NCEES · FE
FE Electrical & Computer practice questions with answers
15 free practice questions for the FE Electrical and Computer Exam (FE), spread across the exam's topics. Try each one, then open the answer to see why every option is right or wrong.
Question 1 · Mathematics
Evaluate \(\ln(e^5)\).
- A\(e^5\)
- B5
- C1
- D0
Show answer and explanation
Answer: B) 5
- AIncorrect. The log undoes the exponential.
- BCorrect. The natural log undoes the exponential.
- CIncorrect. That's \(\ln e\).
- DIncorrect. That's \(\ln 1\).
Question 2 · Mathematics
The indefinite integral \(\int \cos(2x)\,dx\) is:
- A\(\tfrac{1}{2}\cos(2x) + C\)
- B\(2\sin(2x) + C\), multiplying by 2
- C\(\tfrac{1}{2}\sin(2x) + C\)
- D\(\sin(x) + C\)
Show answer and explanation
Answer: C) \(\tfrac{1}{2}\sin(2x) + C\)
- AIncorrect. The antiderivative of cosine is sine.
- BIncorrect. Integration divides by 2.
- CCorrect. Divide by the inner derivative, 2.
- DIncorrect. The argument stays 2x.
Question 3 · Probability and Statistics
To halve the standard error of a sample mean, the sample size must be multiplied by:
- A8
- B2
- C4
- D16
Show answer and explanation
Answer: C) 4
- AIncorrect. Recheck: \(\sqrt{4} = 2\).
- BIncorrect. SE varies with \(1/\sqrt{n}\).
- CCorrect. SE varies with \(1/\sqrt{n}\).
- DIncorrect. That quarters the SE.
Question 4 · Ethics and Professional Practice
An engineer inspecting a contractor's electrical work is offered a job by that contractor. The engineer should:
- AAccept the job and keep inspecting the contractor's work
- BDisclose the offer and step back from the inspection
- CApprove the work first and then accept the job
- DKeep the offer secret until the project ends
Show answer and explanation
Answer: B) Disclose the offer and step back from the inspection
- AIncorrect. That's a conflict.
- BCorrect. It creates a conflict of interest.
- CIncorrect. That's still a conflict.
- DIncorrect. Disclosure is required now.
Question 5 · Engineering Economics
At 4% interest compounded annually, $5,000 invested today grows in 10 years to approximately:
- A$7,401
- B$7,000
- C$6,083
- D$7,697
Show answer and explanation
Answer: A) $7,401
Single-payment compound amount: F = P(1 + i)^n = 5,000(1.04)^10 ≈ $7,401.
- ACorrect. F = P(F/P, 4%, 10) = 5,000 × (1.04)^10 = 5,000 × 1.4802 ≈ $7,401.
- BIncorrect. This uses simple interest: 5,000 × (1 + 0.04 × 10) = $7,000. It leaves out interest earned on interest.
- CIncorrect. This compounds for only 5 periods: 5,000 × (1.04)^5 ≈ $6,083.
- DIncorrect. This compounds for 11 periods instead of 10: 5,000 × (1.04)^11 ≈ $7,697.
Question 6 · Circuit Analysis (DC and AC Steady State)
In DC steady state, a capacitor behaves like:
- AA voltage source of zero
- BA short circuit
- CA resistor equal to its capacitance
- DAn open circuit
Show answer and explanation
Answer: D) An open circuit
- AIncorrect. It's an open circuit.
- BIncorrect. That's an inductor in DC steady state.
- CIncorrect. It's an open circuit.
- DCorrect. No current flows once it's charged.
Question 7 · Circuit Analysis (DC and AC Steady State)
A circuit has an open-circuit voltage of 12 V and a Thevenin resistance of 4 Ω. Its Norton current is:
- A0.33 A
- B48 A
- C3 A
- D16 A
Show answer and explanation
Answer: C) 3 A
- AIncorrect. That's inverted.
- BIncorrect. Divide, don't multiply.
- CCorrect. \(I_N = V_{oc}/R_{th}\).
- DIncorrect. Divide, don't add.
Question 8 · Linear Systems
For \(H(j\omega) = 1/(1 + j\omega)\), the magnitude at ω = 1 rad/s is about:
- A1.414
- B0.5
- C1.0
- D0.707
Show answer and explanation
Answer: D) 0.707
- AIncorrect. That's inverted.
- BIncorrect. Recheck: \(|1 + j| = \sqrt{2}\).
- CIncorrect. That's at ω = 0.
- DCorrect. \(1/\sqrt{2}\), the −3-dB point.
