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ANSWER:
The PLC ladder logic diagram as stated above is shown below.
| Input/output/timer | TYPE | |
| I:1/0 | NORMALLY CLOASE CONTACT | NORMALLY OPEN SWITCH S1 |
| T4:0 | ON DELAY TIMER | 10 sec delay timer TD1 |
| T4:0/DN | normally open contact | Done bit of timer T4:0 |
| O:2/0 | Output Coil | light L1 |
The program works as stated in question.
Rung 000: S1 is pressed i.e., I:1/0 becomes Normally closed and energises the timer T4:0.
Timer's Time delay = Preset value * timer base = 100* 0.1 = 10 seconds
Rung 001: After 10 sec, the accumulated value of timer becomes equal to its preset value. Then the done bit of timer, T4:0/DN becomes active and its NO contact becomes Normally closed which energises the ouptut L1.
when the switch S1 is opened, I:1/0 becomes NO again, timer de-energises and its Done bit becomes Low. Thus output L1 turns off.
LAD 2: X 000 은 EN S1 -TON Timer On Delay Timer 14:0 Time Base 0.1 Preset 100 Accum 0 10 sec TIMER TD1 DN T4:0/DN 0:2/0 001 DONE BIT OF TIMER TD1 L1 002 ENDS
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