FE Electrical and Computer Mathematics Practice Problems

The current FE Electrical and Computer specification assigns 11-17 of the 110 exam questions to Mathematics. 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.

Two feeder branches deliver phasor currents $I_{1} = 12∠2 5^{\circ}$ A and $I_{2} = 9∠ - 4 0^{\circ}$ A into a common node, both referred to the same voltage reference. The magnitude of the total current leaving the node is most nearly:

(A)11.6 A
(B)15.0 A
(C)17.8 A
(D)21.0 A

Show worked solution

Answer: (C)

Added in rectangular form, the two branch currents give a resultant of 17.8 A. The 65 degree phase separation is what holds the total below the 21 A arithmetic sum.

Phasor addition requires rectangular components, since the two branch currents differ in phase by 65 degrees.

I_{1} = 12 (cos 2 5^{\circ} + j sin 2 5^{\circ}) = 10.876 + j 5.071 A

I_{2} = 9 (cos 4 0^{\circ} - j sin 4 0^{\circ}) = 6.894 - j 5.785 A

Component-wise addition leaves a resultant almost aligned with the real axis, the imaginary parts nearly cancelling.

I_{1} + I_{2} = 17.770 - j 0.714 A

Magnitude of the resultant follows from its rectangular components.

∣ I_{1} + I_{2} ∣ = 17.77 0^{2} + 0.71 4^{2} = 17.8 A

FE Reference Handbook — Mathematics: Complex Numbers, Polar and Rectangular Forms

Why the other choices appear

(A)Subtracts the phasors instead of adding them, giving sqrt(225 - 91.29) = 11.56 A.
(B)Treats the two currents as orthogonal, giving sqrt(12^2 + 9^2) = 15.0 A.
(D)Adds the magnitudes arithmetically, 12 + 9 = 21 A, ignoring the 65 degree phase separation.