ECE 211 · Lecture 1: Voltage, Current, Power & the Sign Convention

§1 · Voltage

Voltage is energy per unit charge

The voltage $V_{ab}$ between two points $a$ and $b$ of a circuit is the energy required to move a unit of positive charge from $a$ to $b$ :

V_{ab} = \frac{dw}{dq}, \qquad 1\ \text{V} = \frac{1\ \text{J}}{1\ \text{C}} = 1\ \frac{\text{N}\cdot\text{m}}{\text{C}}

where $w$ is energy in joules and $q$ is charge in coulombs. A voltage is therefore not a quantity of energy; it is a quantity of energy carried per unit of charge, which is the reason a small battery and a large one may share the same voltage.

The order of the subscripts is a reference, not a property

A voltage is always a difference taken between two points, and the order of the subscripts records which point the difference is taken from: $V_{ab} = V_a - V_b$ . Reversing the order negates the value, so the two readings below describe one element and one physical state.

The same 9 V element, read in the two possible orders. The polarity marks record the choice of reference; they are not a property of the element.

\tm{v.ta}{V_{ab}} = V_a - V_b = +9\ \text{V}

Terminal $a$ is nine joules per coulomb above terminal $b$ . A unit of positive charge carried from $b$ to $a$ through the element gains 9 J.

\tm{v.tb}{V_{ba}} = V_b - V_a = -9\ \text{V}

The same element, read in the opposite order. The negative sign carries no additional physical content: it reports only that the order chosen for the reading opposes the polarity of the element.

Two labels, one element

The polarity marks are a reference, chosen before the answer is known. They are not a claim about the element, and they cannot be chosen wrongly; they fix what the sign of the reported number will mean. This point returns, in a stronger form, in §4.

§2 · The reference node

One subscript means "with respect to the reference"

A voltage written with a single subscript abbreviates a difference whose second point is the reference node of the circuit: $V_a$ means $V_{a0}$ , where the node marked $0$ carries the ground symbol and is assigned the value zero by definition.

The reference node is chosen for convenience, and the choice is free. Moving it shifts every node voltage by one common constant and therefore changes no difference between two nodes. A node voltage is consequently not a property of the node alone, and no answer to a circuit problem depends on where the ground symbol has been drawn.

The same chain, with the reference node placed at each of the three nodes in turn. The node voltages change; the voltages across the elements do not.

With the reference at C, the source holds A at $+12$ V, and the 1 A loop current places B at $+8$ V. This is the conventional choice, because it makes every node voltage in the chain positive.

With the reference at B, the node voltages become $+4$ V and $-8$ V. Negative node voltages are not an error and not a sign of trouble; the reference has simply been placed in the middle of the chain.

With the reference at A, every node voltage in the chain is negative. All three readings describe one circuit, and all three give the same current, the same element voltages, and the same power.

The three quantities that a measurement could return are the differences, and these are what the tabs leave untouched:

V_AB = 4 VV_BC = 8 VV_AC = 12 V

The ground symbol is a label, not a drain

Current does not disappear into the reference node, and a node does not become special by being chosen as the reference. Two ground symbols drawn at different places on one diagram denote the same single node, connected by a wire that has been omitted for clarity.

§3 · Current

Current is the rate at which charge passes

Charge is the electrical property of atomic particles, measured in coulombs. Current is the rate at which charge passes a point in the circuit:

i = \frac{dq}{dt}, \qquad 1\ \text{A} = \frac{1\ \text{C}}{1\ \text{s}}

A direction and a sign state one fact twice

A current label consists of an arrow and a number, and the two are redundant: reversing the arrow and negating the number leaves the physical statement unchanged. The two labels below therefore describe one conductor carrying one current.

One conductor, labeled in both directions. Every arrow in this course, and in every circuit diagram, marks the motion of positive charge.

4 A to the rightis the same current as−4 A to the left

The redundancy is the same one met in §1, where reversing the order of the subscripts negated the voltage. In both cases a reference has been chosen, and the sign of the number reports how that choice stands against the physics.

§4 · Power and the passive sign convention

Power is the rate of absorbing or expending energy

Power is the time rate at which energy is transferred, measured in watts. The chain rule expresses it as a product of the two quantities already defined:

p = \frac{dw}{dt} = \underbrace{\frac{dw}{dq}}_{\textstyle v} \cdot \underbrace{\frac{dq}{dt}}_{\textstyle i} = v \cdot i

The derivation also answers the question of which $v$ and which $i$ belong in the product. The factor $dw/dq$ is the energy given up by a unit of positive charge in passing from the $+$ terminal to the $-$ terminal, and $dq/dt$ is the charge per second that passes in exactly that direction. Their product is therefore the energy per second deposited in the element, and it is positive when the element absorbs. This is the entire content of the convention: it is the one pairing of the two references for which $p = vi$ requires no minus sign.

