**Confirmed Errata – Study Guide for PE Power**

**Corrections in Problems section:**

Problem 2.4 g) Option B = $93.60

Problem 3.1 c) ….. It will be routed in a conduit with **3** other conductors….

Problem 4.5 l) Option A = -865 VAR

Problem 8.1 e) Option A = 382 feet

Problem 8.7 s) Option D = 0.029

Problem 8.7 t) Option D = 64.2 Ohm

Problem 9.3 e) Option D = 6000 A

**Corrections in Solutions section:**

Problem 2.4 g) Peak consumption = 50% , Mid-peak consumption = 20%

Problem 3.1 d) Correct Answer = B = 244A

Problem 4.5 j) Zeq = 8/ 6.9⁰

Problem 4.5 k) Correct Answer = B cos(θ) = cos(6.9⁰) = 0.99, P = 240 x 30 x 0.99 = 7147W

Problem 4.5 l) Correct Answer = 865VAR sin(θ) = sin(6.9⁰) = 0.12, Q = 240 x 30 x 0.12 = 865VAR

Problem 8.1 d) Z_eff = 1.2 × 0.75 + 0.063 × sin(41.40°)

Problem 8.1 e) Therefore, total circuit length shall be less than 2 x 191 ft = 382 ft to maintain voltage drop under 3%.

Problem 8.1 f) Z_eff = 0.10 × 0.75 + 0.054 × sin (41.40°)

Problem 8.5 h) Options B and D are both correct because given scenario represents a Single Line-Ground and Double Line-Line faults.

Problem 8.7 s) 𝛾 = √ (8.53 × 10 ⁻⁴ ) = 0.029

Problem 8.7 t) Zc = 64.2 Ohm