Electrical Fundamentals and DC Circuits

FE Exam Weight: Basic Electrical Engineering is 6–9 questions (~7%) on the FE Other Disciplines exam. Nearly every question reduces to Ohm's law, a series/parallel combination, or Kirchhoff's laws. All formulas are in the searchable NCEES FE Reference Handbook, so practice locating and applying them with correct units.

Fundamental Quantities

Charge Q (coulombs) is the basic electrical quantity; one electron carries −1.602×10⁻¹⁹ C. Current I is the rate of charge flow, I = dQ/dt, measured in amperes (1 A = 1 C/s). By convention, current flows in the direction positive charge would move — from + to − in the external circuit — even though electrons physically drift the other way. Voltage V (volts) is the work done per unit charge moving between two points, V = W/Q; it is always a difference between two points (a potential drop or rise).

Ohm's Law and Power

Ohm's law governs any linear (ohmic) resistor: V = IR. Solve for whatever is unknown: I = V/R, R = V/I. Power dissipated as heat in a resistor has three equivalent forms — pick the one matching your known quantities: $P = VI = I^2 R = \frac{V^2}{R}$ Energy is power × time, E = Pt; utilities bill in kilowatt-hours, where 1 kWh = 1,000 W × 3,600 s = 3.6×10⁶ J. A common trap is mixing forms: if you know current and resistance, use I²R; if you know voltage and resistance, use V²/R — don't compute I from V and R only to re-multiply and risk a rounding/unit slip.

Series and Parallel Resistance

Series elements carry the same current; their resistances add: $R_{eq} = R_1 + R_2 + R_3 + \cdots$ The supply voltage divides among them in proportion to resistance. Parallel elements share the same voltage; their resistances combine reciprocally: $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots, \qquad R_{eq}(\text{two}) = \frac{R_1 R_2}{R_1 + R_2}$ The parallel equivalent is always smaller than the smallest branch — a quick sanity check that catches arithmetic errors.

The voltage divider (series) and current divider (parallel): $V_x = V_{total}\frac{R_x}{R_{total}}, \qquad I_1 = I_{total}\frac{R_2}{R_1 + R_2}$ Note the current divider is "flipped": more current flows through the smaller resistor, so the numerator is the other branch's resistance.

Worked combination example. A 12 V source drives a 4 Ω resistor in series with a parallel pair (6 Ω ∥ 12 Ω). The parallel pair: R_p = (6×12)/(6+12) = 72/18 = 4 Ω. Total: R_eq = 4 + 4 = 8 Ω. Source current I = V/R_eq = 12/8 = 1.5 A. Voltage across the parallel pair = I·R_p = 1.5 × 4 = 6 V; current through the 6 Ω branch = 6/6 = 1 A, through 12 Ω = 6/12 = 0.5 A (they sum to 1.5 A ✓). Power from the source = VI = 12 × 1.5 = 18 W.

Kirchhoff's Laws

KCL (current law): at any node, ΣI_in = ΣI_out — charge is conserved, nothing accumulates at a junction. KVL (voltage law): around any closed loop, the algebraic sum of voltage rises and drops is zero — energy is conserved. These two laws, plus Ohm's law, solve any resistive network. Two systematic methods:

In the node-voltage method you pick a ground reference, label the other node voltages, and write KCL (currents expressed via Ohm's law) at each. In the mesh-current method you assign a circulating current to each independent loop and write KVL around each. Both produce a small linear system to solve.

Thévenin, Norton, and Superposition

These theorems let you collapse a complicated linear network into something trivial to analyze when you only care about one load.

Thévenin's theorem: any linear two-terminal network can be replaced by a single voltage source V_th in series with a resistance R_th. V_th is the open-circuit voltage at the terminals; R_th is the resistance seen looking back into the terminals with all independent sources deactivated (voltage sources shorted, current sources opened).

Norton's theorem: the dual — a current source I_N (the short-circuit current at the terminals) in parallel with R_N = R_th. Convert between them with V_th = I_N·R_th.

Superposition: in a linear circuit with several independent sources, the response (any voltage or current) equals the sum of the responses to each source acting alone, with the others deactivated. Useful when sources have different values or types but tedious for many sources.

Worked Thévenin example. A 24 V source feeds a node through 6 Ω, and an 12 Ω resistor connects that node to ground; we want the equivalent driving an external load at the node. V_th (open circuit) = voltage across the 12 Ω from the divider = 24 × 12/(6 + 12) = 24 × 0.667 = 16 V. R_th = 6 Ω ∥ 12 Ω = (6×12)/18 = 4 Ω. So the network looks like a 16 V source behind 4 Ω. Connect a 4 Ω load: by the voltage divider the load gets 16 × 4/(4 + 4) = 8 V, and load current = 8/4 = 2 A.

Maximum power transfer: a load receives maximum power when R_load = R_th; at that match the load power is P_max = V_th²/(4R_th). For the example above with R_th = 4 Ω, the matched-load power is 16²/(4×4) = 256/16 = 16 W. Recognizing the Thévenin form makes both load analysis and the maximum-power condition immediate.