FE Electrical and Computer Circuit Analysis (DC and AC Steady State) Practice Problems

The current FE Electrical and Computer specification assigns 11-17 of the 110 exam questions to Circuit Analysis (DC and AC Steady State). Form A contains 12 original problems in this area. The sample below is published in full, with the same worked-solution format used throughout the book.

A 120 V DC source is connected through a 10 Ω series resistor to a parallel section. One branch of that section is a single 60 Ω resistor; the other branch is a 12 Ω resistor in series with an 18 Ω resistor. The voltage across the 18 Ω resistor is most nearly:

(A)21.6 V
(B)24.0 V
(C)32.0 V
(D)48.0 V

Show worked solution

Answer: (D)

Only 80 V reaches the parallel section, and 48 V of that appears across the 18 Ω resistor.

Reduction starts at the far end, where the 12 Ω and 18 Ω resistors carry a common current and combine before the parallel step.

R_{ser i es} = 12 + 18 = 30 Ω

R_{p} = \frac{( 60 ) ( 30 )}{60 + 30} = 20 Ω

With 30 Ω of total loop resistance, the source drives 4 A, and the 10 Ω resistor absorbs part of the supply voltage before the parallel section.

I_{s} = \frac{120}{10 + 20} = 4 A

V_{p} = (4) (20) = 80 V

Splitting that section voltage between the two series resistors of the composite branch leaves the larger share on the 18 Ω element.

I_{b r an c h} = \frac{80}{30} = 2.67 A

V_{18} = (2.67) (18) = 48.0 V

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

Why the other choices appear

(A)Treats the two branches as being in series (60 + 30 = 90 Ω, total 100 Ω, 1.2 A), giving (1.2)(18) = 21.6 V.
(B)Inverts the current-division ratio, 4 A times 30/90 = 1.33 A, giving (1.33)(18) = 24.0 V.
(C)Solves for the drop across the 12 Ω resistor of the same branch, (2.67)(12) = 32.0 V.