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Question

In a PLC STL, the sequence LD I2.0, AND I2.1, ORN I2.2, = Q2.0 is used. If I2.0 is ON, I2.1 is ON, and I2.2 is OFF, what does Q2.0 become?

This question was previously asked in
RRB NTPC 2025 Graduate Level CBT 1 Question Paper (25-Mar-2026) (Shift 3)
The correct answer is

Q2.0 will be ON

The given instruction sequence in PLC STL (Structured Text Language) is analyzed as follows:

  1. LD I2.0: Load the status of input I2.0 into the accumulator. In this case, since I2.0 is ON, the accumulator will be TRUE.
  2. AND I2.1: Perform a logical AND operation between the current value in the accumulator (which is TRUE) and input I2.1. Since I2.1 is also ON, the result of TRUE AND TRUE is TRUE. The accumulator remains TRUE.
  3. ORN I2.2: Perform a logical OR NOT operation with input I2.2. Since I2.2 is OFF, NOT I2.2 becomes TRUE. Thus, the result of TRUE OR TRUE is TRUE. The accumulator remains TRUE.
  4. = Q2.0: Assign the result in the accumulator (which is TRUE) to output Q2.0. Hence, Q2.0 will be ON.

Hence, based on the logic operations performed, the correct answer is:

Q2.0 will be ON.

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