Free FE ECE practice problems

Ten real questions from our FE Electrical and Computer bank — one from each knowledge area, with the same exam-realistic figures and fully worked solutions our members get. No signup, no email. Try each one first, then reveal the solution.

10 with figures10 knowledge areasFull worked solutions

1
medium
A. DC circuit analysis: KCL/KVL, equivalents, Thevenin & Norton
A voltage divider across a 24 V DC source uses $R_1=200\,\Omega$ (series) and $R_2=300\,\Omega$ (shunt). A load $R_L=600\,\Omega$ is then connected across the output terminals (across $R_2$ ). The output voltage across the load is most nearly:
1. 19.6 V
2. 12 V
3. 9.6 V
4. 14.4 V
Show the answer & worked solution
Answer: B.12 V
Loaded bottom leg: $R_2\|R_L=\dfrac{300\cdot600}{900}=200\,\Omega$ . $V_{out}=V\dfrac{R_2\|R_L}{R_1+R_2\|R_L}=24\cdot\dfrac{200}{400}=12$ V. Distractors: the unloaded divider value (loading ignored), the inverted divider ratio $R_1/(R_1+R_2)$ , and the load wrongly placed in series with $R_2$ instead of in parallel.
2
medium
A. AM/FM/PCM modulation, Fourier methods, multiplexing & digital communications
The AM (DSB-LC) modulator of an industrial telemetry transmitter uses a 100 kHz carrier, amplitude-modulated by a single sinusoidal tone at 16 kHz (the figure shows the carrier and the message tone BEFORE modulation). The sideband frequencies and occupied bandwidth of the transmitted signal are most nearly:
1. sidebands at 68 and 132 kHz; BW = 64 kHz
2. sidebands at 84 and 116 kHz; BW = 32 kHz
3. sidebands at 92 and 108 kHz; BW = 16 kHz
4. sidebands at 16 and 100 kHz; BW = 84 kHz
Show the answer & worked solution
Answer: B.sidebands at 84 and 116 kHz; BW = 32 kHz
Single-tone AM puts one sideband on each side of the carrier: $f_c\pm f_m=100\pm16$ kHz, i.e. 84 and 116 kHz, so the occupied bandwidth is $BW=2f_m=2(16)=32$ kHz (the carrier itself stays at 100 kHz). Distractors double the sideband offset to $f_c\pm2f_m$ , treat $f_m$ as the TOTAL bandwidth and split it across the carrier ( $f_c\pm f_m/2$ ), and assume the signal occupies the band between the message tone and the carrier.
3
medium
A. Block diagrams, Bode plots, stability & controller performance
Two blocks with constant gains $G_1=0.8$ and $G_2=2$ are connected in cascade inside a unity negative-feedback loop, as shown. The closed-loop gain $C/R$ is most nearly:
1. 0.615
2. 0.737
3. 1.6
4. 0.889
Show the answer & worked solution
Answer: A.0.615
Cascade blocks multiply: $G_1G_2=(0.8)(2)=1.6$ . With unity feedback $T=\dfrac{G_1G_2}{1+G_1G_2}=\dfrac{1.6}{2.6}=0.615$ . Distractors leave the loop OPEN (report the product 1.6 itself), close the loop around $G_1$ only ( $G_1G_2/(1+G_1)=0.889$ ), and ADD the cascaded gains instead of multiplying ( $\tfrac{G_1+G_2}{1+G_1+G_2}=0.737$ ).
4
medium
A. Sampling, analog & digital filters, Z-transforms
A 16 kHz sinusoidal tone is sampled at $f_s = 10$ kHz — below the Nyquist rate (see spectrum sketch). The apparent (alias) frequency of the tone in the sampled signal is most nearly:
1. 4 kHz
2. 16 kHz
3. 1 kHz
4. 6 kHz
Show the answer & worked solution
Answer: A.4 kHz
Sampled images appear at $|f-kf_s|$ : $|16-10|=6$ kHz and $|16-20|=4$ kHz. Only 4 kHz lies in the baseband $[0,\ f_s/2=5\ \text{kHz}]$ , so the tone appears at 4 kHz. Distractors stop after one subtraction ( $f-f_s=6$ kHz, which still exceeds $f_s/2$ and must fold again), repeatedly subtract $f_s/2$ as if the spectrum replicated every half sample rate (1 kHz), and report the original 16 kHz tone as if sampling preserved it.
5
medium
A. Electrostatics, magnetostatics, electrodynamics & transmission lines
A transmission line with a characteristic impedance of 50 $\Omega$ is terminated in a purely resistive load of 25 $\Omega$ . The voltage reflection coefficient at the load is most nearly:
