Answer the questions, in sequence, for the jump-to- label program in Figure 2. A

Question

Answer the questions, in sequence, for the jump-to- label program in Figure 2. Assume all switches are turned off after each operation. Switch S3 is turned on. Will output PL1 be energized? Why? Switch S2 is turned on first, then switch S5 is turned on. Will output PL4 be energized? Why? Switch S3 is turned on and output PL1 is energized. Next, switch S2 is turned on. Will output PL1 be energized or de-energized after turning on switch S2? Why? All switches are turned on n order according to the following sequence: S1, S2, S3, S5, S4. Which pilot lights will turn on?

Explanation / Answer

A- Yes after switching S3 the pl1 output will be energized, as the program is executing normally. Here is no use of JMP instruction till the rung 2 is false. Hence the switch S3 will work normally. Pressing the S3 will complete the circuit 3 and the pl1 is energized.

B- No. When we turn ON the switch S2, it will activate the JMP( jump to label instruction) which means that the program will skip the rungs between jmp and lbl(label) instruction. It will start the execution from LBL row onwards. Hence, there is no use of pressing S5 when the JMP instruction is active.

C- Yes, It will remain active. When we press S3 it will energize the pl1. After which, if we will press S2 , it will activate JMP, which means the next adress to be executed is from LBL. Hence pressing S2 will not affect the PL1 output. JMP only provide the adress for the next execution.

D- PL3 & PL2 will be energized only. pressing s1 will activate pl3. pressing S2 will activate JMP. after JMP is activated , the program will skip to LBL and start execution from s4. Hence ther is no use of pressing s3 and s5.

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