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Saudi Council of Engineers FE Exam: Electrical Engineering Part: free practice test

Start with 10 free questions to see what needs more practice. Finish to see your score, weakest units and every option explained.

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Practice test questions with answers and explanations

Try the questions in the test above first, then open each one here to see the correct answer and why every other option is wrong.

01Two sources feed a common node: a 20 V source through a 4 Ω resistor and an 8 V source through a 2 Ω resistor. A third 4 Ω resistor connects the node to ground, and the negative terminals of both sources are grounded. Using KCL, what is the node voltage?

Practice question Unit: Circuits

  1. A12 V

    This comes from leaving out the resistor to ground in the node equation.

  2. B14 V

    This is the average of the two source voltages and ignores the resistor values and the ground branch.

  3. C1 V

    This comes from a sign error on one source voltage when writing the KCL equation.

  4. D9 V ✓ Correct answer

    \frac{V-20}{4}+\frac{V-8}{2}+\frac{V}{4}=0 gives V = 9 V.

Reference: SCE-Qiyas Engineering Fundamentals Exam, Electrical Engineering Standards (2014), Topic T1: Circuits, indicator EE-T1-01; standard theory: KCL at a node

02A 24 V source feeds a resistor R1 = 4 Ω in series with a parallel network. One branch is R2 = 12 Ω, and the other is R3 = 2 Ω in series with R4 = 4 Ω. What power is dissipated in R2?

Practice question Unit: Circuits

  1. A12 W ✓ Correct answer

    The network is 12 \parallel 6 = 4 Ω, the total current is 24/8 = 3 A and the network voltage is 12 V, so P = 12^2/12 = 12 W.

  2. B24 W

    This comes from using the other branch's resistance (6 Ω) instead of 12 Ω in the power calculation.

  3. C48 W

    This comes from placing the full 24 V source voltage across R2 and ignoring R1.

  4. D108 W

    This comes from assuming the whole 3 A flows through R2 and using I^2 R.

Reference: SCE-Qiyas Engineering Fundamentals Exam, Electrical Engineering Standards (2014), Topic T1: Circuits, indicator EE-T1-02

03A balanced delta-connected load has a phase impedance of 30∠30° Ω and is connected to a three-phase supply with a line voltage of 208 V. What is the total real power?

Practice question Unit: Power

  1. A4.33 kW

    This is the total apparent power 3 × 208²/30 = 4.33 kVA; cos 30° has not been applied.

  2. B2.16 kW

    This is half the apparent power, a value that no correct step produces.

  3. C6.49 kW

    This is 3 × 208 × 12.0 × cos 30°, multiplying line quantities by 3 instead of √3.

  4. D3.75 kW ✓ Correct answer

    The phase voltage is 208 V and the phase current is 6.93 A, so P = 3 × 6.93² × 30 cos 30° = 3.75 kW.

Reference: SCE-Qiyas Engineering Fundamentals Exam, Electrical Engineering Standards (2014), Topic T2: Power, indicator EE-T2-01

04The electrostatic potential is V=3x^2y-z (volts, coordinates in metres). What is the electric field \mathbf{E} at the point (1, 1, 0)?

Practice question Unit: Electromagnetics

  1. A6\,\mathbf{a}_x+3\,\mathbf{a}_y-\mathbf{a}_z V/m

    This is the gradient \nabla V itself; it misses the minus sign in \mathbf{E}=-\nabla V.

  2. B-6\,\mathbf{a}_x-3\,\mathbf{a}_y+\mathbf{a}_z V/m ✓ Correct answer

    The field \mathbf{E}=-\nabla V=-(6xy,\,3x^2,\,-1), which at (1, 1, 0) equals (-6,-3,+1) V/m.

  3. C-6\,\mathbf{a}_x-3\,\mathbf{a}_y-\mathbf{a}_z V/m

    The first two components are right, but \partial(-z)/\partial z=-1, so the z-component of E is +1.

  4. D-3\,\mathbf{a}_x-6\,\mathbf{a}_y+\mathbf{a}_z V/m

    This answer swaps the two derivatives: \partial V/\partial x=6xy and \partial V/\partial y=3x^2.

Reference: SCE-Qiyas Engineering Fundamentals Exam, Electrical Engineering Standards (2014), Topic T3: Electromagnetics, indicator EE-T3-02; standard formula: \mathbf{E}=-\nabla V

05Two blocks in series form the forward path: G_1(s)=\dfrac{2}{s+1} and G_2(s)=\dfrac{5}{s+4}, with unity negative feedback. What is the closed-loop transfer function?

Practice question Unit: Control Systems

  1. A\dfrac{10}{s^2+5s-6}

    Wrong: this comes from positive feedback, which makes the constant term 4-10.

  2. B\dfrac{10}{s^2+5s+14} ✓ Correct answer

    Correct: G=\dfrac{10}{(s+1)(s+4)}=\dfrac{10}{s^2+5s+4}, so T=\dfrac{10}{s^2+5s+4+10}=\dfrac{10}{s^2+5s+14}.

