Electrical Study Guide Ch.8 Pt.2

Front 1

1. The resistance of a conductor is affected by temperature change. This is called the__________.

(a) temperature correction factor

(b) temperature coefficient

(c) ambient temperature Factor

(d) none of these

Back 1

(b) temperature coefficient

Front 2

2. The total opposition to current in an alternating current circuit is a spruce and ohms and is called__________.

(a) impedance

(b) conductance

(c) reluctance

(d) resistance

Back 2

(a) impedance

Front 3

3. A 240V, 40A, single phase load is located 150 ft from an existing junction box. The junction boxes located 50 ft from the panel board and its wide with 4 AWG aluminum wire. The total resistance of the two 4 AWG conductors from the panel board to the junction box is approximately______.

(a) 0.03 ohms

(b) 0.04 ohms

(c) 0.05 ohms

(d) 0.08 ohms

Back 3

(c) 0.05 ohms

Hint 3

[Chapter 9, Table 9]

Assume a steel Raceway.

The AC resistance of four AWG aluminum conductor is 0.51 ohms and still come do it for 1,000 ft

R = (Resistance/1,000 ft) × Conductor Length

R = (0.51 ohms/1,000 ft) × (2 wires × 50 ft)

R = 0.051 ohms

Front 4

4. A load is located a 100 ft from a 230V power supply and is wired with 4 AWG aluminum conductors. What size copper conductors can be used to replace the aluminum conductors and not increase the conductor voltage drop?

(a) 8 AWG

(b) 6 AWG

(c) 2 AWG

(d) 1/0 AWG

Back 4

(b) 6 AWG

Hint 4

No calculations are required for this problem. From Chapter 9, Table 9, the AC resistance of 4 AWG aluminum in a steel Raceway (assumed) = 0.51 ohms. 6 AWG copper in a steel Raceway has an AC resistance of 0.49 ohms [Chapter 9, Table 9].

Front 5

5. A 40A, 240V, rated single phase load is 150 ft from a junction box and the junction boxes located 50 ft from a panel board ( for a total of 200 ft). If the voltage at the panel board is 240V, what's the minimum voltage recommended by the NEC at the 40A load?

(a) 117.70V

(b) 228.20V

(c) 232.80V

(d) 236.20V

Back 5

(c) 232.80V

Hint 5

The NEC recommend a maximum of 3% voltage drop on the branch circuit conductors, which calculates out to be: 240V × 0.03 = 7.20V. The minimum voltage at the load is 240V less 7.20V = 232.80V, or 240V × 0.97 = 232.80V.

Front 6

6. A 240V, 40A, single phase load is located 150 ft from an existing junction box. The junction box is located 50 ft from the panel board. When the 40A load is on, the voltage at the junction box is calculated to be 236V. The NEC recommends that the voltage drop for this branch circuit not exceed 3% of the 240V source (7.20V). What size copper conductors can be installed from the junction box to the load and still meet the NEC recommendation for branch circuits?

(a) 6 AWG

(b) 3 AWG

(c) 1 AWG

(d) 1/0 AWG

Back 6

(b) 3 AWG

Hint 6

Cmil = (2 × K × I × D)/ VD

K = 12.90, copper

I = 40A

D = 150 ft

VD = 240V - 236V = 4V already dropped.

VD = 7.20V - 4V = 3.20V

Cmil = (2 × 12.90 × 40A × 150 ft)/3.20V

Cmil = 48,375

[Chapter 9, Table 8 = 3 AWG]

Note: Table 310.15(B)(16), 110.14(C)(1)(a) 3 AWG rated 85A at 60°C, okay for 40A load.

Front 7

7. An existing junction boxes located 65 ft from the panel board and contains 4 AWG aluminum conductors. What size copper conductors can be used to extend the circuit 85 ft and supply of 50A, 208V load?

Note: applied the NEC recommended voltage drop limits.

(a) 10 AWG

(b) 8 AWG

(c) 6 AWG

(d) 4 AWG

Back 7

(d) 4 AWG

Hint 7

Step 1: determine the voltage drop of the existing conductors.

VD = (2 × K × I × D)/Cmil

K = 21.20, aluminum

I = 50A

D = 65 ft

Cmil = 41,740 [Chapter 9, Table 8]

VD = (2 × 21.20 × 50A × 65ft)/41,740 cmil

VD = 3.30V

Step 2: determine the voltage drop permitted for the extension.

Determine the voltage drop permitted for the extension by subtracting the voltage drop of the existing conductors from the permittedd voltage drop. The voltage drop recommended for the extension will be:

Recommended VD = 208V × 0.03

Recommended VD = 6.24V

6.24V - 3.30 = 2.94V

Step 3: determine the extended conductor size.

Cmil = (2 × K × I × D)/VD

K = 12.90 ohms, copper

I = 50A

D = 85 ft

VD = 2.94V

Cmil = (2 × 12.90 ohms × 50A × 85 ft)/2.94V

Cmil = 37,296 [ Chapter 9, Table 8 = 4 AWG]

Note: 4 AWG is rated for 70A at 60° C

[110.14(C)(1)(a) and Table 310.15(B)(16)]. Which is sufficient to carry the load.

Front 8

8. What's the voltage drop of two 3 AWG aluminum conductors that supply a 5hp, 115V, single phase motor that has a nameplate rating of 55A ? The motor is located 95 ft from the power supply.

(a) 3.25V

(b) 4.21V

(c) 6.24V

(d) 7.26V

Back 8

(b) 4.21V

Hint 8

VD = (2 × K × I × D)/Cmil

K= 21.20, aluminum

I = 55A ( for voltage drop calculation, use the nameplate current rating, not the FLC rating)

D = 95 ft

Cmil = 3 AWG, 52,620 cmil, [Chapter 9, Table 8]

VD = 230V × 3% = 6.90V

K = 12.90, copper

I = 50A

D = (52,620 cmil × 6.90V)/(1.732 × 12.90 × 50A)

D = 325 ft

Front 9

10. Two 8 AWG copper conductor Supply a 120V load that's located 225 ft from the panel board. What's the maximum load and amperes that can be applied to these conductors without exiting the NEC recommendation for conductor voltage drop?

(a) 0A

(b) 5A

(c) 10A

(d) 15A

Back 9

(c) 10A

Hint 9

I = (Cmil × VD Allowance)/(2 × K × D)

Cmil = 16,510 [Chapter 9, Table 8]

VD = 120V × 3%

VD = 3.60V

K = 12.90, copper

D = 225 ft

I = (16,510 cmil × 3.60V)/(2 × 12.90 × 225 ft)

I = 10A