123-ME · KVL, KCL & Circuit Elements

KVL, KCL & Circuit Elements

Two conservation laws, one accounting trick, and the parts they audit

Chapter 1 gave you the raw ingredients — charge, current, voltage, power. This chapter gives you the accounting rules that let those ingredients combine into an actual circuit you can solve. Kirchhoff's Current Law (KCL) and Kirchhoff's Voltage Law (KVL) aren't arbitrary rules to memorize — each one is a direct, one-line consequence of a conservation law you already believe: charge doesn't pile up anywhere, and energy doesn't appear from nowhere. Once those two laws are in hand, we meet the cast of components they govern — conductors, independent and dependent sources, resistors — and put everything together on a full series circuit. Budget real time on the KCL/KVL derivations below: every later technique in this course (node-voltage analysis, mesh-current analysis, Thevenin equivalents) is just KCL and KVL applied more cleverly, never a different law.

Where we're going

By the end of this chapter you'll be able to: state and derive KCL and KVL from first principles; identify series vs. parallel connections by inspection; recognize conductors, independent sources, and all four families of dependent sources on a schematic; apply Ohm's law and compute resistor power; and solve a complete single-loop circuit while verifying your answer with a power-balance check.

2.1 KCL: charge conservation at a node

A node is any point (or set of points joined by ideal wire) where two or more circuit elements meet. A node is not a physical object — it has no volume, no capacitance, nowhere to store charge. That single fact is the entire derivation.

Derivation, from charge conservation

Let $q_{\text{node}}(t)$ be the charge sitting at a node at time $t$ . Because a node stores no charge, $q_{\text{node}}(t) = 0$ for all $t$ , so its rate of change is also zero:

\frac{dq_{\text{node}}}{dt} = \sum i_{\text{in}}(t) - \sum i_{\text{out}}(t) = 0

which is exactly Kirchhoff's Current Law:

\sum i_{\text{in}} = \sum i_{\text{out}}

Equivalently, if you define every current reference direction as leaving the node (a common convention that avoids having to sort currents into two columns), KCL reads $\sum i = 0$ with entering currents simply carrying a negative value.

A node with two currents entering (

i_1

i_2

) and one leaving (

i_3

). KCL:

i_1 + i_2 = i_3

, so

i_3 = 3 + 2 = 5\text{ A}

— the node can't accumulate the other

5\text{ A}

anywhere, it must all leave through the one remaining path.

Worked Example 2.1 — KCL at a single node

For the node above, $i_1 = 3\text{ A}$ enters from the left, $i_2 = 2\text{ A}$ enters from the top, and $i_3$ leaves to the right. Find $i_3$ .

Apply KCL with entering currents on one side, leaving currents on the other:

i_1 + i_2 = i_3

Substitute the known values:

i_3 = 3 + 2 = 5\text{ A}

i_3 = 5\text{ A, leaving the node}

A useful corollary falls straight out of KCL: if two elements are connected so that no other path joins the node between them, whatever current enters one must entirely enter the other — they carry the same current. This is the formal definition of a series connection: $i_a = i_b$ for series elements $a, b$ . Elements connected across the same two nodes, by contrast, share voltage rather than current — that's a parallel connection, covered fully in Chapter 2.1, but the KCL/KVL roots of both definitions live right here.

2.2 KVL: energy conservation around a loop

A loop is any closed path through the circuit that starts and ends at the same node. Voltage at a point is potential energy per unit charge — it doesn't depend on the path you took to get there, only on where you are. Walk a test charge all the way around a loop back to its starting point, and it's back at the same potential it started at, no matter which elements it passed through along the way.

Derivation, from energy conservation

Move a charge $q$ around a closed loop through elements with voltage drops $v_1, v_2, \ldots, v_n$ (each measured in the direction of travel). The energy gained or lost crossing element $k$ is $q\,v_k$ . Since the charge returns to its exact starting potential, the total energy change over the full loop is zero:

q\left(v_1 + v_2 + \cdots + v_n\right) = 0 \ \Rightarrow\ \sum v_{\text{loop}} = 0

which is Kirchhoff's Voltage Law: the algebraic sum of voltages around any closed loop is zero.