  3. C\dfrac{10}{s^2+5s+4}

    Wrong: this is the open-loop function, with the feedback not applied.

  4. D\dfrac{10}{s^2+5s+10}

    Wrong: the gain 10 is added to the constant term 4 in the denominator; it does not replace it.

Reference: SCE-Qiyas Engineering Fundamentals Exam, Electrical Engineering Standards (2014), Topic T4: Control Systems, indicator EE-T4-01; standard formula: T=G/(1+G)

06The unmodulated carrier power is 100 W. The carrier is modulated by a sinusoidal tone with a modulation index of 0.8. What is the total power of the modulated wave?

Practice question Unit: Communications

  1. A164 W

    This is 100(1+m^2), forgetting the division by 2.

  2. B140 W

    This is 100(1+m/2); the index in the formula is squared.

  3. C132 W ✓ Correct answer

    Correct: P_t = P_c(1+m^2/2) = 100(1+0.32) = 132 W.

  4. D180 W

    This is 100(1+m), which is wrong.

Reference: SCE-Qiyas Engineering Fundamentals Exam, Electrical Engineering Standards (2014), Topic T5: Communications, indicator EE-T5-01

07An ADC must achieve an SQNR of at least 70 dB for a full-scale sinusoid. What is the minimum number of bits?

Practice question Unit: Signal Processing

  1. A12 bits ✓ Correct answer

    N \ge (70-1.76)/6.02 = 11.34, rounded up to 12 bits, which gives 74.0 dB.

  2. B10 bits

    10 bits give only about 62.0 dB, below the requirement.

  3. C11 bits

    11 bits give about 68.0 dB, just under 70 dB.

  4. D14 bits

    14 bits meets the requirement but is not the minimum; it gives about 86 dB.

Reference: SCE-Qiyas Engineering Fundamentals Exam, Electrical Engineering Standards (2014), Topic T6: Signal Processing, indicator EE-T6-01; standard formula: N >= (SQNR - 1.76)/6.02

08A first-order active low-pass filter has a DC gain magnitude of 10 and a cutoff frequency of 1\,\text{kHz}. What is the gain magnitude at 3\,\text{kHz}?

Practice question Unit: Electronics

  1. A3.33

    This comes from 10/3, ignoring the square root in the magnitude formula.

  2. B7.07

    This is the gain at the cutoff frequency itself (1\,\text{kHz}), not at 3 kHz.

  3. C9.49

    This results from inverting the ratio f/f_c, which is the high-pass formula.

  4. D3.16 ✓ Correct answer

    |A| = 10/\sqrt{1+(3)^2} = 10/\sqrt{10} \approx 3.16.

Reference: SCE-Qiyas Engineering Fundamentals Exam, Electrical Engineering Standards (2014), Topic T7: Electronics, indicator EE-T7-02; standard formula: |A| = A0 / sqrt(1 + (f/f_c)^2)

09In 8-bit two's complement, adding +120 and +100 gives the pattern 11011100. How should this result be interpreted?

Practice question Unit: Digital Systems

  1. AOverflow occurred: the true sum +220 exceeds +127, and the pattern reads as −36 ✓ Correct answer

    Correct: two positives gave a negative pattern; the range is −128 to +127, and 11011100 = 220−256 = −36.

  2. BNo overflow occurred, and the result is +220 because the top bit is part of the magnitude

    +220 cannot be represented in 8-bit two's complement; this reading treats the pattern as unsigned.

  3. CNo overflow occurred, and the result is +92 after the top bit is dropped as a carry-out

    The top bit is the sign bit, not a carry-out, and 92 is a wrong value obtained by dropping it (220−128).

  4. DOverflow occurred, but the pattern represents −92 because the sign is separate from the other bits

    The overflow is real but −92 is wrong; sign-magnitude reading does not apply to two's complement, and the value is −36.

Reference: SCE-Qiyas Engineering Fundamentals Exam, Electrical Engineering Standards (2014), Topic T8: Digital Systems, indicator EE-T8-01

10A cache has a hit time of 2 ns and a hit rate of 95%. Each miss costs an additional 60 ns to fetch the block from main memory. What is the average memory access time?

Practice question Unit: Computer Systems

  1. A57.1 ns

    This swaps the two rates: 0.05 × 2 + 0.95 × 60 = 57.1 ns, weighting the hit time by the miss rate and the miss penalty by the hit rate.

  2. B4.9 ns

    This computes 0.95 × 2 + 0.05 × 60 and forgets that every access, even a miss, first pays the cache lookup time.

  3. C5.0 ns ✓ Correct answer

    Average time = hit time + miss rate × miss penalty = 2 + 0.05 × 60 = 5.0 ns.

  4. D3.0 ns

    This counts only the miss penalty contribution (0.05 × 60) and omits the hit time.

Reference: SCE-Qiyas Engineering Fundamentals Exam, Electrical Engineering Standards (2014), Topic T9: Computer Systems, indicator EE-T9-01

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