FE Electrical and Computer Circuit Analysis (DC and AC Steady State) Practice Problems

Work 3 free FE Electrical and Computer circuit analysis (dc and ac steady state) practice problems with complete solutions, governing equations and checks. Attempt the setup before reading the answer.

Free sample problem

A 120 V DC source is connected through a 10 Ω series resistor to a parallel section. One branch of that section is a single 60 Ω resistor; the other branch is a 12 Ω resistor in series with an 18 Ω resistor. The voltage across the 18 Ω resistor is most nearly:

(A)21.6 V
(B)24.0 V
(C)32.0 V
(D)48.0 V

Show worked solution

Answer: (D)

Only 80 V reaches the parallel section, and 48 V of that appears across the 18 Ω resistor.

Reduction starts at the far end, where the 12 Ω and 18 Ω resistors carry a common current and combine before the parallel step.

R_{series} = 12 + 18 = 30\ \Omega

R_p = \frac{(60)(30)}{60 + 30} = 20\ \Omega

With 30 Ω of total loop resistance, the source drives 4 A, and the 10 Ω resistor absorbs part of the supply voltage before the parallel section.

I_s = \frac{120}{10 + 20} = 4 \text{ A}

V_p = (4)(20) = 80 \text{ V}

Splitting that section voltage between the two series resistors of the composite branch leaves the larger share on the 18 Ω element.

I_{branch} = \frac{80}{30} = 2.67 \text{ A}

V_{18} = (2.67)(18) = 48.0 \text{ V}

FE Reference Handbook — Electrical and Computer Engineering: Resistors in Series and Parallel, Voltage and Current Division

Why the other choices appear

(A)Treats the two branches as being in series (60 + 30 = 90 Ω, total 100 Ω, 1.2 A), giving (1.2)(18) = 21.6 V.
(B)Inverts the current-division ratio, 4 A times 30/90 = 1.33 A, giving (1.33)(18) = 24.0 V.
(C)Solves for the drop across the 12 Ω resistor of the same branch, (2.67)(12) = 32.0 V.

Two more free problem types

Try a different setup before reading the solution. These additional original questions are separate from the sold practice forms, so solving them does not reveal another question from a full practice form.

A loaded DC voltage divider

An ideal 24 V DC source feeds a 10-kilohm series resistor. From the output node to ground are a 20-kilohm divider resistor and a 20-kilohm load in parallel. What is the steady output voltage?

(A)6 V
(B)8 V
(C)12 V
(D)16 V

Show worked solution

Answer: (C)

The loaded output is 12 V; the load lowers it from the unloaded 16 V.

The load is in parallel with the lower divider resistor. Two equal 20-kilohm resistors have a 10-kilohm equivalent resistance.

R_p=\frac{20(20)}{20+20}=10\ \mathrm{k}\Omega

The source therefore sees two equal 10-kilohm resistances in series.

V_{out}=24\frac{10}{10+10}=12\ \mathrm{V}

The source current is 1.2 mA. At the node, each 20-kilohm branch carries 0.6 mA, which sums to the source current. The unloaded 16 V ignores the extra branch.

FE Reference Handbook: Electrical and Computer Engineering — series-parallel resistance, voltage division and Kirchhoff current law.

Why the other choices appear

(A)This uses an incorrect equivalent resistance for the two parallel branches.
(B)This gives three equal series resistors, although the lower resistors are parallel.
(D)This is the unloaded divider voltage and neglects the load.

AC impedance: magnitude is not a scalar sum

An ideal 120 V rms sinusoidal source feeds a series circuit with resistance 30 ohms and net inductive reactance 40 ohms. What is the rms current magnitude?

(A)1.71 A
(B)2.40 A
(C)3.00 A
(D)4.00 A

Show worked solution

Answer: (B)

The rms current magnitude is 2.40 A. Current lags source voltage by about 53.13 degrees.

Resistance and reactance are perpendicular components of complex impedance, not two real resistances to add.

Z=30+j40\ \Omega,\qquad |Z|=\sqrt{30^2+40^2}=50\ \Omega

Divide the rms voltage magnitude by the impedance magnitude.

|I|=120/50=2.40\ \mathrm{A},\qquad \angle I=-\tan^{-1}(40/30)

The resistor and reactance voltage magnitudes are 72 V and 96 V. Their vector magnitude sqrt(72 squared + 96 squared) is the stated 120 V. Rms quantities stay rms throughout.

FE Reference Handbook: Electrical and Computer Engineering — AC phasors, impedance and Ohm law.

Why the other choices appear

(A)This divides by 30 + 40 and incorrectly treats complex impedance as a scalar sum.
(C)This ignores resistance and divides only by reactance.
(D)This ignores reactance and divides only by resistance.

Download all 14 additional questions, diagrams and complete solutions (free PDF). Download the editable practice review sheet (CSV).

Original questions and diagrams by EngineerProof, September 30, 2026. Personal study and unchanged redistribution with attribution are permitted. The questions are independent practice, not official NCEES material. Report a correction.

The complete FE Electrical and Computer practice exam contains 12 problems in this area and 110 problems overall, with an answer key, full solutions, and a score sheet by knowledge area.

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