Electricity and Magnetism is 5-8 questions on the FE Mechanical exam. For most mechanical candidates it is the least familiar area on the paper, and the one most often written off as a loss.
That is a poor trade. The scope is narrow, the questions are shallow relative to an electrical engineer's version of them, and five to eight questions is more than dynamics is worth on the FE Civil exam. A weekend on circuits recovers most of these points.
Exam weight: NCEES lists Electricity and Magnetism at 5-8 questions of the 110-question FE Mechanical exam. Work each problem below on paper first, then reveal the worked solution — reading a solution you have not attempted builds recognition, not recall.
What NCEES Tests in Electricity and Magnetism
The specification covers charge, current, voltage, resistance and power, along with DC circuit analysis, AC circuit fundamentals, and motors and generators. Expect equivalent resistance of a series-parallel network, a voltage divider, power dissipated in a resistor, and an RMS-versus-peak conversion.
The motor and generator items are conceptual more often than computational: efficiency, power factor, and the relationship between electrical input and mechanical output. These sit naturally alongside the mechanical power questions elsewhere on the exam.
5 Free Electricity and Magnetism Practice Problems
Each electricity magnetism problem below comes from the FEExamClick FE Mechanical question bank, with worked solutions, and matches the style, difficulty and format of the real exam. Attempt each one under a three-minute limit — roughly the pace the exam demands.
Problem 1 — A. Electrical fundamentals
A 120-V heater is designed to use a 20-Ω resistor. The power (W) consumed by the heater when in use is most nearly:
Answer: A) 720
Refer to the Power Absorbed by a Resistive Element section (p. 363) in the Electrical and Computer Engineering chapter of the FE Reference Handbook.
Power dissipated by a resistor:
$$P = \frac{V^2}{R}$$
$$P = \frac{(120)^2}{20} = \frac{14{,}400}{20}$$
$$P = 720 \text{ W}$$
Problem 2 — D. Motors and generators
A 6-pole induction motor operates on a 50 Hz supply with a slip of 4%. The rotor speed is most nearly:
Answer: C) 960 rpm
Synchronous speed:
\(n_s = \frac{120f}{p} = \frac{120 \times 50}{6} = 1000\) rpm
Rotor speed:
\(n = n_s(1 - s) = 1000(1 - 0.04) = 1000 \times 0.96 = 960\) rpm
Problem 3 — B. DC circuit analysis
In the resistor circuit shown below, the equivalent resistance \(R_{\text{eq}}\) (Ω) at terminals a-b is most nearly:
Answer: B) 8.75
Refer to the DC Circuit Analysis section in the Electrical and Computer Engineering chapter of the FE Reference Handbook.
Step 1 – Read the topology from the figure. Looking in at terminals a–b, the left 5-Ω resistor carries the whole current. The path then splits into the 5-Ω and 15-Ω resistors, which rejoin before terminal b. Those two share both of their nodes, so they are in parallel, and that pair is in series with the left 5 Ω.
Step 2 – Combine the parallel pair (product over sum):
\(R_p = \dfrac{R_1 R_2}{R_1 + R_2} = \dfrac{(5)(15)}{5 + 15} = \dfrac{75}{20} = 3.75\ \Omega\)
Step 3 – Add the series resistor:
\(R_{eq} = 5 + 3.75 = 8.75\ \Omega\)
Why the other options fail:
• 25.0 Ω adds all three in series, \(5+5+15\). This is the most common slip: the two right-hand resistors are not end-to-end, they sit on separate branches that start and finish at the same two nodes.
• 10.0 Ω assumes the parallel combination equals the smaller resistor (5 Ω), giving \(5+5\). The combination is always less than the smaller one, not equal to it.
• 5.0 Ω counts only the series resistor, treating the parallel section as a short circuit. It is not a short — it contributes 3.75 Ω.
Sanity check worth memorising: a parallel combination is always smaller than the smallest resistor in it. Here \(3.75 < 5\), so \(R_{eq}\) must land between 5 Ω (the series resistor alone) and 10 Ω (the series resistor plus a full 5 Ω). Only 8.75 Ω falls in that window.
Correct answer: (B)
Problem 4 — C. AC circuit analysis
A series RLC circuit has R = 5 ohms, L = 10 mH, and C = 10 µF. At resonance, the quality factor Q is most nearly:
Answer: D) 6.3
At resonance:
\(\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{10 \times 10^{-3} \times 10 \times 10^{-6}}} = \frac{1}{\sqrt{10^{-7}}} = \frac{1}{3.16 \times 10^{-4}} = 3162\) rad/s
Quality factor for series resonance:
\(Q = \frac{\omega_0 L}{R} = \frac{3162 \times 0.01}{5} = \frac{31.62}{5} = 6.32 \approx 6.3\)
Problem 5 — B. DC circuit analysis
In the circuit shown, a 48-V source feeds a network with R1 = 6 Ω in series with two parallel branches: R2 = 8 Ω and (R3 = 4 Ω in series with R4 = 12 Ω). The current (A) through R2 is most nearly:
Answer: B) 2.82
Refer to the DC Circuit Analysis section in the Electrical and Computer Engineering chapter of the FE Reference Handbook.
