123-ME · Charge, Current, Voltage, Power & Energy

Charge, Current, Voltage, Power & Energy

Slides 1.2–1.3 · the four quantities everything else in this course is built from

This guide has eleven chapters, and every one of them is downstream of this one. Circuits are built in layers: this chapter gives you the four raw quantities — charge, current, voltage, power — and the sign convention that keeps their bookkeeping honest. Chapter 2 adds Kirchhoff's two laws, which are just charge conservation and energy conservation wearing circuit clothes. Chapters 3–4 use those laws to simplify whole networks (series/parallel, dividers, node/mesh analysis, Thévenin/Norton). Chapters 5–6 add two new elements — capacitors and inductors — whose defining feature is that they store energy rather than dissipate it, which forces the idea of a circuit changing over time. Chapters 7–9 take that same energy-storage idea and push it to sinusoidal steady state: AC, phasors, impedance, complex power. Chapter 10 closes with op-amps, an active element built from all of the above. Nothing later works if $i = dq/dt$ and $p = vi$ aren't reflexive by the time you get there — budget real time here even though it looks like the "easy" chapter. It is the one chapter every other chapter cites.

Notation, fixed from here on

Lowercase italics (

v

i

q

) for anything that can vary with time; uppercase (

V

I

) for a constant (DC) value. Double-subscript voltage

v_{ab}

means "potential at

a

relative to

b

," and flipping the subscripts flips the sign:

v_{ba} = -v_{ab}

. This is Buchanan's Spring 2026 convention and it is used consistently for the rest of the guide.

1. Charge and current: current is a rate

Charge $q$ is the fundamental quantity — every electron carries a fixed charge of $-1.602\times10^{-19}$ C, and everything electrical ultimately reduces to how much charge moved, and how fast. Current is not a separate substance flowing through a wire — it is a bookkeeping rate: how much charge crosses a fixed cross-section of the conductor per unit time.

Current

i

is defined at a cross-section, not "in" the wire generally — it's how much charge

q

passes through area

A

per second.

That definition, written as a derivative, and its inverse (recovering charge from a known current by integrating):

Current & charge

i(t) = \frac{dq(t)}{dt} \qquad\qquad q(t) = \int_{t_0}^{t_1} i(t)\,dt + q(t_0)

1 ampere $=$ 1 coulomb/second, by definition. The integral form is just the derivative form solved for $q$ — the $+\,q(t_0)$ is not optional bookkeeping, it's the charge that was already there before you started the clock.

Why a derivative and not, say, "charge over time"? Because $i(t)$ can change instant to instant — a $\frac{\Delta q}{\Delta t}$ average would blur out exactly the transient behaviour (charging curves, AC waveforms) that later chapters care about. The derivative is the instantaneous rate, the same idea as instantaneous velocity being $dx/dt$ rather than $\Delta x/\Delta t$ .

Worked example — recovering current from a charge curve

A capacitor's stored charge is measured as $q(t) = 0$ for $t < 0$ and $q(t) = 2 - 2e^{-100t}$ C for $t \geq 0$ . Find $i(t)$ , and describe its behaviour.

Differentiate: $i(t) = \dfrac{dq}{dt} = \dfrac{d}{dt}\left(2 - 2e^{-100t}\right) = 200\,e^{-100t}$ A

At $t=0$ : $i(0) = 200$ A (largest, because $q(t)$ is rising fastest right at $t=0$ ).

As $t\to\infty$ : $i \to 0$ (the charge curve flattens out at $q=2$ C, so the rate of charge arrival dies off).

i(t) = 200e^{-100t}

A for

t\geq 0

— a current spike that decays away as the capacitor finishes charging. This exact shape reappears in Chapter 5 as an RC charging transient; file it away.

2. Voltage: potential difference, and why it needs two subscripts

Voltage is potential energy per unit charge — the work done moving one coulomb of charge from one point to another. Because it's always a difference between two points, a single voltage number is meaningless without saying between which two terminals, and in which order. That's what the double subscript is for: $v_{ab}$ is the potential at $a$ relative to $b$ ; swap the order and you flip the sign, $v_{ba} = -v_{ab}$ .

Reference polarity (which terminal you called

+

) is assumed, not measured. Solve the circuit; if

v_{ab}

comes out negative, terminal

b

is actually the higher-potential one — the arithmetic tells you, you don't have to guess up front.

An ideal voltage source holds $v_{ab}$ fixed regardless of what's connected to it (zero internal resistance); an ideal current source holds the current through it fixed regardless of the voltage across it. Both are idealizations you'll refine in Chapter 3 when real (non-ideal) sources with internal resistance show up.

3. Power: where $p = vi$ actually comes from

Power is not a new physical primitive — it falls straight out of the definitions above. Voltage is energy per unit charge ( $v = dw/dq$ ); current is charge per unit time ( $i = dq/dt$ ). Multiply them and the charge cancels, leaving energy per unit time, which is exactly what power means:

Deriving

p = vi

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

Units check the same way: $\dfrac{\text{J}}{\text{C}} \times \dfrac{\text{C}}{\text{s}} = \dfrac{\text{J}}{\text{s}} = \text{W}$ . This is the whole reason volts $\times$ amps gives watts — it isn't a coincidence of unit naming, it's the coulombs cancelling.

