Free FE ECE practice problems

Ten real questions from our FE Electrical and Computer bank, drawn from 10 of its 17 study topics, with fully worked solutions and the same figures our members get. No signup, no email. Try each one first, then reveal the solution.

10 with figures10 topicsFull worked solutions

1
medium
A. DC circuit analysis: KCL/KVL, equivalents, Thevenin & Norton
A voltage divider made of $R_1=120\,\Omega$ and $R_2=300\,\Omega$ is supplied from a 24 V DC source, with its output taken across $R_2$ . A load $R_L=600\,\Omega$ is then connected across the output terminals, as shown. The voltage across the load is most nearly:
Figure — drag or pinch to move around
1. 15 V
2. 17.1 V
3. 20 V
4. 21.2 V
Show the answer & worked solution
Answer: A.15 V
The load is in parallel with $R_2$ , so the lower leg is $R_2\|R_L=\dfrac{300\cdot600}{300+600}=200\,\Omega$ , in series with $R_1$ (Resistors in Series and Parallel, FE Reference Handbook 10.6 p. 363). The source current is $I=\dfrac{24}{120+200}=0.075$ A, and Ohm's law across the lower leg gives $V_{out}=I\,(R_2\|R_L)=0.075\cdot200=15$ V. Distractors: the unloaded divider value, loading ignored ( $24\cdot300/420=17.1$ V); the load placed in series with $R_2$ instead of in parallel ( $24\cdot900/1020=21.2$ V); and the load taken as replacing $R_2$ rather than paralleling it ( $24\cdot600/720=20$ V).
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 spectra of the message tone and of the unmodulated carrier. The bandwidth occupied by the transmitted signal is most nearly:
Figure — drag or pinch to move around
1. 16 kHz
2. 32 kHz
3. 84 kHz
4. 116 kHz
Show the answer & worked solution
Answer: B.32 kHz
Amplitude modulation multiplies the carrier by $[1+a\,m_n(t)]$ , so a single tone produces the sum and difference frequencies: the transmitted signal holds the carrier at 100 kHz and one sideband on each side at $f_c\pm f_m=100\pm16$ kHz, i.e. 84 and 116 kHz. The occupied bandwidth is the width of that band, $116-84=2f_m=2(16)=32$ kHz: the handbook's "bandwidth of $x_{AM}(t)$ is $2W$ " with $W=f_m$ for one tone (Amplitude Modulation (AM), p. 380). Distractors: 16 kHz takes the message bandwidth $W=f_m$ itself and leaves out the factor 2 (the single-sideband result); 84 kHz reads modulation as simple addition and measures the band from the 16 kHz tone up to the 100 kHz carrier; and 116 kHz measures the band from zero up to the highest frequency present, the upper sideband.
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:
Figure — drag or pinch to move around
1. 0.615
2. 0.737
3. 0.889
4. 1.6
Show the answer & worked solution
Answer: A.0.615
Cascade blocks multiply, so the forward path is $G_1G_2=(0.8)(2)=1.6$ . The handbook's classical feedback model gives $Y/R=G_1G_2/(1+G_1G_2H)$ (Control Systems, p. 231); with unity feedback $H=1$ , so $T=\dfrac{G_1G_2}{1+G_1G_2}=\dfrac{1.6}{2.6}=0.615$ . Distractors: leave the loop open and report the forward product $G_1G_2=1.6$ itself; close the loop around $G_1$ alone, then multiply by $G_2$ ( $G_1G_2/(1+G_1)=0.889$ ); and add the cascaded gains instead of multiplying them ( $\tfrac{G_1+G_2}{1+G_1+G_2}=0.737$ ).
4
medium
A. Sampling, analog & digital filters, Z-transforms
An 18 kHz sinusoidal tone is sampled at $f_s = 10$ kHz (see the spectrum sketch). The apparent (alias) frequency of the tone in the sampled signal is most nearly:
Figure — drag or pinch to move around
1. 2 kHz
2. 3 kHz
3. 8 kHz
4. 18 kHz
Show the answer & worked solution
Answer: A.2 kHz
Sampled images appear at $|f-kf_s|$ : $|18-10|=8$ kHz and $|18-20|=2$ kHz. Only 2 kHz lies in the baseband $[0,\ f_s/2=5\ \text{kHz}]$ , so the tone appears at 2 kHz. Distractors stop after one subtraction ( $f-f_s=8$ 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 (3 kHz), and report the original 18 kHz tone as if sampling preserved it.
5
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:
Figure — drag or pinch to move around
1. $8,593
2. $13,333
3. $20,593
4. $25,333
Show the answer & worked solution
Answer: C.$20,593
The capital recovery factor (FE Reference Handbook 10.6 p. 235) converts a present amount into equal end-of-year payments: $(A/P,\ 6\%,\ 15)=\dfrac{i(1+i)^n}{(1+i)^n-1}=\dfrac{0.06\,(1.06)^{15}}{(1.06)^{15}-1}=0.10296$ , printed as 0.1030 in the handbook's 6% factor table. The annual payment is $200,000 × 0.10296 = $20,593. Distractors: $P/n$ , spreading the principal evenly and ignoring interest ($13,333); the sinking-fund factor $(A/F)$ read in place of $(A/P)$ ($8,593), which treats the loan as a future sum; and simple interest on the full principal for all 15 years, spread evenly, $P(1+in)/n$ ($25,333), which charges interest on money already repaid.
6
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 feedback resistor $R_f=47$ k $\Omega$ . With $v_{in}=0.2$ V applied, the output voltage is most nearly:
Figure — drag or pinch to move around
