# Quizzes on the Code

Choose the best answer: 1. What is the full-load current (FLC) rating of a 50 hp, 460V, 3-phase, Design B, squirrel-cage induction motor (motor nameplate FLA is 61.5A)? (a) 60A (b) 65A (c) 68A (d) 70A 2. What is the locked-rotor current (LRC) rating of a 50 hp, 460V, 3-phase, Design B, squirrel-cage induction motor? (a) 363A (b) 385A (c) 400A (d) 515A 3. What size time delay fuse (TDF) is required

1. What is the full-load current (FLC) rating of a 50 hp, 460V, 3-phase, Design B, squirrel-cage induction motor (motor nameplate FLA is 61.5A)?

(a) 60A
(b) 65A
(c) 68A
(d) 70A

2. What is the locked-rotor current (LRC) rating of a 50 hp, 460V, 3-phase, Design B, squirrel-cage induction motor?

(a) 363A
(b) 385A
(c) 400A
(d) 515A

3. What size time delay fuse (TDF) is required to start and protect a 50 hp, 460V, 3-phase, Design B, squirrel-cage induction motor per the following?

Part 1. Size the minimum (round down) TDF per Table 430-152.

(a) 110A
(b) 125A
(c) 150A
(d) 175A

Part 2. Size the next (round up) TDF per Sec. 430-52(c)(1), Ex. 1.

(a) 110A
(b) 125A
(c) 150A
(d) 175A

Part 3. Size the maximum TDF per Sec. 430-52(c)(1), Ex. 2(b).

(a) 110A
(b) 125A
(c) 150A
(d) 175A

Part 4. Start the motor with the smallest TDF per Table 430-151.

(a) 80A
(b) 90A
(c) 100A
(d) 110A

1. (b). Except for sizing the overload protection, Sec. 430-6(a)(1) requires the FLC in amps from Table 430-150 to be used for sizing the elements of a motor circuit.

Stallcup’s Code Loop: Sec. 430-6(a)(1) and Table 430-150.

2. (a). Sec. 430-7(a)(9) mentions the Design letters of motors as B, C, D, or E. Refer to Table 430-151 for the LRC in amps to be used as starting currents of motors.

3. Part 1. (a). Use Table 430-152 and Sec. 430-52(c)(1) to select the minimum size TDF rating. However, you can use any size TDF that will start and run the motor.

Step 1: First, find the FLA per Sec. 430-6(a)(1) and Table 430-150.

50 hp = 65A

Step 2: Then, find the proper percentage per Sec. 430-52(c)(1) and Table 430-152.

Minimum size = 175%

Step 3: Then, calculate the minimum size per Sec. 430-52(c)(1).

65A x 175% = 113.75A

Step 4: Now, select the proper size TDF per Sec. 240-6(a).

Solution: The minimum size time delay fuse is 110A.

Stallcup’s Code Loop: Sec. 430-6(a)(1), Table 430-150, Sec. 430-52(c)(1), Table 430-152, and Sec. 240-6(a).

Part 2. (b). Sec. 430-52(c)(1), Ex. 1 allows the next size TDF to be used above 113.75A (round up), which is 125A. Note: The round up method is the size most often used.

Part 3. (b). Table 430-152 and Sec. 430-52(c)(1), Ex. 2(b) are used to select the maximum size TDF rating. However, you can use any size TDF that will start and run the motor.

Step 1: First, find the FLA per Sec. 430-6(a)(1) and Table 430-150.

50 hp = 65A

Step 2: Then, find the proper percentage per Sec. 430-52(c)(1), Ex. 2(b) and Table 430-152.

Maximum size = 225%

Step 3: Then, calculate the maximum size per Sec. 430-52(c)(1).

65A x 225% = 146.25A

Step 4: Now, select the proper size TDF per Sec. 240-6(a).

Solution: The maximum size time delay fuse is 125A.

Stallcup’s Code Loop: Sec. 430-6(a)(1), Table 430-150, Sec. 430-52(c)(1), Ex. 2(b), Table 430-152, and Sec. 240-6(a).

Part 4. (a). Based on the answer in question (2), we know the LRC rating of this motor is 363A. We also know a TDF will normally hold five times its rating. Since 80A x 5 = 400A is greater than 363A, an 80A TDF will typically allow a motor with an LRC of 363A to start and run. (See Secs. 430-57, Ex. and 430-55.)

Note: Minimum and maximum sizes are used as a starting point to select the OCPD (smaller sizes can be selected).

TAGS: Quizzes