
Recall The two facts every example below reusesKVL says the algebraic sum of voltage changes around a closed loop is zero ::: ∑ k V k = 0 V = I R
We also fix ONE sign habit for the entire page so nothing is ambiguous:
Every KVL problem you will ever meet falls into one of these cells . The examples below are labelled with the cell they cover, and together they hit all of them .
#
Cell (case class)
What makes it tricky
Covered by
A
Single source, single loop, all drops positive
baseline — get signs right
Ex 1
B
Two sources aiding (same direction)
rises add
Ex 2
C
Two sources opposing
one source becomes a drop
Ex 3
D
Wrong current guess → negative answer
interpreting the sign
Ex 4
E
Degenerate: resistor = 0 Ω (short)the drop vanishes
Ex 5
F
Degenerate: open branch / no load I = 0 Ex 5
G
Two loops sharing a branch (mesh)shared resistor carries I 1 − I 2
Ex 6
H
Real-world word problem (battery + cable)translate words → loop
Ex 7
I
Exam twist: reverse traversal / find an unknown sourceequation × (−1) invariance
Ex 8
A 12 V R 1 = 4 Ω R 2 = 2 Ω I
Forecast: total resistance is 4 + 2 = 6 Ω I 2 A
Step 1 — Assume I −
Why this step? We must commit to a direction so "exit + −
Step 2 — Cross the battery, exit its + + 12
Why? Exiting +
Step 3 — Cross R 1 with the current → − I R 1 = − 4 I
Why? Current enters a resistor's + −
Step 4 — Cross R 2 − I R 2 = − 2 I
Step 5 — Sum to zero and solve.
12 − 4 I − 2 I = 0 ⇒ 12 = 6 I ⇒ I = 2 A
Verify: V R 1 = I R 1 = 2 ⋅ 4 = 8 V V R 2 = 2 ⋅ 2 = 4 V 8 + 4 = 12 V V = A ⋅ Ω
A 10 V 5 V head-to-tail (both pushing current the same way) through one resistor R = 3 Ω I
Forecast: if they cooperate, does the driving voltage add to 15 V I
Step 1 — Assume I +
Why? "Aiding" means both lift you as you walk — see both amber arrows pointing up in the figure.
Step 2 — Write the two rises: + 10 + 5 Why? Exit + + V
Step 3 — Cross the resistor with the current: − 3 I
Step 4 — Sum and solve.
10 + 5 − 3 I = 0 ⇒ 15 = 3 I ⇒ I = 5 A
Verify: the resistor drop is 3 ⋅ 5 = 15 V 10 + 5
A 10 V 4 V fight each other , with R = 3 Ω I
Forecast: if they oppose, does the net push become 10 − 4 = 6 V I
Step 1 — Assume I 10 V
Step 2 — 10 V + + 10
Step 3 — Resistor with current → − 3 I
Step 4 — The 4 V enter its + − − 4
Why this is the key step? An opposing source is a drop for our walk, not a rise. The amber arrow for this source points down against our travel in the figure.
Step 5 — Sum and solve.
10 − 3 I − 4 = 0 ⇒ 6 = 3 I ⇒ I = 2 A
Verify: resistor drop = 3 ⋅ 2 = 6 V = 10 − 4 charged (current pushed into its +
Same circuit as Ex 3, but suppose you (deliberately) guessed the current counterclockwise . Show the answer self-corrects.
Forecast: will the magnitude still be 2 A
Step 1 — Assume I
Why? We test the parent's claim that a wrong guess only flips a sign.
Step 2 — Now the 4 V + + 4 10 V − − 10 − 3 I
Step 3 — Sum and solve.
4 − 3 I − 10 = 0 ⇒ − 6 = 3 I ⇒ I = − 2 A
Verify: the magnitude is 2 A minus sign tells us the real current runs clockwise — exactly what Ex 3 found.
Common mistake "A negative current means I made a mistake."Why it feels right: negatives look like errors.
Why it's wrong: KVL is self-correcting — a negative answer just means the true flow opposes your guessed arrow. The number is still correct.
Worked example Statement (E — a resistor becomes0 Ω R 2 R 2 = 0 Ω I
Forecast: with less resistance, is the current larger ? Where does the battery's voltage go now?
