Basic Physical Quantities and Laws

A 10 V supply feeds several resistors, yet one node sits at about 4.84 V. Has current been “used up” along the way? Why can two waveforms both labeled 5 V produce different heating? Answering these questions requires separating charge, energy, reference directions and connections. We will establish the quantities, then combine Ohm’s law, KCL, KVL and power accounting in one complete circuit.

From Charge to Circuit Quantities

Charge, electric field and potential difference

Charge is denoted by q or Q and measured in coulombs, C. The elementary charge magnitude e = 1.602176634 × 10⁻¹⁹ C is an exact SI defining constant. An electron carries −e, not +e. A charge magnitude of 1 C corresponds to about 6.24 × 10¹⁸ elementary charges. BIPM: SI defining constants

The electric field is a vector: E = F/q for a positive test charge. Its units are N/C, equivalently V/m. The magnitude of the Coulomb force between two stationary point charges in vacuum is k|q₁q₂|/r², where k is approximately 8.99 × 10⁹ N·m²/C². Doubling the distance reduces the magnitude to one quarter; directions still depend on geometry and charge signs.

Under electrostatic conditions, u_AB = φ_A − φ_B = ∫ from A to B E·dl. Voltage is always between two points. Moving positive charge q from A to B lets the field perform work q·u_AB, reducing potential energy by the same amount. Although 1 V = 1 J/C, an arbitrary movement of 1 C does not necessarily “consume” 1 J: the path direction and voltage matter.

A circuit’s 0 V is a chosen reference node. It need not be Earth and does not mean an absence of charge. Adding 100 V to every node potential leaves all pairwise voltages, and the ideal circuit currents, unchanged. When a chassis, person or another circuit is involved, its actual potential relative to that reference still matters.

Current is a rate; voltage is an energy difference

Instantaneous current is i = dq/dt; average current over an interval is Δq/Δt. Thus 1 A = 1 C/s. A constant 2 mA for 3 s transports 6 mC through a cross-section. Conduction electrons in metal drift opposite to conventional current. Other media may conduct using positive ions, so current should not be defined as electron flow alone.

A heating resistor dissipates electrical energy; it does not progressively consume current. Every cross-section of a steady series branch carries the same current. A region receiving more charge than it releases would accumulate charge and alter the electric field, so that assumed steady state could not persist.

The pressure–flow analogy helps distinguish voltage and current, but circuit response does not wait for one electron to travel from source to load. Signal propagation involves electromagnetic fields and the surrounding medium. Electron drift velocity does not determine switching response time.

Resistance and the scope of Ohm’s law

An ideal linear resistor obeys u = Ri, with positive R independent of current in the chosen model. Conductance is G = 1/R, measured in siemens, S. A uniform conductor of constant cross-section approximately follows R = ρL/A: more length increases resistance, while more cross-sectional area reduces it. Resistivity ρ has units Ω·m and depends on material and temperature.

For a general device, the ratio u/i at one operating point does not establish a linear I–V curve. The differential resistance du/di for a small perturbation may differ from u/i. A diode usually cannot be represented by one fixed R, while a resistor made from semiconductor material may be approximately ohmic over a range. The boundary is the device behavior, not merely its material category.

Reference Directions and Power

Choose voltage polarities and current arrows before writing equations. Under the passive sign convention, reference current enters the positive voltage terminal and instantaneous absorbed power is p = ui. Positive p means absorption; negative p means delivery. The same rule applies to a discharging source, whose absorbed power is negative.

A result of −0.2 A means 0.2 A opposite to the chosen arrow. Keep that reference and its negative value consistently; reversing the arrow while retaining the negative sign changes the interpretation twice. Also, 1 W = 1 J/s and 1 kWh = 3.6 × 10⁶ J. Energy and power are different quantities: battery Wh and device W cannot simply be added.

Equations from Connections

Nodes determine series and parallel connections

Points joined by ideal wire form one node. Two components are in series only if their shared intermediate node has no other current branch; they then carry the same current and their resistances add. Two components are parallel only if both ends connect to the same pair of nodes; they then share voltage and their conductances add.

For two positive resistors in parallel, R_eq = R₁R₂/(R₁ + R₂), less than either resistance. A bridge network may resist repeated series–parallel reduction. Return to node and loop equations instead of treating elements sharing only one node as parallel.

KCL: charge conservation at a node

At an ideal node storing no net charge, total incoming current equals total outgoing current. More generally, a finite region obeys “inflow − outflow = rate of charge accumulation.” Lumped circuits represent storage with elements such as capacitors and their branch currents; KCL does not imply that charge can never accumulate anywhere.