The passive sign convention

The reference current arrow is drawn into the terminal marked $+$ . Under this pairing,

p = +v \cdot i

is the power absorbed by the element: positive means absorbed, and negative means supplied.

One physical situation, four labelings

The circuit below is fixed: 2 A flow downward through the 3 Ω resistor, whose upper terminal is 6 V above its lower terminal, and the resistor absorbs 12 W. Only the two reference labels are under the reader's control. Flipping both leaves the convention and the product unchanged; flipping one reverses the convention and reverses the sign of the product, and the two reversals cancel.

The resistor carries the labels; the source and the physical current are fixed. The arrow and the polarity marks are each flipped by the buttons below.

passive labeling

v = +6.00 V i = +2.00 A p_abs = +v · i = 12.00 W

The absorbed power reads $+12$ W in all four cases, because all four describe the same circuit in the same state. The reading to take from this is that a reference direction cannot be assumed incorrectly. An assumption that opposes the physics is reported by a negative value, and the physical conclusion is unchanged. What must not vary is the relationship between the two labels, because that relationship determines which formula computes absorbed power.

The reverse convention, and why it is not mixed in

Labeling the arrow out of the $+$ terminal is the active sign convention, and it is equally self-consistent: under that pairing, $p = v \cdot i$ is the power supplied by the element, and the absorbed power is $-v \cdot i$ . Sources are occasionally drawn this way, because a source usually supplies and a positive number then reads more naturally.

The two conventions may not be mixed within one power balance. If some elements are accounted as absorbers and others as suppliers, the terms of the sum no longer measure the same quantity, and the sum is meaningless. For this reason one convention is fixed for the whole course, and it is the passive one.

Conservation of power

Energy is neither created nor destroyed inside a circuit, so the power absorbed by every element, summed over the circuit, is zero:

\sum_k p_k = 0

Under one convention this is a genuine arithmetic check, and it is available on every problem in the course. Each supplying element contributes a negative term, each absorbing element a positive term, and the two totals must agree in magnitude. The check is applied to a worked circuit in §9.

§5 · The ideal sources

Passive and active elements

A passive element cannot generate energy; the resistor is the example, and later the capacitor and the inductor. An active element can generate energy, and the sources are the examples. Sources are classified in two ways at once: by the variable they fix (voltage or current), and by whether that value stands alone (independent) or is set by a quantity elsewhere in the circuit (dependent).

An ideal source fixes one variable and surrenders the other

An ideal voltage source holds its terminal voltage at the stated value whatever current is drawn from it, and it therefore exercises no control over that current. An ideal current source is its dual: it forces the stated current whatever voltage appears across it, and it exercises no control over that voltage. In each case the missing variable is determined not by the source but by whatever is connected to it.

The 10 V source fixes $v$ and leaves $I$ to the load; the 2 A source fixes $i$ and leaves $V$ to the load.

Each ideal source is therefore described by one straight line in the plane of its terminal current against its terminal voltage: a vertical line for a voltage source, which fixes $v$ at every $i$ , and a horizontal line for a current source, which fixes $i$ at every $v$ . That line is the complete external description of the element, and it is the object compared when two networks are declared equivalent in Lecture 2 and replaced by one another in Lecture 4.

The description also fixes what may not be connected. Because an ideal source insists on its own variable and concedes nothing, any second element that insists on a different value of the same variable produces a circuit with no solution. The four configurations in which this occurs are collected in Lecture 2.

§6 · Dependent sources

The value is set by a quantity elsewhere in the circuit

A dependent source is drawn as a diamond, and its value is a stated multiple of a voltage or a current measured somewhere else in the same circuit. Two choices are therefore made independently: what the source delivers, and what controls it. The four combinations are named by reading the controlling quantity first and the delivered quantity second.

The four dependent sources. The controlling quantity, written $V_x$ or $I_x$ , is a voltage or current elsewhere in the circuit, and is generally an unknown of the problem.

Source	Delivers	Controlled by	Gain K
VCVS	v = K · V_x	a voltage	V/V (none)
CCVS	v = K · I_x	a current	V/A (Ω)
CCCS	i = K · I_x	a current	A/A (none)
VCCS	i = K · V_x	a voltage	A/V (S)

The gain of a dependent source therefore carries units unless the delivered and controlling quantities are of the same kind. A CCVS has a gain in ohms and a VCCS a gain in siemens, and both are frequently mistaken for resistances of the branch they occupy, which they are not.