Step 1 — KVL: 12 − 4 I − 0 ⋅ I = 0 Why? A 0 Ω I R = I ⋅ 0 = 0
Step 2 — Solve: 12 = 4 I ⇒ I = 3 A
Verify: R 1 entire 12 V V R 1 = 3 ⋅ 4 = 12 V V R 2 = 0 V = 12 zero voltage across itself while still passing current.
Worked example Statement (F — the branch is open, no load)Disconnect one wire so the loop is broken (an open switch). What is I
Forecast: with the loop broken, can charge flow at all? Where is the 12 V
Step 1 — No closed path ⇒ I = 0
Why? Current needs a complete loop; the open gap has infinite resistance.
Step 2 — Apply KVL around the loop treating the gap as an unknown voltage V gap
12 − I R 1 − I R 2 − V gap = 0. I = 0 12 − 0 − 0 − V gap = 0 ⇒ V gap = 12 V
Verify: all 12 V none across the resistors (no current ⇒ no I R
Two loops share a middle resistor. Left loop: 12 V R 1 = 2 Ω R 3 = 4 Ω R 3 = 4 Ω R 2 = 2 Ω 6 V I 1 I 2 I 1 , I 2 Mesh Analysis for the general method.
Forecast: the shared resistor carries I 1 − I 2 I 1 I 2
Step 1 — Left-loop KVL. Walking I 1 12 − 2 I 1 − 4 ( I 1 − I 2 ) = 0. Why the ( I 1 − I 2 ) In the shared branch I 1 I 2 net current there is I 1 − I 2
Step 2 — Right-loop KVL. Walking I 2 − 6 − 4 ( I 2 − I 1 ) − 2 I 2 = 0. Why the − 6 Given orientation, we exit the right source's −
Step 3 — Tidy both equations.
6 I 1 − 4 I 2 = 12 − 4 I 1 + 6 I 2 = − 6.
Step 4 — Solve the pair. Multiply the first by 3 2 18 I 1 − 12 I 2 = 36 , − 8 I 1 + 12 I 2 = − 12. 10 I 1 = 24 ⇒ I 1 = 2.4 A 6 ( 2.4 ) − 4 I 2 = 12 ⇒ 14.4 − 4 I 2 = 12 ⇒ I 2 = 0.6 A
Verify: shared branch current = I 1 − I 2 = 1.8 A − 6 − 4 ( 0.6 − 2.4 ) − 2 ( 0.6 ) = − 6 + 7.2 − 1.2 = 0
A car battery reads 12.6 V R lamp = 1.4 Ω R wire = 0.1 Ω
Forecast: the cables "steal" a little voltage. Will the lamp see the full 12.6 V
Step 1 — Translate the sentence into a single loop: source 12.6 V R wire R lamp Voltage Divider situation.
Why? Wires + lamp are in series, so one loop and one current I
Step 2 — KVL: 12.6 − I ( 0.1 ) − I ( 1.4 ) = 0
Step 3 — Solve: 12.6 = 1.5 I ⇒ I = 8.4 A
Step 4 — Voltage at the lamp: V lamp = I R lamp = 8.4 × 1.4 = 11.76 V
Verify: cable drop = 8.4 × 0.1 = 0.84 V 11.76 + 0.84 = 12.6 V
A loop has an unknown source V x 9 V R = 3 Ω I = 1 A aid . Find V x counterclockwise .
Forecast: if 9 V 3 V R 1 A V x
Step 1 — Clockwise KVL: V x + 9 − 3 ( 1 ) = 0 Why? Both sources exit + I R = 3 V
Step 2 — Solve: V x = 3 − 9 = − 6 V Interpretation: the minus means V x opposing the 9 V 6 V
Step 3 — Reverse traversal (counterclockwise). Every sign flips:
− V x − 9 + 3 ( 1 ) = 0.
Step 4 — Solve again: − V x = 9 − 3 = 6 ⇒ V x = − 6 V identical .
Why? Reversing the walk multiplies the whole equation by − 1
Verify: plug V x = − 6 − 6 + 9 − 3 = 0 6 − 9 + 3 = 0
Recall Which cell has a resistor drop of exactly zero, and why?The 0 Ω I R = I ⋅ 0 = 0
Recall In a two-loop circuit, what current flows in the shared branch?The difference of the two mesh currents ::: I 1 − I 2
Recall What does a negative current in your final answer mean?Your assumed arrow was backwards ::: the real current flows the other way; the magnitude is correct.
