(Solved): This is the full question. void ISR CNInterrupt (void) I CNcountert+; CNIF=0 ) uint8 ...

This is the full question.

void ISR CNInterrupt (void) I CNcountert+; $\angle \mathrm{CNIF}=0$ ) uint8 cNcounter; void main () CNcounter $=0$; $\mathrm{T}2\mathrm{CON}=0\times 0000$; PR2 $=99;$ T2IF $=0$; $\overline{\mathrm{CN}}3\mathrm{IE}=1;\_\mathrm{CNIF}=0;\_\mathrm{CNIP}=1;\_\mathrm{CNIE}=1;$ T2 CONbits. TON $\overline{1}=1$; while (1) i if $(-T2IF=1)\uparrow$ IATA $1=0$ - IATA1 $=1$ IT2 $\mathrm{IE}=0$ \} j) Compute the percentage of CPU time taken by the above ISR. Assume the following: - Each statement consumes one instruction cycle. - The RA1 is already configured as input.