The Jump Instruction

In our schematic programs, the "jump" instruction loaded the PC with a 32-bit address.

How does a 32-bit instruction specify a 32-bit address? Some of the instruction's bits must be used for the op-code. Here is the assembly language form of the jump instruction.

j target # after a delay of one machine cycle, # PC 
Here is the machine language form of the instruction:
6 26 000010 00000000000000000000000000 -- fields of the instructuion opcode target -- meaning of the fields
There is room in the instruction for a 26-bit address. The 26-bit target address field is transformed into a 32-bit address. This is done at run-time, as the jump instruction is executed.
Instructions always start on an address that is a multiple of four (they are word-aligned). So the low order two bits of a 32-bit instruction address are always "00". Shifting the 26-bit target left two places results in a 28-bit word-aligned address (the low-order two bits become "00".)
After the shift, we need to fill in the high-order four bits of the address. These four bits come from the high-order four bits in the PC. These are concatenated to the high-order end of the 28-bit address to form a 32-bit address.
For example, here is the machine language for the instruction that jumps to location 0x5B145188 . Say that the instruction is located at address 0x56767250 .
QUESTION 5:
While this is going on, what address is in the PC?