Free practice problems & solutions for the PE Electrical Engineering (Power) exam

Question 1.

Calculate the maximum percentage voltage drop in a single-phase 120 Volt, 20 Amp circuit with #12 uncoated copper wires in steel conduit, which feeds a 750 VA load operating at 95% power factor, if the length of the circuit is 75′, and there are no other items connected to that circuit.

1.0 %
1.5 %
4.5 %
2.2 %

Answer 1: b: 1.5%

Solution:

Apply the formula for voltage drop:

D = (2 x (W – 1)^-1 x (R x F / 100 + X x SIN(ACOS(F / 100))) x L x S) / (10 x V²)

Where:

D = Voltage drop as a percentage of the circuit voltage (V)

W= Number of wires in the circuit = 2 (for single phase)

R= Resistance of conductor per 1,000ft (Ω) – See NEC Table 9 = 1.98

X= Reactance of conductor per 1,000ft (Ω) – See NEC Table 9 = 0.068

F= Power factor of load (%) = 95

L= Length of conductor (ft) = 75

S= Apparent power of load (VA) = 750

V= circuit voltage (V) = 120

I = Line current (A)

So now we have:

D = (2 x (2 – 1)^-1 x (1.98 x 95 / 100 + 0.068 x SIN(ACOS(95 / 100))) x 75 x 750) / (10 x 120²)

D = (2 x 1 x (1.88 + 0.068 x 0.312) x 56,250) / (144,000)

D = 1.5 %

Question 2.

Which statement below most accurately describes the function of a K rated transformers?

K rated transformers are used to supply loads with a high power factor
K rated transformers are used in situations where a higher efficiency is required
K rated transformers are used to supply loads where a low inrush current is required
K rated transformers are used to supply non-linear loads

Answer 2: d. K rated transformers are used to supply non-linear loads

K rated transformers are designed to handle non-linear loads, i.e. loads with a significant amount of harmonic distortion. In standard transformers, loads with high harmonic distortion can cause overheating due to the excessive harmonic currents that are generated in such applications. the higher the K factor, the more non-linear loads the transformer can handle. K rated transformers do not need to be derated as much as standard transformers due to the presence of harmonics and are equipped with double-size neutral bus due to the higher currents that can be generated in the neutral due to these type of loads.

Question 3.

Which of the following locations require the use of GFI type receptacles?

In a cabinet underneath a sink to supply a garbage disposal
In location containing hazardous materials
Above a kitchen cabinet and adjacent to a sink.
In an electrical or mechanical room

Answer 3: c. Above a kitchen cabinet and adjacent to a sink

GFI (ground fault interruption) receptacles are required to be installed within 6′ of faucets unless the receptacle is enclosed by a cabinet. GFI receptacles are also required in other locations where there is a likelyhood of exposure of the receptacle or utilization equipment to water. This type of receptacle causes the circuit breaker to trip automatically when a ground fault current is detected in order to protect persons from harmful ground faults that are likely to occur in the presence of water.

Question 4.

Calculate the available fault current on the secondary side of a 1,500 VA, 4,160-480 Volt utility transformer with an impedance of 5.75%. Assume an infinite available fault current on the primary side of the transformer. Assume that there are no motors that will contribute to the fault current value.

31,300 Amps
22,500 Amps
54,600 Amps
12,500 Amps

Answer 4: a. 31,300 Amps.

Solution:

Use the formula for calculation available fault current for a 3-phase transformer:

F = 100,000 x SQRT(3) / 3 x (S / (V x Z x (1 – P / 100))) + 4 x M + 5 x N

Where:

F= Maximum available fault current at the secondary terminals of transformer

S= Transformer apparent power rating = 1,500 KVA

V= Voltage of transformer secondary = 480 Volts

Z= Transformer percent impedance = 5.75 %

P= Maximum variance of transformer impedance = 0 (since not given)

M = Total full-load-amps of induction motors that can contribute to the fault (A) = 0

N= Total full-load-amps of synchronous motors that can contribute to the fault (A) = 0

F = 100,000 x SQRT(3) / 3 x (1,500 / (480 x 5.75 x (1 – 0 / 100))) + 4 x 0 + 5 x 0

F = 31,300 Amps