Output addressing scheme in PLC Programming - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to assign output address %Q0.1 to the motor start coil.

PLC Programming

MotorStart := [1];

Drag options to blanks, or click blank then click option'

A%Q1.0

B%I0.1

C%M0.1

D%Q0.1

Attempts:

3 left

💡 Hint

Common Mistakes

Using input address %I0.1 instead of output %Q0.1

Using memory bit %M0.1 which is not an output

Using wrong output byte like %Q1.0

The output address %Q0.1 correctly refers to the first output byte, bit 1, which is used for output coils like MotorStart.

2fill in blank

medium

Complete the code to turn on the output at address %Q2.3.

PLC Programming

Outputs[[1]] := TRUE;

Drag options to blanks, or click blank then click option'

A2.3

B%Q2.3

C3.2

DQ2.3

Attempts:

3 left

💡 Hint

Common Mistakes

Using %Q2.3 inside array index which is invalid

Swapping byte and bit numbers like 3.2

Using Q2.3 without % which is not standard

When using an array or bit addressing, the index is given as '2.3' meaning byte 2, bit 3. The %Q prefix is not used inside the array index.

3fill in blank

hard

Fix the error in the output assignment to %Q1.7.

PLC Programming

OutputBit := [1];

Drag options to blanks, or click blank then click option'

A%I1.7

B%Q1.7

C%M1.7

DQ1.7

Attempts:

3 left

💡 Hint

Common Mistakes

Using input address %I1.7 instead of output

Using memory bit %M1.7 instead of output

Omitting % in address

The correct output address uses %Q prefix. %I is input, %M is memory bit, and Q1.7 without % is invalid.

4fill in blank

hard

Fill both blanks to create a mask for output byte 3 to set bit 5.

PLC Programming

Mask := 1 [1] 5;

Drag options to blanks, or click blank then click option'

A<<

B>>

C+

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using right shift >> which moves bits the wrong way

Using subtraction instead of addition

Mixing up operators order

To set bit 5, shift 1 left by 5 bits (1 << 5).

5fill in blank

hard

Fill all three blanks to create a dictionary mapping output names to addresses with condition bit 2 set.

PLC Programming

outputs = { [1]: [2] for [3] in output_list if ([2] & (1 << 2)) != 0 }

Drag options to blanks, or click blank then click option'

Aname

Baddress

Coutput

Daddr

Attempts:

3 left

💡 Hint

Common Mistakes

Using same variable for key and loop variable

Using wrong variable names causing errors

Not matching variable names in condition

We map each output name to its address. The loop variable is 'output'. The condition checks if bit 2 of address is set.

Practice

1. What does the output addressing scheme in a PLC primarily specify?

easy

A. Which physical output device the PLC controls

B. The input device connected to the PLC

C. The programming language used in the PLC

D. The power supply voltage for the PLC

Solution

1. Step 1: Understand output addressing purpose

   Output addressing tells the PLC which output device to activate or control.

2. Step 2: Differentiate from inputs and other settings

   It does not specify inputs, programming language, or power supply.

3. Final Answer:

   Which physical output device the PLC controls -> Option A

4. Quick Check:

   Output addressing = physical output device [OK]

Hint: Output addressing points to the device PLC controls [OK]

Common Mistakes:

• Confusing output with input addressing
• Thinking it defines programming language
• Mixing it with power supply details

2. Which of the following is a correct format for output addressing in a PLC?

easy

A. I:3/7

B. O:2/5

C. M:1/0

D. T:4/2

Solution

1. Step 1: Identify output address format

   Output addresses usually start with 'O' followed by module and point numbers, e.g., O:2/5.

2. Step 2: Recognize other prefixes

   'I' is for inputs, 'M' for memory bits, 'T' for timers, so they are incorrect for output addressing.

3. Final Answer:

   O:2/5 -> Option B

4. Quick Check:

   Output address starts with 'O' [OK]

Hint: Output addresses start with 'O:' followed by module/point [OK]

Common Mistakes:

• Choosing input or memory addresses as output
• Mixing timer addresses with outputs
• Ignoring the prefix letter

3. Given the output address O:1/3, which physical output does it refer to?

medium

A. Module 1, output point 3

B. Module 3, output point 1

C. Input module 1, point 3

D. Memory bit 1, point 3

Solution

1. Step 1: Parse the output address format

   O:1/3 means output module 1, point 3 (the third output on module 1).

2. Step 2: Exclude other interpretations

   It is not a memory bit, not an input module, and the format is module/point (not point/module).

3. Final Answer:

   Module 1, output point 3 -> Option A

4. Quick Check:

   O:1/3 = module 1, point 3 [OK]

Hint: Format is O:module/point, read left to right [OK]

Common Mistakes:

• Swapping module and point numbers
• Confusing output with input addresses
• Ignoring the prefix 'O:'

4. Identify the error in this output address: O:2-4

medium

A. Output point cannot be 4

B. Module number cannot be 2

C. Incorrect separator; should use '/' instead of '-'

D. Prefix 'O' is invalid

Solution

1. Step 1: Check the separator in the address

   Output addresses use a slash '/' to separate module and point, not a dash '-'.

2. Step 2: Validate other parts

   Module 2 and point 4 are valid numbers, and 'O' prefix is correct for output.

3. Final Answer:

   Incorrect separator; should use '/' instead of '-' -> Option C

4. Quick Check:

   Use '/' between module and point [OK]

Hint: Use '/' not '-' between module and point [OK]

Common Mistakes:

• Using dash instead of slash
• Thinking module or point numbers are invalid
• Misunderstanding the prefix letter

5. You have a PLC with 3 output modules. Module 0 has 8 points, module 1 has 8 points, and module 2 has 4 points. Which output address correctly refers to the last output point of module 2?

hard

A. O:1/7

B. O:2/4

C. O:3/0

D. O:2/3

Solution

1. Step 1: Understand module points numbering

   Points start at 0, so module 2 with 4 points has points 0,1,2,3.

2. Step 2: Identify last point of module 2

   The last point is 3, so address is O:2/3. O:2/4 is invalid (point 4 does not exist).

3. Step 3: Exclude other options

   O:3/0 is invalid (module 3 does not exist). O:1/7 is last point of module 1, not module 2.

4. Final Answer:

   O:2/3 -> Option D

5. Quick Check:

   Last point = points - 1 = 3 [OK]

Hint: Last point = total points minus one [OK]

Common Mistakes:

• Counting points starting at 1 instead of 0
• Using invalid module numbers
• Confusing module and point numbers