Kirchhoff's current law (KCL), Kirchhoff's voltage law (KVL): Question bank

Before we start, every symbol used on this page, defined from zero:

$Q$ = electric charge = the total amount of "electrical stuff" (in coulombs, C). Think of it as a quantity of water. See Conservation of Charge.
$I$ = current = the amount of charge $Q$ flowing past a point each second, so $I = \frac{d Q}{d t}$ (in amperes, A). Picture water passing a mark on a pipe.
$V$ = potential difference (voltage) = the "height" you drop or climb in the hilly-landscape picture below (in volts, V). See Electric Potential.
$R$ = resistance = how hard a component fights the current, linked by Ohm's Law $V = I R$ (in ohms, $Ω$ ).
$ε$ = EMF (electromotive force) = the fixed voltage "push" a battery supplies from its $-$ terminal to its $+$ terminal (in volts, V). It is the height of the pump in the picture.
$r$ = internal resistance = the small hidden resistance inside a real battery (in ohms, $Ω$ ).

Worked mini-derivations — see every sign choice

Before the traps, two complete step-by-step derivations so you have seen why each sign and step happens in a real circuit.

Worked example KVL around a single loop — full working

Setup:

ε = 12 V

R_{1} = 2 Ω

R_{2} = 4 Ω

in one loop. Find

I

Assume a direction — say clockwise. Why: KVL needs a reference direction; a wrong guess only returns a negative number, which we then reinterpret.
Walk the loop clockwise starting at the battery's $-$ terminal, adding each signed change:
- Through the battery $- \to +$ : you climb the pump → $+ ε$ .
- Through $R_{1}$ in the current direction: you go down the slide → $- I R_{1}$ .
- Through $R_{2}$ in the current direction: another slide → $- I R_{2}$ . $+ ε - I R_{1} - I R_{2} = 0$ Why it sums to zero: you ended at the same point you left — same height (KVL). Follow the arrows in the figure to see each rise and drop.
Solve. $I = \frac{ε}{R _{1} + R _{2}} = \frac{12}{6} = 2 A$ . Positive, so our clockwise guess was right.

True or false — justify

A resistor uses up current, so less current comes out than went in.

False. A resistor dissipates energy, not charge. By KCL the same current enters and leaves a series resistor; only the voltage falls across it.

KCL only works in steady state (DC).

False in general. KCL as

\sum I = 0

needs

\frac{d Q}{d t} = 0

at the node (

Q

= stored charge); if the node can store charge (a capacitor plate) you must include

\frac{d Q}{d t}

. For ordinary wire junctions it holds because a bare junction can't hoard charge.

KVL says the sum of the magnitudes of all voltages in a loop is zero.

False. It says the sum of signed changes is zero. Rises are

+

, drops are

-

(see the pump-and-slide figure); magnitudes never sum to zero unless the loop is trivial.

If a current comes out negative from your equations, you made an algebra mistake.

False. A negative answer just means your assumed direction was backward; the magnitude is correct, so flip the arrow.

Two resistors in parallel across a battery carry the same current.

False. They share the same voltage (they connect the same two nodes), but the currents differ:

I = V / R

, so the smaller resistor carries the larger current.

KVL is a statement of conservation of charge.

False. KVL comes from energy conservation (potential is single-valued, a round trip returns to the same height). Charge conservation is what gives you KCL.

For an ideal battery, the terminal voltage always equals its EMF

ε

True for an ideal battery (zero internal resistance

r

). A real battery drops

I r

inside, so terminal voltage

= ε - I r

and falls as current rises.

You can apply KVL to a loop even if that loop passes through a gap where no wire exists.

True in circuit terms only if you account for the potential difference across the gap (e.g. an open switch or capacitor). A "loop" for KVL is any closed potential path; an open air gap simply carries the full voltage difference across it.

Spot the error

Figure — Kirchhoff's current law (KCL), Kirchhoff's voltage law (KVL)

"I'll add the currents into the node and set them equal to zero:

I_{1} + I_{2} + I_{3} = 0

, all three flowing in."

If all three genuinely flow in, they can't all be positive and sum to zero (unless all are zero). Real junctions must have at least one current leaving; the mistake is assigning every arrow inward (look at the left node in the figure).

"Walking my loop I hit a resistor against the current, so I write

- I R

Wrong sign. Against the current you climb the slide backwards → it's a rise: write

+ I R

. In the current direction it's the drop

- I R

"The battery gives a rise of

ε

no matter which way I walk through it."

The sign depends on your travel direction. From

-

+

it's

+ ε

(rise); from

+

-

it's

- ε

(drop). Direction is chosen once for the whole loop, then applied rigidly.

"KCL failed — the currents into my node don't balance, so charge is being created."

More likely you mislabelled a direction or missed a branch. KCL is exact; an apparent imbalance signals a bookkeeping error, not a physics violation.

"These two resistors are in parallel, so I add their resistances."

Parallel resistors add as reciprocals:

1/ R_{e q} = 1/ R_{1} + 1/ R_{2}

. You add resistances directly only in series. See Series and Parallel Resistors.

"I chose clockwise for the loop and clockwise for the branch current, so the current is definitely clockwise."

The loop direction is just a bookkeeping convention; it doesn't fix the physical current. Only solving the equations (and the sign of the result) tells you the true direction.