Step 1 – Reduce the two branches. R₃ and R₄ are end-to-end on the same branch, so they add; that branch is then in parallel with R₂:
\(R_{34} = 4 + 12 = 16\ \Omega\)
\(R_p = \dfrac{(8)(16)}{8 + 16} = \dfrac{128}{24} = 5.33\ \Omega\)
Step 2 – Total resistance and source current. R₁ carries the whole current, so it is in series with that pair:
\(R_T = 6 + 5.33 = 11.33\ \Omega \qquad I_T = \dfrac{48}{11.33} = 4.24\ \text{A}\)
Step 3 – Voltage across the parallel section. R₁ drops part of the source voltage first — this is the step most often skipped:
\(V_p = I_T R_p = (4.24)(5.33) = 22.6\ \text{V}\)
Step 4 – Current through R₂:
\(I_2 = \dfrac{V_p}{R_2} = \dfrac{22.6}{8} = 2.82\ \text{A}\)
Faster route (current divider). Split \(I_T\) between the two branches using the opposite branch's resistance in the numerator:
\(I_2 = I_T\,\dfrac{R_{34}}{R_2 + R_{34}} = 4.24\left(\dfrac{16}{24}\right) = 2.82\ \text{A}\)
Check: the other branch carries \(I_{34} = 4.24 - 2.82 = 1.41\) A, and the smaller resistance (8 Ω) correctly takes the larger share of the current.
Why the other options fail:
• 6.00 A puts the full 48 V across R₂, \(48/8\). It ignores that R₁ drops voltage before the parallel section — only 22.6 V actually reaches R₂.
• 4.24 A is the total current \(I_T\) leaving the source. That current then splits between the two branches, so it is larger than the current in R₂ alone.
• 1.41 A is the current-divider ratio applied upside down, \(I_T\,R_2/(R_2+R_{34})\). That expression actually returns the current in the other branch. Remember: the branch with the smaller resistance must carry the larger current.
Correct answer: (B)
Using the FE Reference Handbook for Electricity and Magnetism
The Electrical and Computer Engineering chapter is larger than what you need, so anchor on three things: the resistor combination rules, the impedance definitions for inductors and capacitors, and the RMS relationships. Search the handbook for impedance and for RMS during practice so you land on those pages directly rather than reading around them.
Four Mistakes That Cost Points
- Adding parallel resistors as if they were in series. Two 10-ohm resistors in parallel give 5 ohms, not 20. Under time pressure the series habit takes over, and both answers appear among the options.
- Mixing peak and RMS values. AC power questions expect RMS unless stated otherwise, and the ratio is the square root of two. Feeding a peak voltage into a power expression overstates the result by a factor of two.
- Treating impedance magnitude as an algebraic sum. Resistance and reactance combine in quadrature, not by addition. The arithmetic sum is always one of the four options.
- Losing the milli and micro prefixes. Capacitance in microfarads and current in milliamps are the norm here, and a factor of a thousand is the most common single error in this area. Convert to base units before substituting.
Frequently Asked Questions
How many electricity questions are on the FE Mechanical exam?
NCEES specifies 5-8 questions from Electricity and Magnetism out of 110. It is comparable in weight to Measurements, Instrumentation and Controls.
How deep does the electrical content go?
Not deep. The exam tests whether a mechanical engineer can analyse a simple circuit, reason about AC quantities, and understand motor efficiency. It does not test transmission lines, semiconductor physics or three-phase analysis at the level an FE Electrical candidate would face.
Is it worth studying if I have not touched circuits since sophomore year?
Yes. Five to eight questions is a meaningful block, and the required scope is small enough to rebuild in a weekend. Start with series-parallel resistance and Ohm's law, then power, then RMS. That sequence covers the majority of what appears.
Do I need to know three-phase power?
Basic exposure is enough. Questions involving motors may reference three-phase supply, but the calculation asked is usually power or efficiency rather than a phase analysis. Focus your time on single-phase AC fundamentals first.
Keep Going
These topics feed into each other on the exam:
- FE Mechanical Measurements, Instrumentation, and Controls practice problems — 5-8 questions on the exam
- FE Mechanical Mathematics practice problems — 6-9 questions on the exam
- FE Mechanical Mechanical Design and Analysis practice problems — 10-15 questions on the exam
Done with electricity magnetism? Browse every knowledge area from the free FE Mechanical practice problem hub, see what the full bank covers on the FE Mechanical exam prep page, or plan your schedule with the FE study timeline. Or test yourself against the clock with the free FE Mechanical practice exam: 20 timed questions in the CBT format.