FE Other Disciplines (NCEES Fundamentals of Engineering) — All Questions
28 questions
A resistor of 10 ohms carries a current of 2 A. What is the voltage across it?
- a.5 V
- b.12 V
- c.20 V✓
- d.0.2 V
Ohm's law gives V = I·R = 2 A x 10 ohm = 20 V. Voltage, current, and resistance are directly linked; doubling current at fixed resistance doubles the voltage drop.
Two 10-ohm resistors are connected in parallel. What is their equivalent resistance?
- a.20 ohms
- b.0.05 ohms
- c.10 ohms
- d.5 ohms✓
For two equal resistors in parallel, R_eq = R/2 = 10/2 = 5 ohms. In general 1/R_eq = 1/R1 + 1/R2; parallel combinations always yield a resistance smaller than the smallest branch.
A 4-ohm resistor carries a current of 3 A. What power does it dissipate?
- a.48 W
- b.12 W
- c.1.33 W
- d.36 W✓
Power P = I^2*R = 3^2 x 4 = 9 x 4 = 36 W. The equivalent forms P = VI and P = V^2/R give the same result.
What power is dissipated by a 6-ohm resistor with 12 V across it?
- a.2 W
- b.24 W✓
- c.48 W
- d.72 W
P = V^2/R = 12^2/6 = 144/6 = 24 W. When the voltage is known, P = V^2/R avoids first computing the current.
Three resistors of 2, 3, and 5 ohms are connected in series. What is the total resistance?
- a.3.3 ohms
- b.0.97 ohms
- c.10 ohms✓
- d.1 ohm
Series resistances add directly: 2 + 3 + 5 = 10 ohms. The same current flows through every series element.
A 6-ohm and a 3-ohm resistor are connected in parallel. What is the equivalent resistance?
- a.4.5 ohms
- b.18 ohms
- c.9 ohms
- d.2 ohms✓
R_eq = (R1*R2)/(R1 + R2) = (6 x 3)/(6 + 3) = 18/9 = 2 ohms. A parallel combination is always smaller than the smallest branch.
In a series circuit, 12 V is applied across a 4-ohm and an 8-ohm resistor. What is the voltage across the 8-ohm resistor?
- a.12 V
- b.4 V
- c.6 V
- d.8 V✓
Voltage divider: V8 = 12 x 8/(4 + 8) = 12 x 8/12 = 8 V. Voltage divides in proportion to each series resistance.
Two 4 uF capacitors are connected in series. What is the equivalent capacitance?
- a.0.5 uF
- b.4 uF
- c.8 uF
- d.2 uF✓
Capacitors in series combine like resistors in parallel: 1/C = 1/4 + 1/4, so C = 2 uF. Series capacitance is smaller than either capacitor.
A 2 uF and a 3 uF capacitor are connected in parallel. What is the total capacitance?
- a.1 uF
- b.6 uF
- c.1.2 uF
- d.5 uF✓
Parallel capacitors add directly: 2 + 3 = 5 uF. They share the same voltage, and their stored charges sum.
How much energy is stored in a 100 uF capacitor charged to 10 V? (E = (1/2)CV^2)
- a.0.5 mJ
- b.5 mJ✓
- c.10 mJ
- d.1 mJ
E = (1/2)CV^2 = 0.5 x 100x10^-6 x 10^2 = 0.5 x 100x10^-6 x 100 = 5x10^-3 J = 5 mJ. Capacitors store energy in their electric field.
What charge is stored on a 10 uF capacitor charged to 5 V? (Q = CV)
- a.15 uC
- b.2 uC
- c.0.5 uC
- d.50 uC✓
Q = CV = 10 uF x 5 V = 50 uC. Capacitance is the ratio of stored charge to voltage.
A 1 kohm resistor is in series with a 100 uF capacitor. What is the time constant of the RC circuit?
- a.1 s
- b.0.01 s
- c.0.1 s✓
- d.10 s
tau = RC = 1000 x 100x10^-6 = 0.1 s. After one time constant the capacitor charges to about 63.2% of the supply voltage.
A 2 H inductor is in series with a 4-ohm resistor. What is the time constant of the RL circuit?
- a.0.25 s
- b.2 s
- c.8 s
- d.0.5 s✓
tau = L/R = 2/4 = 0.5 s. A larger inductance or a smaller resistance slows the current's exponential rise.
How much energy is stored in a 2 H inductor carrying 3 A? (E = (1/2)LI^2)
- a.18 J
- b.3 J
- c.9 J✓
- d.6 J
E = (1/2)LI^2 = 0.5 x 2 x 3^2 = 0.5 x 2 x 9 = 9 J. Inductors store energy in their magnetic field.