Question 9 · Signal Processing
An ideal symmetric square wave contains:
- AAll harmonics equally
- BOnly even harmonics
- COnly the fundamental
- DOnly odd harmonics
Show answer and explanation
Answer: D) Only odd harmonics
- AIncorrect. Odd harmonics with decreasing amplitude.
- BIncorrect. Odd harmonics only.
- CIncorrect. It has many harmonics.
- DCorrect. Its Fourier series has odd terms only.
Question 10 · Electronics
An emitter-follower (common-collector) amplifier has a voltage gain of about:
- Aβ
- B1
- C−100
- D0
Show answer and explanation
Answer: B) 1
- AIncorrect. That's the current gain.
- BCorrect. It's used as a buffer.
- CIncorrect. That's typical of a common-emitter stage.
- DIncorrect. It passes the signal.
Question 11 · Power Systems
A single-phase, two-wire 240 V feeder supplies a 40 A resistive load. Each conductor has a resistance of 0.25 Ω, and reactance can be neglected. The total voltage drop along the feeder is closest to:
- A10.0 V
- B17.3 V
- C20.0 V
- D34.6 V
Show answer and explanation
Answer: C) 20.0 V
In a single-phase two-wire feeder the current passes through both conductors, so the drop is I × 2R = 40 × 0.5 = 20 V.
- AIncorrect. 40 A × 0.25 Ω = 10 V counts only one conductor. In a two-wire circuit the current flows out on one conductor and back on the other.
- BIncorrect. √3 × 40 × 0.25 ≈ 17.3 V is the line-to-line drop formula for a balanced three-phase feeder, which does not apply to a single-phase circuit.
- CCorrect. The loop resistance is 2 × 0.25 = 0.5 Ω, so the drop is 40 A × 0.5 Ω = 20 V (about 8.3% of 240 V).
- DIncorrect. √3 × 40 × 0.5 ≈ 34.6 V applies a three-phase √3 factor to the loop resistance, which counts the drop twice over.
Question 12 · Control Systems
Root locus branches begin, at zero gain, at the:
- AOpen-loop poles
- BOpen-loop zeros
- COrigin always
- DClosed-loop zeros
Show answer and explanation
Answer: A) Open-loop poles
- ACorrect. They end at the open-loop zeros or infinity.
- BIncorrect. They end there.
- CIncorrect. They start at the open-loop poles.
- DIncorrect. They start at the open-loop poles.
Question 13 · Communications
A baseband digital signal (before any carrier modulation) with symbol rate R uses raised-cosine pulse shaping with roll-off factor α (0 ≤ α ≤ 1). What is its required transmission bandwidth?
- AR(1 + α)
- B(R/2)(1 + α)
- CR/2
- DαR/2
Show answer and explanation
Answer: B) (R/2)(1 + α)
For baseband raised-cosine signaling, B = (R/2)(1 + α): the Nyquist bandwidth R/2 plus the excess αR/2. α = 0 gives R/2, and α = 1 doubles it to R.
- AIncorrect. R(1 + α) is the bandwidth after the signal is modulated onto a carrier (double-sideband passband). Both sidebands are sent, which doubles the baseband value.
- BCorrect. The raised-cosine spectrum starts at the Nyquist bandwidth R/2 and extends a further fraction α of it, so B = (R/2)(1 + α).
- CIncorrect. R/2 is the Nyquist minimum bandwidth. It applies only to the ideal α = 0 (sinc) pulse; any roll-off above zero adds bandwidth.
- DIncorrect. αR/2 is only the excess bandwidth added by the roll-off, beyond the Nyquist bandwidth. The total bandwidth is R/2 + αR/2 = (R/2)(1 + α).
Question 14 · Digital Systems
Decimal 255 in hexadecimal is:
- A100
- BF0
- CFF
- DEE
Show answer and explanation
Answer: C) FF
- AIncorrect. That's 256.
- BIncorrect. That's 240.
- CCorrect. 15 × 16 + 15.
- DIncorrect. That's 238.
Question 15 · Computer Systems
A byte-addressable memory with 16-bit addresses can address:
- A16 KB
- B64 KB
- C1 MB
- D256 bytes
Show answer and explanation
Answer: B) 64 KB
- AIncorrect. That's \(2^{14}\).
- BCorrect. \(2^{16}\) bytes.
- CIncorrect. That's \(2^{20}\).
- DIncorrect. That's \(2^8\).
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Original questions written from the official exam outline. Not affiliated with NCEES; never copied from the real exam.