Electronics Engineer (ECE) board exam reviewer
At least 70% in every subject. A subject at 60–69% may be retaken in a removal exam; below 60% in any subject fails. (RA 9292). Practise each subject with worked solutions, then sit a timed mock exam and see where you stand against that rule.
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Subjects, weights and topics
Sample questions
1. Find the determinant of the 2×2 matrix with rows [3 2] and [1 4].
- 14
- 10
- −10
- 5
Show the answer and solution
Answer: B. 10
det = 3(4) − 2(1) = 12 − 2 = 10.
2. Solve 3^x = 597 for x.
- 6.392
- 8.328
- 5.818
- 199
Show the answer and solution
Answer: C. 5.818
x = log(597)/log(3) = 5.818.
3. Which power device combines a MOSFET-like insulated gate drive with the low conduction loss of a bipolar output stage?
- TRIAC
- GTO thyristor
- Power diode
- IGBT
Show the answer and solution
Answer: D. IGBT
The insulated-gate bipolar transistor has a MOS gate (voltage drive, negligible gate current) and bipolar conduction for low on-state voltage at high current.
4. A unity-feedback system has open-loop G(s) = 38/(s + 1). What is the steady-state error to a unit step?
- 97.4%
- 0%
- 2.63%
- 2.56%
Show the answer and solution
Answer: D. 2.56%
Type 0 system, unit step: ess = 1/(1 + Kp) = 1/(1 + 38) = 2.56%.
5. How many grams of O₂ are needed to burn 44 g of propane completely? (C₃H₈ = 44 g/mol, O₂ = 32 g/mol)
- 32 g
- 96 g
- 224 g
- 160 g
Show the answer and solution
Answer: D. 160 g
The balanced equation is C₃H₈ + 5O₂ → 3CO₂ + 4H₂O. 1 mol of propane needs 5 mol of O₂ = 5 × 32 = 160 g.
6. What is the effective annual rate of 14% compounded monthly?
- 15.17%
- 14%
- 14.93%
- 1.167%
Show the answer and solution
Answer: C. 14.93%
i = (1 + j/m)^m − 1 = (1 + 14%/12)^12 − 1 = 14.93%.
7. A source emits three symbols with probabilities 0.5, 0.25 and 0.25. What is its entropy?
- 1.0 bits/symbol
- 1.58 bits/symbol
- 1.5 bits/symbol
- 2.0 bits/symbol
Show the answer and solution
Answer: C. 1.5 bits/symbol
H = Σ p log₂(1/p) = 0.5(1) + 0.25(2) + 0.25(2) = 1.5 bits/symbol (1.58 would require equal probabilities).
8. What is the TE₁₀ cutoff frequency of an air-filled rectangular waveguide whose broad wall is 47.55 mm?
- 1.58 GHz
- 6.31 GHz
- 3.15 GHz
- 2.1 GHz
Show the answer and solution
Answer: C. 3.15 GHz
TE₁₀ cutoff: fc = c/(2a) = 3×10⁸/(2 × 47.55 mm) = 3.15 GHz.
An independent, free reviewer: not affiliated with the PRC or the Boards. Subject weights and passing rules are from the PRC examination programs and the laws (RA 9292, RA 7920); check the current PRC program before you file. Code questions follow the Philippine Electrical Code 2017 Part 1; check them against your own copy of the Code.