Energy is then just power integrated back up over time — the inverse relationship, exactly parallel to the $i \leftrightarrow q$ pair above:

Energy from power

w = \int_{t_0}^{t_1} p(t)\,dt \qquad \text{(joules)}

3.1 The passive sign convention — the one rule that decides absorbing vs. delivering

$p=vi$ only tells you a number; whether that number means "this element is absorbing energy" or "supplying it" depends entirely on how the reference arrow for $i$ sits relative to the reference $+$ terminal for $v$ . This is the single most exam-tested idea in the whole chapter, and it's a two-case rule:

Passive configuration — the "normal" case for a resistor.

Active configuration — the "normal" case for a source doing its job.

Passive sign convention

\text{current enters }+\text{ terminal:}\quad p = vi \;(>0 \Rightarrow \text{absorbed})

\text{current exits }+\text{ terminal:}\quad p = -vi \;(>0 \Rightarrow \text{delivered})

Same physical situation, two bookkeeping conventions: a battery being charged has current forced into its $+$ terminal (passive, absorbing — it's storing chemical energy). The same battery discharging naturally has current flowing out of its $+$ terminal (active, delivering).

This is arrow bookkeeping, not physics

The circuit doesn't know or care which convention you used — the sign of

p

automatically comes out right either way, as long as you're consistent about which formula matches which arrow direction. Get the arrow direction wrong relative to the formula and you'll flip every subsequent sign in a bigger circuit. When in doubt, redraw the reference arrows before writing

p=vi

p=-vi

— don't try to keep it in your head.

Worked example — three elements, three verdicts

Classify each as absorbing or delivering power, and find the magnitude.

Element A: $v_a = 12$ V, $i_a = 2$ A, passive configuration (current into $+$ ).

$p_a = v_a i_a = 12 \times 2 = 24$ W, absorbed.

Element B: $v_b = 12$ V, $i_b = 1$ A, also passive.

$p_b = 12 \times 1 = 12$ W, absorbed.

Element C: $v_c = 12$ V, $i_c = -3$ A, active configuration.

$p_c = -v_c i_c = -(12)(-3) = 36$ W $> 0$ , so — watch the double negative — it comes out positive in the active formula, meaning $36$ W is genuinely delivered.

If you'd (incorrectly) used the passive formula here,

p = vi = 12\times(-3) = -36

W, which you'd read as "

-36

W absorbed" — same physical conclusion (36 W delivered) but only by accident of a double sign flip. Match the formula to the arrow, every time.

A absorbs 24 W, B absorbs 12 W, C delivers 36 W.

Quick check

3.2 Energy with time-varying $v$ and $i$

When voltage and current are constant, $w = p\,t$ . When they vary with time, you have to integrate $p(t) = v(t)i(t)$ — this is exactly where the $i=dq/dt$ derivative-and-integral pattern from §1 reappears for power and energy.

Worked example — energy from a decaying current

A device is held at a constant $v(t) = 12$ V while its current decays as $i(t) = 2e^{-t}$ A (passive configuration). Find the total energy absorbed over $t \in [0,\infty)$ .

$p(t) = v(t)i(t) = 12 \times 2e^{-t} = 24e^{-t}$ W

$w = \displaystyle\int_0^\infty 24e^{-t}\,dt = 24\left[-e^{-t}\right]_0^\infty = 24(0-(-1)) = 24$ J

w = 24

J absorbed, total, even though the process technically never finishes — the tail of the exponential contributes a vanishing amount, so the integral still converges to a finite number.

4. Build the meter: power, energy, and a real electricity bill

Every "electricity bill" problem in the exercises (P1.19, P1.26, Tutorial 1 P4/P5) is the same three-step chain: pick $v$ and $i$ , get instantaneous power $p=vi$ , then integrate power over time to get energy, then multiply by a utility's $/\text{kWh}$ rate to get cost. The widget below makes that chain live: drag $V$ and $I$ and watch the meter accumulate energy and cost in real time (sped up — 1 real second below stands for the "elapsed time" slider's unit).

Power & energy meter

Voltage

V

= 120 V Current

I

= 5.0 A Utility rate = 0.15 $/kWh Run time = 1.0 h

instantaneous power

p = VI

: —

energy over run time: —

energy in kWh: —

cost at this rate: —

equivalent constant current at 120 V: —

Bar height tracks cumulative energy

w = p \times t_{\text{elapsed}}

as the run plays out; the final height matches the "energy over run time" readout.

Quick check

5. Practice problems

P1.1 — Recovering charge from a decaying currentcharge/current

A device has $i(t) = 4e^{-20t}$ A for $t \geq 0$ , and an initial stored charge of $q(0) = 1$ C.

