Before the start of the PLC code including the functions below, the content of the memory byte MB8 was "1011'0100". If the content of the memory byte MBO is "1010'1010", and MB1 is "0001'1000", what will be the content of MB8 if input 10.0 is enabled? Hint: the INV function is a bit wise inversion function. %10.0 "TAG_IN" H H OR Byte INV Byte EN ENO EN ENO H %MBO %MB2 %MB2 "TAG VALUE1" IN1 OUT "TAG_RESULT "TAG RESULTIN %MB8 "TAG_RESULT %MB1 OUT-XNOR "TAG_VALUE2" IN2 O a. 1010'1010 O b. 1011'0100 ○ c. 0001'1000 ○ d. 0000'0001 Before the start of the PLC code including the functions below, the content of the memory byte MB8 was "1011'0100". If the content of the memory byte MBO is "1010'1010", and MB1 is "0001'1000", what will be the content of MB8 if input 10.0 is enabled? %10.0 "TAG_IN" OR Byte INV Byte EN ENO EN ENO H %MBO %MB2 "TAG VALUE1" IN1 OUT TAG RESULT %MB2 "TAG_RESULTIN %MB8 "TAG_RESULT OUT-XNOR" %MB1 "TAG_VALUE2" IN2 a. None of the choices O b. 1011'0100 ○ c. 1010'1010 O d. 1111'1110 O e. 0001'1000 Before the start of the PLC code including the functions below, the content of the memory byte MB8 was "1011'0100". If the content of the memory byte MBO is "1010'1010", and MB1 is "0001'1000", what will be the content of MB8 if input 10.0 is enabled? %10.0 "TAG_IN" OR Byte INV Byte EN ENO EN ΕΝΟ %MBO "TAG_VALUE1" %MB2 %MB2 %MB8 IN1 OUT "TAG RESULT "TAG RESULTIN "TAG_RESULT OUT-XNOR" %MB1 "TAG_VALUE2" IN2 Oa. None of the choices O b. 0001'1000 O c. 1011'0100 O d. 1010'1010 O e. 0000'0001

In the circuit shown in the figure below, C1=43µF, C2=33µF, C3=22µF, and a voltage Vab=43V is applied accross points a and b. After C1 is fully charged, the switch is thrown to the right. What is the final voltage on C3? Express your answer in units of V(Volts) using one decimal place. a C2 C3 b

Determine the voltage dropped on R3

Can you explain how you got the 25 for I when calculating p3?