1. -0.33
2. 0.33
3. 0.50
4. -0.50
Show the answer & worked solution
Answer: A.-0.33
$\Gamma=\dfrac{Z_L-Z_0}{Z_L+Z_0}=\dfrac{25-50}{25+50}=-0.333$ — negative because the load is BELOW the line impedance (the reflected voltage inverts). Distractors flip the sign by subtracting line-minus-load ( $(Z_0-Z_L)/(Z_0+Z_L)$ ), normalize by $Z_0$ alone ( $(Z_L-Z_0)/Z_0$ ), and combine both errors ( $(Z_0-Z_L)/Z_0$ ).
6
medium
A. Power theory, transmission, transformers, motors & generators
An industrial plant draws 400 kW at 0.72 lagging power factor from its 480 V three-phase service. The rating of the shunt capacitor bank required to raise the power factor to 0.92 lagging is most nearly:
1. 215 kvar
2. 121 kvar
3. 170 kvar
4. 386 kvar
Show the answer & worked solution
Answer: A.215 kvar
$\theta_1=\cos^{-1}0.72=43.9^\circ$ , $\theta_2=\cos^{-1}0.92=23.1^\circ$ . $Q_C=P(\tan\theta_1-\tan\theta_2)=400(0.9639-0.426)=215$ kvar. Distractors cancel ALL of the reactive power ( $P\tan\theta_1$ , correcting to unity), use $\sin\theta$ instead of $\tan\theta$ , and report the remaining $Q_2=P\tan\theta_2$ instead of the difference.
7
medium
A. Time value of money, cost analysis & decision making
An engineer borrows $200,000 to be repaid in 15 equal end-of-year payments at 6% per year. The annual payment is most nearly:
1. $8,593
2. $13,333
3. $20,593
4. $12,000
Show the answer & worked solution
Answer: C.$20,593
$(A/P,0.06,15)=\dfrac{i(1+i)^n}{(1+i)^n-1}=0.10296$ , so $A=$ $200,000 $\times\ 0.10296=$ $20,593. Distractors: $P/n$ ignoring interest, the interest-only payment $P\cdot i$ , and the sinking-fund factor (A/F) used in place of A/P.
8
medium
A. Diodes, transistors, op-amps, instrumentation & power electronics
The ideal inverting op-amp circuit shown (±15 V supplies) has input resistor $R_1=10$ k $\Omega$ and a 47 k $\Omega$ feedback resistor. With $v_{in}=0.2$ V applied, the output voltage is most nearly:
1. -0.0426 V
2. 0.94 V
3. -0.94 V
4. 1.14 V
Show the answer & worked solution
Answer: C.-0.94 V
$v_{out}=-\dfrac{R_f}{R_1}v_{in}=-\dfrac{47}{10}\times0.2=-0.94$ V (the virtual ground forces the input current $v_{in}/R_1$ through $R_f$ ). Distractors drop the minus sign, apply the NON-inverting gain $1+R_f/R_1$ , and invert the resistor ratio ( $-R_1/R_f$ ).
9
medium
A. Transient/frequency response, resonance, Laplace & transfer functions
The RC network shown is a first-order low-pass filter with $R = 1.6$ kΩ and $C = 0.1$ µF, so $H(s)=\dfrac{1}{RCs+1}$ . Its corner (−3 dB) frequency is most nearly:
1. 6250 Hz
2. 39269.9 Hz
3. 1989.4 Hz
4. 994.7 Hz
Show the answer & worked solution
Answer: D.994.7 Hz
$RC=1.6\,k\Omega\times0.1\,\mu F=0.16$ ms, so $\omega_c=1/(RC)=6250$ rad/s and $f_c=\omega_c/(2\pi)=994.7$ Hz. Distractors report $\omega_c$ in rad/s as if it were Hz (skip the $2\pi$ ), multiply by $2\pi$ instead of dividing ( $(2\pi)^2$ too high), and use $\pi$ instead of $2\pi$ (doubling $f_c$ ).
10
medium
A. Topologies, models, routing/switching, and network security
The figure shows a network in which every node connects by its own link to a single central device. This topology and its key vulnerability are best described as:
1. Star — a central-hub failure disconnects every node
2. Bus — a fault on the shared backbone halts all traffic
3. Ring — a single cable break can split the loop
4. Full mesh — every node links directly to every other node
Show the answer & worked solution
Answer: A.Star — a central-hub failure disconnects every node
Every node homing to one central device is a star, whose center is its single point of failure. Distractors name other shapes' failure modes — a ring break, a bus-backbone fault, or a mesh's all-to-all links — none of which matches a hub-and-spoke drawing.

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