(a) Find $q(t)$ for $t \geq 0$ . (b) What is the total charge that has ever passed into the device, as $t \to \infty$ ? (c) Differentiate your answer to (a) as a check — does it return the given $i(t)$ ?

$q(t) = \displaystyle\int_0^t 4e^{-20\tau}\,d\tau + q(0) = \left[-\tfrac{4}{20}e^{-20\tau}\right]_0^t + 1 = -0.2e^{-20t} + 0.2 + 1$

$q(t) = 1.2 - 0.2e^{-20t}$ C

As $t\to\infty$ , $e^{-20t}\to 0$ , so $q \to 1.2$ C.

$\dfrac{dq}{dt} = \dfrac{d}{dt}\left(1.2 - 0.2e^{-20t}\right) = -0.2(-20)e^{-20t} = 4e^{-20t}$ A — matches the given $i(t)$ . ✓

This differentiate-back check is the fastest way to catch an arithmetic slip in the integration — it takes ten seconds and directly re-derives the problem statement.

(a)

q(t) = 1.2 - 0.2e^{-20t}

C. (b)

q(\infty) = 1.2

C. (c) Verified — the derivative reproduces

i(t)

P1.2 — Solar panel and load: power balance and daily savingspassive sign convention

A small solar panel feeds a resistive load. The panel's terminals read $v_p = 18$ V, and its reference current $i_p = 2.5$ A is defined as leaving the panel's $+$ terminal. The load sees the same current entering its own $+$ terminal, with $v_R = 18$ V across it.

(a) Is the panel absorbing or delivering power, and how much? (b) Is the load absorbing or delivering, and how much? (c) Verify energy conservation for the pair. (d) If the panel runs like this for 5 hours a day, find the daily energy delivered in kWh, and its dollar value at a utility rate of $0.15/kWh.

Current leaving the $+$ terminal is the active configuration:

$p_p = -v_p i_p = -(18)(2.5) = -45$ W. Negative in the active formula means power is delivered: the panel supplies 45 W.

Current entering the $+$ terminal is the passive configuration:

$p_R = v_R i_R = (18)(2.5) = 45$ W, absorbed by the load.

$p_p + p_R = -45 + 45 = 0$ W. ✓ — everything the panel supplies, the load absorbs, exactly, as it must in a two-element loop.

This is the same check used in the slides' worked series-circuit example (Ch. 1.7): sum of all element powers, with correct signs, is always zero.

Power delivered is constant at 45 W, so energy over 5 h:

$w = p \times t = 45\ \text{W} \times 5\ \text{h} = 225\ \text{Wh} = 0.225\ \text{kWh}$

Value at $0.15/kWh: $0.225 \times 0.15 = \mathdollar 0.0338$ — about 3.4¢ per day.

Cross-check the unit conversion:

225\ \text{Wh} \div 1000 = 0.225\ \text{kWh}

, matching the direct kWh statement above.

(a) Panel delivers 45 W. (b) Load absorbs 45 W. (c) Sum

=0

, balanced. (d)

0.225

kWh/day, worth

\approx \mathdollar 0.034

/day at

\mathdollar 0.15

/kWh.

P1.3 — A day of piecewise power: energy and cost from a scheduleenergy & cost

A window air conditioner runs on this daily schedule: off from midnight to 8 a.m. (0 W); 900 W from 8 a.m. to 5 p.m.; 1500 W from 5 p.m. to 10 p.m.; and a 300 W fan-only mode from 10 p.m. to midnight.

(a) Total energy consumed in one day, in kWh. (b) Daily cost at $0.14/kWh. (c) Average power over the full 24 hours — compute it two independent ways and confirm they agree.

Segment energies: $8\text{h}\times 0\text{W} = 0$ ; $9\text{h}\times 900\text{W} = 8100$ Wh; $5\text{h}\times1500\text{W} = 7500$ Wh; $2\text{h}\times300\text{W} = 600$ Wh.

Total: $0 + 8100 + 7500 + 600 = 16{,}200$ Wh $= 16.2$ kWh.

Cost $= 16.2\ \text{kWh} \times \mathdollar 0.14/\text{kWh} = \mathdollar 2.268 \approx \mathdollar 2.27$ .

Method 1 (energy over time): $\bar{p} = w/t = 16{,}200\ \text{Wh} / 24\ \text{h} = 675$ W.

Method 2 (duration-weighted average of the four power levels):

$\bar p = \dfrac{8(0)+9(900)+5(1500)+2(300)}{24} = \dfrac{16{,}200}{24} = 675$ W. ✓ Both methods agree, as they must — they're the same sum, grouped differently.

(a)

16.2

kWh. (b)

\approx \mathdollar 2.27

. (c)

\bar p = 675

W, confirmed two ways.

Charge, current, voltage, power, energy, and the passive sign convention are the alphabet; the next chapter gives you the grammar — Kirchhoff's current and voltage laws, which let you write down equations for a whole network instead of one element at a time.