JESTION 1 Given that the PC contains 511, the address of the first instruction, and the partial list of opcodes as shown in Figure 1: | Address Content Instruction Meaning Load AC from memory Store AC to memory Add to AC from memory Subtract memory from AC 511 1824 0001 512 5822 0010 513 2823 0101 0110 822 21AB Figure 1. Partial List Opcodes 823 FFDE 824 AB12 Original memory contents i. Illustrate the execution of the instructions from memory content 512 and 513 in the diagram (511, 512 and 513). ii. Show the contents of the registers (Program Counter – PC, Accumulator – AC, and Instruction Register – IR).

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