"Algebraic" is the key word: each term's sign depends on whether you enter that element's $+$ or $-$ terminal first, while traversing in your chosen direction. Enter $+$ first → that term is $+v$ ; enter $-$ first → that term is $-v$ . Pick a traversal direction, walk the loop once, and add up signed terms.

Starting at the source's

-

terminal and traversing clockwise: entering

V_s

+

terminal first gives

+20\text{V}

; entering

R_1

-

terminal first (current flows from

+

-

through a resistor, per the passive sign convention) gives

-12\text{V}

; entering

R_2

-

terminal first gives

-8\text{V}

. Sum:

20 - 12 - 8 = 0

. ✓

Worked Example 2.2 — KVL around a loop

For the loop above, verify KVL by writing the traversal equation starting at the source and going clockwise.

List each element's contribution in traversal order. The source's $+$ terminal is met first (source term positive); both resistors are entered at their $-$ terminal (drop in the direction of current, so each is subtracted):

+V_s - V_{R_1} - V_{R_2} = 0

Substitute the known drops:

20 - 12 - 8 = 0

0 = 0 \ \checkmark \text{ — KVL confirmed}

The same corollary as before, mirrored: elements connected across the identical pair of nodes must have the same voltage between them (walk the zero-length loop between the two connection points and KVL forces it) — that's the formal root of the parallel definition, $v_a = v_b$ .

2.3 Circuit elements: conductors, sources, resistors

Conductors: the two extremes

An ideal conductor (wire) has zero resistance, so by Ohm's law ( $v = iR$ ) its voltage drop is exactly zero no matter how much current flows through it. Two special cases show up constantly in circuit reduction:

Short circuit — two nodes tied by ideal wire, forced to the same voltage. Current can flow freely; only its value is set by the rest of the circuit.

Open circuit — no connection at all. No current can flow through the gap, regardless of the voltage the rest of the circuit tries to place across it.

Two failure modes, not synonyms

"Short" and "open" are opposite extremes, easy to mix up under exam pressure. Short: $v=0$ always, $i$ can be anything. Open: $i=0$ always, $v$ can be anything. When you reduce a circuit and a branch turns out to carry zero current, don't call it "shorted" — it's open in effect.

Independent sources

An independent voltage source maintains a fixed voltage across its terminals no matter what's connected to it or how much current is drawn — a 12 V source stays at 12 V whether it delivers 1 mA or 10 A. An independent current source is the dual: it maintains a fixed current through itself no matter what voltage that requires across its terminals. Both are drawn as a circle; the voltage source shows $+/-$ marks, the current source shows an arrow for the reference direction. Both can be DC (constant) or time-varying, e.g. $5\cos(2\pi t)\text{ V}$ .

Dependent (controlled) sources

A dependent source's output isn't fixed — it's set by some other voltage or current elsewhere in the circuit, the control variable. There are exactly four combinations of (output type) × (control-variable type), each with its own Greek-letter gain constant in this course's notation, and each drawn as a diamond rather than a circle to visually flag "this one depends on something else":

The four dependent-source families. Diamond outline signals "controlled"; the label inside shows

+/-

for a voltage output or an arrow for a current output.

\alpha, \gamma

carry units that convert volts→volts or volts→amps;

\beta,\delta

convert amps→volts or amps→amps. Compare against the plain circle of an independent source below — the shape alone tells you whether a source's value is fixed or slaved to the rest of the circuit.

Name	Controlled by	Produces	Formula
VCVS (voltage-controlled voltage source)	a voltage $v_{ctrl}$	a voltage	$v_{out}=\alpha v_{ctrl}$
CCVS (current-controlled voltage source)	a current $i_{ctrl}$	a voltage	$v_{out}=\beta i_{ctrl}$
VCIS (voltage-controlled current source)	a voltage $v_{ctrl}$	a current	$i_{out}=\gamma v_{ctrl}$
CCIS (current-controlled current source)	a current $i_{ctrl}$	a current	$i_{out}=\delta i_{ctrl}$

Resistors and Ohm's law

A resistor converts electrical energy to heat; the relationship between its voltage and current is linear:

v = iR \qquad\text{equivalently}\qquad i = Gv,\ \ G=\frac{1}{R}

$R$ is measured in ohms ( $\Omega$ ), $G$ (conductance) in siemens (S). Physically, resistance comes from a material property (resistivity $\rho$ ) and geometry:

R = \frac{\rho L}{A}

where $L$ is length and $A$ is cross-sectional area — longer and thinner means more resistance, exactly like a narrow, long pipe resists fluid flow more than a short, wide one. Materials range from conductors ( $\rho \sim 10^{-8}\,\Omega\cdot\text{m}$ , e.g. copper) through semiconductors to insulators ( $\rho > 10^{21}\,\Omega\cdot\text{m}$ , e.g. glass).

Since a resistor always has current entering its $+$ terminal under the passive sign convention (that's what "resistor" means physically — it can only absorb, never supply), its power is always positive:

p = vi = i^2R = \frac{v^2}{R} \ \ (\text{always} \ge 0 \text{ — a resistor only absorbs})

Worked Example 2.3 — resistance from geometry

Find the resistance of a copper wire, diameter $2.05\text{ mm}$ , length $10\text{ m}$ ( $\rho_{Cu} = 1.72\times10^{-8}\ \Omega\cdot\text{m}$ ).

Cross-sectional area:

A = \frac{\pi d^2}{4} = \frac{\pi (2.05\times10^{-3})^2}{4} = 3.30\times10^{-6}\text{ m}^2

Apply $R = \rho L / A$ :

R = \frac{(1.72\times10^{-8})(10)}{3.30\times10^{-6}}

R \approx 0.052\ \Omega

Check independently: dimensional check — $[\Omega\cdot\text{m}][\text{m}]/[\text{m}^2] = \Omega\cdot\text{m}^2/\text{m}^2 = \Omega$ . Units are consistent, and a few centimeters of hookup-wire-gauge copper reading a fraction of an ohm matches everyday experience (copper wire is nearly a short). ✓

Worked Example 2.4 — resistance from rated power

A heater element is rated 1500 W at 120 V. Find its resistance.

Find the operating current from $P = Vi$ :

i = \frac{P}{V} = \frac{1500}{120} = 12.5\text{ A}

Apply Ohm's law:

R = \frac{V}{i} = \frac{120}{12.5}

R = 9.6\ \Omega

Check independently: via $R = V^2/P = 120^2/1500 = 14400/1500 = 9.6\ \Omega$ — same answer from the power-based formula rather than the current-based one, confirming no arithmetic slip. ✓

2.4 Putting it together: a full series circuit

Chapter 1.7 exists to show that KVL, Ohm's law, KCL, and power all click together on one circuit — the smallest possible loop is exactly one source and one resistor, and it's worth seeing solved start to finish, including the balance check that catches sign errors before they propagate into a bigger circuit.

Worked Example 2.5 — full series-circuit solve with power balance

A $10\text{ V}$ source drives a single $5\ \Omega$ resistor in series. Find the resistor's voltage and current, and verify power conservation.

KVL around the one-element loop: $-V_s + V_R = 0 \Rightarrow V_R = V_s = 10\text{ V}$ .

Ohm's law: $i_R = V_R/R = 10/5 = 2\text{ A}$ . KCL (series) forces $i_s = i_R = 2\text{ A}$ .

Power: resistor absorbs $P_R = V_R i_R = 10\times2 = 20\text{ W}$ . The source, with current exiting its $+$ terminal (non-passive/active configuration), supplies $P_s = -V_s i_s = -20\text{ W}$ .

P_s + P_R = -20 + 20 = 0 \ \checkmark \text{ — power conserved}

Why the balance check matters

$P_s + P_R = 0$ isn't a coincidence to verify once and forget — it's KVL and Ohm's law combined into a single number, so it catches almost any arithmetic or sign mistake in one line. Make it a habit on every circuit you solve, not just the practice problems below.

2.5 Interactive: KCL/KVL Explorer

The two panels below let you rebuild both derivations by hand. In KCL mode, four currents of fixed magnitude meet at a node — click any arrow to flip its direction, and watch the live equation tell you whether charge is conserved. In KVL mode, step around a loop one element at a time and watch the running signed sum build toward zero.

