FE Other Disciplines Basic Electrical Engineering Practice Problems

The current FE Other Disciplines specification assigns 6-9 of the 110 exam questions to Basic Electrical Engineering. Form A contains 7 original problems in this area. The sample below is published in full, with the same worked-solution format used throughout the book.

A 24 V DC source drives a 150 $Ω$ resistor in series with a parallel combination of a 470 $Ω$ resistor and a 330 $Ω$ resistor. The voltage across the 150 $Ω$ resistor is most nearly:

(A)3.79 V
(B)5.81 V
(C)6.55 V
(D)10.5 V

Show worked solution

Answer: (D)

Because the parallel pair reduces to 194 $Ω$ , the 150 $Ω$ series element claims about 10.5 V of the 24 V supply.

Parallel branches reduce first, since only their equivalent shares the loop with the 150 $Ω$ element.

R_{p} = \frac{470 \times 330}{470 + 330} = 193.9 Ω

Loop current is fixed by the total series resistance seen by the source.

R_{T} = 150 + 193.9 = 343.9 Ω

I = \frac{24}{343.9} = 0.0698 A

Series element voltage follows from that shared current, with the balance appearing across the parallel pair.

V_{150} = 0.0698 \times 150 = 10.5 V

V_{p} = 24 - 10.5 = 13.5 V

FE Reference Handbook — Electrical and Computer Engineering: Resistors in Series and Parallel, Voltage Division

Why the other choices appear

(A)Adds the two branch resistors as if in series, 470 + 330 = 800 ohm, giving 24(150)/950 = 3.79 V.
(B)Ignores the 330 ohm branch entirely, giving 24(150)/(150 + 470) = 5.81 V.
(C)Uses the arithmetic mean of the branch resistors, 400 ohm, giving 24(150)/550 = 6.55 V.