A series loop contains a 12 V battery, R₁ = 4 Ω, and R₂ = 8 Ω. Applying KVL clockwise (entering the battery at its negative terminal), what equation do you write and what current results?
A+12 − 4I − 8I = 0, yielding I = 1 A
B−12 + 4I + 8I = 0, yielding I = 1 A (same result, different traversal)
C+12 + 4I + 8I = 0, yielding I = −1 A
DI = 12/4 = 3 A from R₁ alone
Traversing clockwise: entering the battery at − and exiting at + is a voltage rise (+12 V). Each resistor traversed in the direction of current is a drop (−4I, −8I). KVL: +12 − 4I − 8I = 0 → I = 1 A. Option B is also valid (counterclockwise traversal multiplies through by −1 but gives the same answer). Option C incorrectly makes the source a drop. Option D uses only one resistor, ignoring KVL as a loop equation.
Question 2 Multiple Choice
A student applies KVL to a loop and solves for current I, getting I = −0.5 A. What does this mean?
AThe actual current flows opposite to the assumed direction — the analysis is still valid
BAn error was made; current cannot be negative in circuit analysis
CThe loop must be rewritten with the current flowing the other direction before proceeding
DKVL was applied with the wrong sign convention and must be redone
A negative current in KVL analysis simply means the actual current flows opposite to the direction you assumed when writing the equation. The magnitude and all resulting voltages are correct — just reverse the arrow. This is a feature of linear circuit analysis, not an error. You never need to redo the loop; interpret I = −0.5 A as 0.5 A flowing the other way.
Question 3 True / False
KVL is a consequence of energy conservation: moving a charge around a closed loop returns it to its original potential energy, so the net voltage change must be zero.
TTrue
FFalse
Answer: True
Voltage at a point is the potential energy per unit charge relative to a reference. Moving a unit charge around any closed loop and returning to the start involves zero net energy change — otherwise energy would be created or destroyed by repeating the loop. The algebraic sum of voltage rises and drops equaling zero is precisely this energy-conservation statement.
Question 4 True / False
The direction you choose to traverse a loop in KVL changes the numerical value of the current you solve for.
TTrue
FFalse
Answer: False
The choice of traversal direction is arbitrary and affects only the signs of each term in the KVL equation. If you reverse traversal direction, every term flips sign, but the resulting equation is just multiplied by −1 — it yields the same current. The physical current is determined by the circuit, not by the analyst's traversal choice.
Question 5 Short Answer
Explain what KVL is really saying physically, and why the algebraic sum of voltages around any closed loop must equal zero.
Think about your answer, then reveal below.
Model answer: KVL expresses conservation of energy for electric charge. Voltage measures potential energy per unit charge. If you move a unit charge around a closed loop through sources and resistors and return to the starting point, it must have the same energy as when it left — no energy was created or destroyed. Voltage rises (energy gained from sources) must exactly cancel voltage drops (energy lost to resistors and other loads), so their algebraic sum is zero.
The analogy: if you hike a loop trail and return to your starting elevation, your net change in altitude is zero, regardless of the hills and valleys along the way. KVL is the electrical version of this: potential is the altitude, and closing the loop guarantees zero net change.