KCL / KVL Explorer

Click an arrow to flip its direction. Find the one arrangement where charge is conserved.

2.6 Self-checks

Self-check

At a node, $5\text{ A}$ enters and $3\text{ A}$ leaves through one branch. A second branch also leaves the node. What current flows in that second leaving branch?

Self-check

A dependent source's output voltage is given by $v_{out} = \beta i_{ctrl}$ . Which family of dependent source is this?

Self-check

Two nodes are connected directly by an ideal wire (a short circuit). Which statement is correct?

2.7 Practice problems

Problem 2.1

KCL, multi-branch node

At a node, four currents meet: $i_1 = 6\text{ A}$ entering, $i_4 = 1\text{ A}$ entering, $i_2 = 2\text{ A}$ leaving, and $i_3$ (unknown) leaving. Find $i_3$ .

Method 1 — entering/leaving form. KCL: $\sum i_{in} = \sum i_{out}$ , so $i_1 + i_4 = i_2 + i_3 \Rightarrow 6 + 1 = 2 + i_3 \Rightarrow i_3 = 5\text{ A}$ .

Method 2 — all-leaving reference form (independent check). Redefine every current as leaving the node: $i_1 \to -6$ , $i_4 \to -1$ , $i_2 \to +2$ , $i_3 \to +i_3$ . KCL: $-6 - 1 + 2 + i_3 = 0 \Rightarrow i_3 = 5\text{ A}$ . Both conventions agree.

i_3 = 5\text{ A, leaving the node}

Problem 2.2

KVL, series loop, power balance

A single loop has a $24\text{ V}$ source in series with $R_1 = 6\ \Omega$ and $R_2 = 2\ \Omega$ . Find the loop current $i$ , the voltage across each resistor, and verify power conservation.

Method 1 — KVL + Ohm's law. KVL: $-24 + iR_1 + iR_2 = 0 \Rightarrow i(6+2) = 24 \Rightarrow i = 3\text{ A}$ . Then $V_{R_1} = iR_1 = 18\text{ V}$ , $V_{R_2} = iR_2 = 6\text{ V}$ . Check: $18 + 6 = 24\text{ V} = V_s$ . ✓

Method 2 — power balance (independent check). $P_{R_1} = i^2R_1 = 9\times6 = 54\text{ W}$ , $P_{R_2} = i^2R_2 = 9\times2 = 18\text{ W}$ ; total absorbed $= 72\text{ W}$ . Source (current exits its $+$ terminal): $P_s = -V_si = -24\times3 = -72\text{ W}$ (delivers $72\text{ W}$ ). $P_s + P_{R_1} + P_{R_2} = -72+54+18=0$ ✓ — confirms $i=3\text{ A}$ independently of the KVL algebra above (a sign or arithmetic slip in Method 1 would not have balanced here).

i = 3\text{ A}, \quad V_{R_1} = 18\text{ V}, \quad V_{R_2} = 6\text{ V}

Problem 2.3

VCVS, dependent source loop

A single loop contains a $20\text{ V}$ independent source, a $4\ \Omega$ resistor with voltage $v_x$ across it, and a VCVS $4v_x$ , all in series (same polarity convention as Worked Example 2.2 — source $+$ first, then drops). Find $v_x$ , the loop current $i$ , and verify power conservation.

Method 1 — KVL with the control-variable substitution. KVL: $-20 + v_x + 4v_x = 0 \Rightarrow 5v_x = 20 \Rightarrow v_x = 4\text{ V}$ . Ohm's law on the resistor: $i = v_x/R = 4/4 = 1\text{ A}$ .

Method 2 — power balance (independent check). Source delivers $P_s = -20\times1 = -20\text{ W}$ . Resistor absorbs $P_R = v_x i = 4\times1 = 4\text{ W}$ . Dependent source's own drop is $4v_x = 16\text{ V}$ , and it absorbs $P_{dep} = 16\times1 = 16\text{ W}$ (same current flows through every series element, including the dependent source, by KCL). $P_s+P_R+P_{dep} = -20+4+16=0$ ✓, confirming $v_x=4\text{ V}$ without re-using the KVL equation from Method 1.

v_x = 4\text{ V}, \qquad i = 1\text{ A}