KVL: consistent potential changes around a loop

In an applicable lumped circuit model, the algebraic sum of branch voltages around a closed loop is zero. Traversing a resistor in its current direction gives −IR; traversing an ideal source from its negative to positive terminal gives +V. Completing the loop returns to the starting potential. OpenStax: Kirchhoff’s rules

This does not make every electric field conservative. Changing magnetic flux produces induced emf, with ∮E·dl = −dΦ_B/dt. Inductors, transformers and magnetic coupling require those induced effects in the model. When propagation delay and spatial distribution matter, transmission-line or field models are needed; low wire resistance alone does not establish an equipotential connection.

Complete Example: 2 Ω in Series with 3 Ω ∥ 5 Ω

Use an ideal DC model. After R₁, node A splits into R₂ and R₃, with the bottom wire at 0 V and source top at +10 V. These connections impose I₁ = I₂ + I₃; the three resistors do not carry the same current.

First reduce the parallel pair: R₂₃ = 3×5/(3+5) = 15/8 Ω. Total resistance is 2 + 15/8 = 31/8 Ω, so I₁ = 80/31 A ≈ 2.581 A.

Node A is at U_A = 150/31 V ≈ 4.839 V. The branch currents are I₂ = U_A/3 = 50/31 A and I₃ = U_A/5 = 30/31 A. The lower resistance carries more current.

Alternatively, write the node equation directly: (10 − U_A)/2 = U_A/3 + U_A/5. The left side is incoming current through R₁; the right side contains both outgoing branches. Both methods must give the same U_A.

Power accounting catches sign errors that arithmetic alone may hide. The three resistors absorb 12800/961, 7500/961 and 4500/961 W, totaling 800/31 W. The source delivers 10×80/31 = 800/31 W. Under the passive convention, source absorption is negative and all element powers sum to zero. This is an arithmetic teaching circuit, not a construction plan specifying real resistor power ratings.

from fractions import Fraction as F
r1, r2, r3, vs = F(2), F(3), F(5), F(10)
r_parallel = 1 / (1/r2 + 1/r3)
i1 = vs / (r1 + r_parallel)
va = vs - i1*r1
i2, i3 = va/r2, va/r3
assert (i1, i2, i3, va) == (F(80,31), F(50,31), F(30,31), F(150,31))
assert i1 - i2 - i3 == 0
assert vs - i1*r1 - i2*r2 == 0
assert vs - i1*r1 - i3*r3 == 0
assert i1*i1*r1 + i2*i2*r2 + i3*i3*r3 == vs*i1
print("Node, both loops and power balance pass exact rational checks")

Where does current divide?

Increase R3: its current falls, but node A rises and changes R2 current too. The branches do not each see an independently fixed source voltage. Check both KCL and power.

Waveforms and RMS

Ideal constant DC does not vary with time, though real supplies may carry ripple. A periodic signal has a mean, amplitude, period and shape. “5 V signal” alone cannot determine power or interface compatibility.

RMS is not a universal “amplitude × 0.707” conversion. Its definition is V_rms = √[(1/T)∫u²(t)dt]. For a fixed ideal resistor, average power is V_rms²/R, so RMS matches heating, not average voltage. Only a zero-offset sinusoid A sin(ωt+φ) has V_rms = A/√2. OpenStax: simple AC circuits

For ideal pulses between 0 and A with high-state duty D, the mean is AD and RMS is A√D. At A = 5 V and D = 25%, the mean is 1.25 V but RMS is 2.5 V. Average power in 100 Ω is 62.5 mW, not the 15.625 mW obtained by squaring the mean voltage.

For a sinusoid with DC offset, u = U₀ + A sin(ωt), V_rms² = U₀² + A²/2. A general AC load requires averaging u(t)i(t). The product of voltage RMS and current RMS is not necessarily active power: phase and waveform correlation also matter.

Why square before averaging?

Both plots share a time scale. Shorter pulses reduce area, not height. Power follows the lower area: at D=25%, RMS is half the peak, not a quarter. The two power calculations coincide at D=0 or 1.

$$U_{rms}=A\sqrt{D},\overline{P}=\frac{A^{2}D}{R}$$

Units and Next Steps

Case matters: MΩ and mΩ differ by 10⁹. Use consistent units inside equations, then convert the result. If the order of magnitude seems wrong, check prefixes, directions and connections first.

Continue with DC circuit analysis to apply the same conservation laws to node analysis, superposition and terminal equivalents. For tolerance, temperature drift and power limits, see Resistors.