Why these elements exist

Dependent sources model devices whose output responds to an input elsewhere: transistors, operational amplifiers, and integrated circuits generally. The small-signal model of a field-effect transistor, for example, represents the drain current as $i_d = g_m v_{gs}$ , a current controlled by the gate-to-source voltage, which is a VCCS with gain $g_m$ in siemens. Nothing further about the device is required in order to analyze the circuit; the element law is enough.

A dependent source is not a known value

The controlling quantity is usually one of the unknowns being solved for, so a dependent source contributes an equation rather than a number. It cannot be evaluated before the circuit is solved, and it must not be treated as an independent source of the same symbol. The systematic handling of dependent sources is developed in Lecture 2.

§7 · The resistor and Ohm's law

Resistance opposes the passage of current

Resistance is measured in ohms and takes any value in the range $0 \le R \le \infty$ , the two limiting cases being the short circuit and the open circuit. Ohm's law relates the voltage across a resistor to the current through it, and is stated here under the passive sign convention, with the positive mark placed where the labeled current enters:

\tm{o.r}{v} = \tm{o.i}{i} \cdot \tm{o.r}{R}, \qquad R = \frac{v}{i}, \qquad 1\ \Omega = 1\ \frac{\text{V}}{\text{A}}

The resistor labeled under the passive sign convention. Had the arrow been drawn the other way, the law would read $v = -iR$ ; the physics would be unchanged.

The two limits are a dual pair

Short: R = 0 fixes v = 0, and leaves i to the rest of the circuitOpen: R = ∞ fixes i = 0, and leaves v to the rest of the circuit

Each limit fixes one variable at zero and determines nothing about the other. The unknown in each case is set by the network behind the terminals.

A resistor always absorbs

Combining $p = vi$ with $v = iR$ gives two further forms, both of which are used constantly:

p = v \cdot i = \frac{v^2}{R} = i^2 R

Because $R > 0$ and both remaining forms are squares, the power absorbed by a resistor is never negative. This is the precise sense in which the resistor is a passive element: it converts electrical energy to heat, and it returns none of it.

§8 · Kirchhoff's voltage law

The voltages around any closed loop sum to zero

\sum_{\text{loop}} v = 0

The law follows from the definition of voltage. A voltage is a difference of node potentials, and a loop returns to the node it started from, so the differences accumulated around it must cancel. The bookkeeping rule that turns this statement into an equation is the sign convention for traversal: the sign written for an element is the sign of the terminal at which the loop enters it.

Four elements in one loop. The direction of traversal is selected by the tabs; the polarity marks belong to the elements and do not move.

Traversing clockwise from the lower-left corner, the loop enters $V_1$ at its $-$ mark, $V_2$ and $V_3$ at their $+$ marks, and $V_4$ at its $-$ mark:

\tm{k.v1}{-V_1} \tm{k.v2}{\;+\; V_2} \tm{k.v3}{\;+\; V_3} \tm{k.v4}{\;-\; V_4} = 0

Traversing counterclockwise, every element is entered at the opposite terminal, so every term changes sign:

\tm{k.v1}{+V_1} \tm{k.v2}{\;-\; V_2} \tm{k.v3}{\;-\; V_3} \tm{k.v4}{\;+\; V_4} = 0

This is the clockwise equation multiplied by $-1$ . The two are one equation, and the direction of traversal is therefore free.

Starting the same clockwise traversal at a different node rotates the terms of the sum without changing any of them, and addition is commutative. Neither the starting node nor the direction affects the equation, so both may be chosen for convenience.

What the law does not depend on

The polarity marks on the elements are references, assigned once when the circuit is labeled. The traversal direction is a separate choice, made when the equation is written. Confusing the two is the most common source of sign errors in KVL, and keeping them separate removes it: mark the polarities first, then walk the loop and read the marks.

§9 · Worked example: a power balance

Find the power absorbed or delivered by each element

Two voltage sources and one resistor form a single loop. The circuit is solved for its one current, and the power of each element is then evaluated under the passive sign convention.

The problem

Given a 10 V source, a 10 Ω resistor, and a 5 V source in one loop, determine the power absorbed or delivered by each of the three elements. The ground symbol fixes the reference node and is not itself an element.

Step 1: Assign a reference current and mark the polarities

One loop carries one current, so a single symbol $I$ suffices, and its direction is assumed clockwise. Under the passive sign convention, the positive mark on the resistor is placed where the assumed current enters it. Both steps are performed before any equation is written.