"A wire has no resistor, so I can ignore it in KVL."

Fine only for an ideal wire (

V = I R = 0

because

R = 0

). It still connects nodes at equal potential — ignoring the wire is right, but never ignore the nodes it joins.

Why questions

Why can't charge pile up at an ordinary node?

A bare junction has no capacity to store charge

Q

; any imbalance would build a huge electric field instantly that pushes charge back out. So in effect

\frac{d Q}{d t} = 0

and in must equal out.

Why is KVL equivalent to saying the electrostatic field is conservative?

A conservative field means

\oint E \cdot d l = 0

— the work per charge around any closed loop is zero, which is exactly the sum of potential changes being zero. See Electric Potential.

Why do we get to assume a current direction before we know it?

The equations are linear and honest: a wrong guess produces a negative number of the same magnitude. Assuming lets us write signs consistently; the algebra self-corrects.

Why do parallel branches share the same voltage?

Their endpoints are the same two nodes. Potential is a property of a point, so the difference between those two points is one number that every branch between them must feel.

Why does the series-resistance formula

R_{e q} = R_{1} + R_{2}

fall out of KVL?

One current

I

passes through both; KVL gives

ε = I R_{1} + I R_{2} = I (R_{1} + R_{2})

, so the pair behaves like a single resistor

R_{1} + R_{2}

. See Series and Parallel Resistors.

Why do we need both laws to solve a two-loop circuit, not just one?

KCL alone relates unknown currents but doesn't pin their values; KVL alone doesn't know how currents split at nodes. Together they give enough independent equations for all unknowns — the basis of Mesh and Nodal Analysis.

Why does a balanced Wheatstone Bridge have zero current in its middle branch?

When the two node potentials at the bridge's midpoints are equal, KVL across the middle branch gives zero voltage, so by Ohm's law zero current — no round-trip height difference to push charge.

Edge cases

What does KCL say at a node with only two connections in series?

The current in equals the current out — the same

I

everywhere along a series path. A "node" with two elements is really just a point on one wire.

What happens to KVL if a loop contains a capacitor with no current flowing (steady state)?

The capacitor carries a voltage but no current; KVL still holds using that capacitor voltage as one of the signed terms. The resistor drops are zero because

I = 0

Zero resistance in a branch — what does KVL force?

The voltage across that branch is

I R = 0

, so its two nodes sit at equal potential. Any current can flow; the branch is a short.

Infinite resistance (open circuit) in a branch — what does KCL force?

No current can flow through it, so that branch's current is zero. The full loop voltage appears across the open gap (KVL), pushing nothing through.

An ideal battery short-circuited by an ideal wire — what do the laws predict?

KVL gives

ε = I R = I \cdot 0 = 0

, a contradiction unless

ε = 0

. This is the idealisation breaking down: real batteries have internal resistance

r

, giving finite

I = ε / r

A node where all currents are momentarily zero — is KCL satisfied?

Yes, trivially:

\sum 0 = 0

. KCL is a balance, and empty balance is still balance.

A loop of pure ideal wire (no source, no resistor) — what current flows?

KVL sums to

0 = 0

, giving no constraint that forces current. With no EMF to drive it, the physical current is zero; nothing pumps it.

If you traverse the same loop in the opposite direction, does KVL change?

No. Every rise becomes a drop and every drop a rise, so the whole equation is multiplied by

- 1

0 = 0

still. The physics is direction-independent.

Edge cases — where the laws themselves break

When does KVL start to fail?

When the circuit is comparable to the signal wavelength

λ

(high frequency / fast edges). Then

\oint E \cdot d l = - \frac{d Φ _{B}}{d t} \neq = 0

(a changing magnetic flux

Φ_{B}

induces EMF), so the loop voltages no longer sum to zero — you must use Faraday's law instead.

When does KCL start to fail, and what fixes it?

In fast, non-quasi-static regimes charge can momentarily build up (e.g. between capacitor plates). Maxwell's fix is the displacement current

I_{d} = ε_{0} \frac{d Φ _{E}}{d t}

(from a changing electric flux

Φ_{E}

); adding it to the real current restores

\sum I = 0

in the full Ampère–Maxwell sense.

Why is the displacement current needed at a capacitor?

No charge physically crosses the gap between the plates, yet current flows in the wires. The changing electric field between the plates carries a displacement current

I_{d} = ε_{0} \frac{d Φ _{E}}{d t}

that exactly continues the conduction current, keeping the KCL-style balance intact.

For a slow (DC or low-frequency) circuit, why can we safely ignore all this?

The quasi-static condition holds: magnetic flux through loops and displacement current are negligible, so plain KVL (

\sum V = 0

) and KCL (

\sum I = 0

) are excellent approximations — exactly the regime of this chapter.

Connections

Ohm's Law — every "why" above quietly uses $V = I R$ .
Conservation of Charge — the reason no current "vanishes" in a resistor.
Conservation of Energy — why a round trip nets zero voltage.
Electric Potential — single-valuedness underlies every KVL trap.
Series and Parallel Resistors — the parallel/series confusion items.
Wheatstone Bridge — the balanced-bridge edge case.
Mesh and Nodal Analysis — why both laws are needed together.