What is the inductive reactance of a 0.1 H inductor at 60 Hz? (XL = 2*pi*f*L)
- a.0.6 ohms
- b.6 ohms
- c.377 ohms
- d.37.7 ohms✓
XL = 2*pi*f*L = 2*pi*(60)(0.1) = 37.7 ohms. Inductive reactance rises with frequency, so inductors impede high-frequency currents.
What is the capacitive reactance of a 100 uF capacitor at 60 Hz? (Xc = 1/(2*pi*f*C))
- a.26.5 ohms✓
- b.377 ohms
- c.0.0377 ohms
- d.60 ohms
Xc = 1/(2*pi*f*C) = 1/(2*pi*60*100x10^-6) = 1/0.0377 = 26.5 ohms. Capacitive reactance falls as frequency rises.
A series RLC circuit has L = 1 mH and C = 1 uF. What is its resonant frequency? (f = 1/(2*pi*sqrt(LC)))
- a.50.3 kHz
- b.5.03 kHz✓
- c.159 Hz
- d.1.59 kHz
sqrt(LC) = sqrt(10^-3 x 10^-6) = sqrt(10^-9) = 3.16x10^-5; f = 1/(2*pi*3.16x10^-5) = 5.03 kHz. At resonance the inductive and capacitive reactances cancel.
A sinusoidal voltage has a peak value of 170 V. What is its RMS value? (Vrms = Vpeak/sqrt(2))
- a.85 V
- b.170 V
- c.120 V✓
- d.240 V
Vrms = 170/sqrt(2) = 170/1.414 = 120 V. This is why standard 120 V mains has a peak near 170 V.
A series AC circuit has a resistance of 3 ohms and a reactance of 4 ohms. What is the magnitude of its impedance?
- a.7 ohms
- b.3.5 ohms
- c.5 ohms✓
- d.1 ohm
|Z| = sqrt(R^2 + X^2) = sqrt(3^2 + 4^2) = sqrt(25) = 5 ohms. Resistance and reactance combine as perpendicular components (a 3-4-5 triangle).
An ideal transformer has 1000 primary turns and 100 secondary turns. If 120 V is applied to the primary, what is the secondary voltage?
- a.120 V
- b.24 V
- c.1200 V
- d.12 V✓
Vs/Vp = Ns/Np, so Vs = 120 x (100/1000) = 12 V. A step-down turns ratio lowers voltage while raising available current.
An ideal step-down transformer delivers 12 V at 2 A to its load from a 120 V source. What primary current does it draw?
- a.0.2 A✓
- b.0.02 A
- c.20 A
- d.2 A
Ideal transformers conserve power: Vp*Ip = Vs*Is, so Ip = (12 x 2)/120 = 0.2 A. Stepping voltage down steps current up.
An AC load draws 5 A at 120 V with a power factor of 0.8. What is the real (average) power consumed?
- a.600 W
- b.480 W✓
- c.750 W
- d.384 W
Real power P = V*I*cos(theta) = 120 x 5 x 0.8 = 480 W. The power factor is the cosine of the phase angle between voltage and current.
In a single loop, a 12 V source drives current through two resistors. If one drops 5 V, what is the voltage across the other by Kirchhoff's voltage law?
- a.5 V
- b.12 V
- c.7 V✓
- d.17 V
KVL requires the drops to sum to the source: 12 - 5 = 7 V across the second resistor. The algebraic sum of voltages around any closed loop is zero.
At a circuit node, 3 A and 2 A flow in while one branch carries current out. By Kirchhoff's current law, what is the outgoing current?
- a.5 A✓
- b.6 A
- c.2.5 A
- d.1 A
KCL requires current in to equal current out: 3 + 2 = 5 A leaving the node. Charge is conserved, so none accumulates at a node.
A 10 A current enters two equal parallel resistors. How much current flows through each?
- a.5 A✓
- b.2.5 A
- c.10 A
- d.0 A
Equal parallel branches split the current equally: 10/2 = 5 A each. In a current divider, more current takes the lower-resistance path.
What current flows through a 12-ohm resistor connected across a 24 V source?
- a.12 A
- b.288 A
- c.0.5 A
- d.2 A✓
Ohm's law: I = V/R = 24/12 = 2 A. Current is proportional to voltage and inversely proportional to resistance.
Three identical 30-ohm resistors are connected in parallel. What is their equivalent resistance?
- a.10 ohms✓
- b.15 ohms
- c.30 ohms
- d.90 ohms
For n equal resistors in parallel, R_eq = R/n = 30/3 = 10 ohms. Adding parallel paths lowers the total resistance.
A resistor carries an AC current with RMS value 2 A through 10 ohms. What average power does it dissipate?
- a.40 W✓
- b.20 W
- c.10 W
- d.80 W
Average power in a resistor uses RMS values: P = Irms^2*R = 2^2 x 10 = 40 W. RMS is defined precisely so this DC-like power formula holds.