3.1 DC Circuit Analysis

Why DC analysis anchors the exam

The NCEES FE Electrical and Computer exam has 110 questions in a 6-hour appointment, scored against a psychometric cut score rather than a fixed percentage. The Circuit Analysis (DC and AC) knowledge area is one of the larger blocks, and DC fundamentals also feed the Electronics, Power, Linear Systems, and Control items. Every formula below is in the searchable NCEES FE Reference Handbook that you use on-screen, so the tested skill is not memorization but finding and applying the right relation quickly and tracking signs and units.

Most DC questions are short once the model is correct, so build a reliable setup routine instead of hunting for clever tricks.

Ohm's law and the two Kirchhoff laws

Ohm's law relates voltage, current, and resistance: V = IR. Power dissipated in a resistor is P = VI = I^2 R = V^2 / R, in watts. These three power forms let you solve for power knowing any two of V, I, R.

Kirchhoff's Voltage Law (KVL): the algebraic sum of voltages around any closed loop is zero. Walk the loop in one direction and assign a consistent sign to each rise (+) or drop (-).

Kirchhoff's Current Law (KCL): the algebraic sum of currents entering a node equals the sum leaving; it expresses charge conservation and is the basis of the node-voltage method.

Declare a reference current direction and a ground (datum) node before writing equations. A negative result simply means the true direction is opposite your assumption — never "fix" the sign by re-guessing.

Series, parallel, and dividers

Resistors in series carry the same current and add directly: R_eq = R_1 + R_2 + ... Resistors in parallel share the same voltage and their conductances add: 1/R_eq = 1/R_1 + 1/R_2 + ... For exactly two resistors in parallel, the product-over-sum shortcut is fastest: R_eq = R_1 R_2 / (R_1 + R_2); two equal resistors in parallel give half the value.

Voltage divider (series elements across a source): V_x = V_s * R_x / R_total. Current divider (two parallel resistors): I_1 = I_total * R_2 / (R_1 + R_2) — note the opposite resistor appears in the numerator, a classic trap.

Capacitors combine the opposite way to resistors — series capacitors use reciprocals.

Worked node-voltage example

Find node voltage V_A in a circuit where a 10 V source connects through a 2 ohm resistor to node A, node A connects through a 4 ohm resistor to ground, and node A also connects through a 4 ohm resistor to ground (a second branch). Take the bottom rail as ground.

Write KCL at node A (currents leaving A sum to zero):

(V_A - 10)/2 + V_A/4 + V_A/4 = 0.

Multiply through by 4: 2(V_A - 10) + V_A + V_A = 0, so 2V_A - 20 + 2V_A = 0, giving 4V_A = 20 and V_A = 5 V. The current from the source is (10 - 5)/2 = 2.5 A.

Thevenin, Norton, and source transformation

Any linear two-terminal network reduces to a Thevenin equivalent (an ideal voltage source V_th in series with R_th) or its dual, the Norton equivalent (a current source I_N in parallel with the same R_th), where I_N = V_th / R_th. Procedure:

1. V_th = open-circuit voltage across the terminals.
2. I_N = short-circuit current through the terminals.
3. R_th = V_th / I_N, or deactivate all independent sources (short voltage sources, open current sources) and find the resistance looking back in.

With dependent sources present you cannot simply zero them; apply a 1 A or 1 V test source at the terminals and compute R_th = V_test / I_test.

Superposition, max power transfer, and mesh analysis

Superposition: in a linear circuit the response equals the sum of responses to each independent source acting alone. Zero the others (voltage sources -> short, current sources -> open). It is handy when sources differ in type or frequency, but it does not apply to power, which is nonlinear in V and I.

Maximum power transfer: for a fixed Thevenin source, the load receives maximum power when R_L = R_th, giving P_max = V_th^2 / (4 R_th). At the match each of source and load dissipates half the total, so efficiency is only 50%. In AC the condition becomes the complex conjugate Z_L = Z_th*.

Worked check: with V_th = 20 V, R_th = 10 ohm, the matched load R_L = 10 ohm gives load current 20/(10+10) = 1 A and P = I^2 R_L = 1^2 x 10 = 10 W, matching P_max = 400/40 = 10 W.

The node-voltage method applies KCL at each non-reference node; the mesh-current method applies KVL around each independent loop. Mesh analysis suits planar circuits dominated by voltage sources; node analysis suits circuits with current sources and many parallel branches. Pick whichever yields fewer simultaneous equations.