1. -1.14 V
2. -0.94 V
3. -0.165 V
4. -0.0426 V
Show the answer & worked solution
Answer: B.-0.94 V
The + input is grounded, so the ideal op-amp holds the − input at 0 V (a virtual ground): the input current $v_{in}/R_1$ flows on through $R_f$ and the output swings negative. The handbook's inverting case of the two-source circuit, $v_0=-(R_2/R_1)\,v_a$ with $R_2=R_f$ (Operational Amplifiers, p. 386), gives $v_{out}=-\dfrac{R_f}{R_1}v_{in}=-\dfrac{47}{10}\times0.2=-0.94$ V, well inside the ±15 V supplies. Distractors: keep the minus sign but use the non-inverting magnitude, $-(1+R_f/R_1)v_{in}=-1.14$ V; ignore the virtual ground and send the input current through $R_1$ and $R_f$ in series, $-v_{in}R_f/(R_1+R_f)=-0.165$ V; and invert the resistor ratio, $-(R_1/R_f)v_{in}=-0.0426$ V.
7
medium
A. Electrostatics, magnetostatics, electrodynamics & transmission lines
A lossless 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:
Figure — drag or pinch to move around
1. -0.50
2. -0.33
3. 0.11
4. 0.67
Show the answer & worked solution
Answer: B.-0.33
The handbook defines the reflection coefficient at the load (Lossless Transmission Lines, p. 374): $\Gamma=\dfrac{V^-}{V^+}=\dfrac{Z_L-Z_0}{Z_L+Z_0}=\dfrac{25-50}{25+50}=-0.333$ — negative because the load is BELOW the line impedance, so the reflected voltage is inverted (subtracting line-minus-load would wrongly give a positive value). Distractors: $Z_L-Z_0$ divided by $Z_0$ alone instead of by the sum $Z_L+Z_0$ (-0.50); $\Gamma$ squared (0.11): $|\Gamma|^2$ is the fraction of incident POWER reflected, not the voltage ratio asked for; and the transmission coefficient $1+\Gamma=2Z_L/(Z_L+Z_0)$ (0.67), the fraction of the incident voltage that appears across the load.
8
medium
A. Transient/frequency response, resonance, Laplace & transfer functions
The first-order RC low-pass filter shown has $R_1 = 1.5$ kΩ and $C = 0.1$ µF, and it drives a load resistor $R_2 = 10$ kΩ connected across its output. The corner (−3 dB) frequency of the loaded filter is most nearly:
Figure — drag or pinch to move around
1. 138 Hz
2. 1060 Hz
3. 1220 Hz
4. 7670 Hz
Show the answer & worked solution
Answer: C.1220 Hz
With the source replaced by a short, the capacitor sees $R_1$ and the load $R_2$ in parallel — the handbook's first-order RC low-pass cell (Analog Filter Circuits, p. 384): $R_P=\dfrac{R_1R_2}{R_1+R_2}=\dfrac{1.5\times10}{1.5+10}=1.304$ kΩ, so $\omega_c=\dfrac{1}{R_P C}=\dfrac{1}{(1304.3\ \Omega)(0.1\times10^{-6}\ \text{F})}=7667$ rad/s and $f_c=\dfrac{\omega_c}{2\pi}=1220$ Hz. Distractors: $R_1$ and $R_2$ added in series instead of combined in parallel, the $R_S=R_1+R_2$ the handbook uses in its RC high-pass and RL low-pass cells (138 Hz); the load left out, so the capacitor is charged through $R_1$ alone (1060 Hz, the unloaded filter); and $\omega_c = 1/(R_P C)$ in rad/s reported as if it were hertz (7670 Hz).
9
medium
A. Topologies, models, routing/switching, and network security
The figure shows the wiring of a five-station local-area network. The topology shown, and the single failure that would cut off every station, are best described as:
Figure — drag or pinch to move around
1. Star — failure of the central device
2. Ring — a break in any one link of the loop
3. Star — a break in any one station's cable
4. Bus — a break in the shared backbone cable
Show the answer & worked solution
Answer: A.Star — failure of the central device
Each station has its own cable to one central device (a hub or switch): that is a star. All traffic between stations passes through the center, so the central device is the single point of failure — lose it and every station is cut off. Distractors: a break in one station's cable is a star fault too, but it cuts off only that one station; the ring and bus options misread the drawing — there is no closed loop of neighbor-to-neighbor links and no single backbone cable that every station taps.
10
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:
Figure — drag or pinch to move around
1. 121 kvar
2. 170 kvar
3. 215 kvar
4. 386 kvar
Show the answer & worked solution
Answer: C.215 kvar
The real power stays at 400 kW; only the reactive power changes. From the power triangle (handbook AC Power, Complex Power, p. 368), $Q=P\tan\theta$ with $\theta=\cos^{-1}\text{pf}$ : $\theta_1=\cos^{-1}0.72=43.9^\circ$ and $\theta_2=\cos^{-1}0.92=23.1^\circ$ , so $Q_C=P(\tan\theta_1-\tan\theta_2)=400\ \text{kW}\times(0.9639-0.426)=215$ kvar. The 480 V service voltage does not enter. Distractors: all of the reactive power cancelled, $P\tan\theta_1$ (386 kvar) — that corrects to unity, not to 0.92; $\sin\theta$ used where $\tan\theta$ belongs, $P(\sin\theta_1-\sin\theta_2)$ (121 kvar); and the reactive power left after correction, $Q_2=P\tan\theta_2$ (170 kvar), reported instead of the change the bank supplies.

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