Step 2: Apply KVL around the loop

Traversing clockwise from the lower-left corner, the loop enters the 10 V source at its $-$ terminal, the resistor at its $+$ mark, and the 5 V source at its $+$ terminal:

\tm{e.vs10}{-10} + \tm{e.r10 e.ic}{10\,I} + \tm{e.vs5}{5} = 0

Step 3: Solve

10\,I = 5 \;\Longrightarrow\; \boxed{I = \tfrac{1}{2}\ \text{A}}

The result is positive, so the assumed clockwise direction agrees with the physics. The same value is obtained directly from Ohm's law, since the resistor carries the difference of the two sources: $I = (10 - 5)/10 = \tfrac{1}{2}$ A.

Step 4: Evaluate the power of each element, and check

Each element is taken in turn. The current leaves the 10 V source through its $+$ terminal, which is the active orientation, so its absorbed power is $-vi$ . The current enters the 5 V source at its $+$ terminal, which is the passive orientation, so its absorbed power is $+vi$ . Hovering over a row highlights the corresponding element:

Element	p = v · i	Power
10 V source	−(10 V)(½ A)	−5 W
10 Ω	(½ A)² · 10 Ω	+2.5 W
5 V source	+(5 V)(½ A)	+2.5 W
Total		0 W ✓

The 5 V source absorbs power. A source is not required to supply, and here it is being charged by the larger source, exactly as a battery is charged. The sum is zero, which confirms the solution.

§10 · Your turn

Problem 1: reading a labeled element

The element $E$ below is part of a larger circuit, and its two references have already been assigned. Determine whether the labeling is passive or active, and find the power absorbed by $E$ . The problem should be attempted on paper before the hints are opened, in order.

The element $E$ with $V = 12$ V and $I = 3$ A as labeled. Only the labels are given; nothing else about $E$ is known or needed.

Hint 1: Read the two labels against each other

The question is not what $E$ is, but where the arrow points relative to the $+$ mark. Follow the arrow to the terminal it enters, and compare that terminal with the terminal carrying the $+$ mark.

Hint 2: Choose the formula the labeling requires

If the arrow enters the $+$ terminal, the absorbed power is $+vi$ . If it enters the $-$ terminal, the labeling is active and the absorbed power is $-vi$ .

Solution

The arrow enters the right-hand terminal, which carries the $-$ mark, so the labeling is active:

p_{\text{abs}} = -v \cdot i = -(12\ \text{V})(3\ \text{A}) = \boxed{-36\ \text{W}}

The element delivers 36 W to the rest of the circuit, so $E$ is a source, or an element being discharged.

Had the arrow been drawn in the opposite direction, the labeling would be passive, and the measured current would be $i = -3$ A. The absorbed power would then be $p_{\text{abs}} = +v \cdot i = (12)(-3) = -36$ W, which is the same answer. ✓ The labeling governs the formula, and the pair of labels governs the value; the physical conclusion is independent of both.

Problem 2: a negative current, and the balance that follows

The two sources in the loop below oppose each other. Find the current $I$ in the assumed direction, and then the power absorbed or delivered by each of the four elements.

The 5 V source and the 20 V source both carry their $+$ marks at the top, so around the loop they oppose. The current $I$ is assumed clockwise.

Hint 1: Mark the polarities first

Under the passive sign convention, the positive mark on each resistor is placed where the assumed current $I$ enters it: at the left end of the 3 Ω resistor, and at the right end of the 2 Ω resistor, since the assumed current runs leftward along the bottom.

Hint 2: Walk the loop clockwise

Starting from the lower-left corner, the loop enters the 5 V source at its $-$ terminal and the 20 V source at its $+$ terminal. Write the four terms in order and set the sum to zero.

Hint 3: Expect a negative answer

The 20 V source opposes the 5 V source and is the larger of the two, so the current cannot in fact run clockwise. The assumption is not thereby invalidated; it is reported by the sign of the result.

Solution

KVL clockwise from the lower-left corner:

\tm{q2.vs5}{-5} + \tm{q2.r3 q2.ip}{3I} + \tm{q2.vs20}{20} + \tm{q2.r2 q2.ip}{2I} = 0 \;\Longrightarrow\; 5I = -15 \;\Longrightarrow\; \boxed{I = -3\ \text{A}}

The current is 3 A counterclockwise. Every power is now evaluated with $I = -3$ A retained as a signed quantity, which keeps the sign convention doing the work:

Element	p_abs	Power	Interpretation
20 V source	+(20)(−3)	−60 W	supplies 60 W
5 V source	−(5)(−3)	+15 W	absorbs, being charged
3 Ω	(−3)² · 3	+27 W	absorbs
2 Ω	(−3)² · 2	+18 W	absorbs
Total		0 W ✓

The 20 V source supplies exactly what the other three elements absorb, and the balance confirms the negative current. Note that the 5 V source appears with $-vi$ because the assumed current enters it at the $-$ terminal, while the 20 V source appears with $+vi$ because the assumed current enters it at the $+$ terminal.