Universal shift register behavior — CLEAR inactive and mode selects HIGH: In a typical universal shift register (for example, 74xx194/74xx195), the mode control inputs S1 and S0 select the operation, while CLEAR is an asynchronous reset that is usually active LOW. If CLEAR is held HIGH (inactive) and both S1 = 1 and S0 = 1, what operation takes place on the next active clock edge?

Introduction / Context:
Universal shift registers such as the 74xx194/195 families support multiple modes selected by two control lines, commonly labeled S1 and S0. These modes typically include hold (no change), shift right, shift left, and parallel load. An additional CLEAR input usually provides an asynchronous reset (often active LOW). Understanding how CLEAR and the mode selects interact is essential for correct design and troubleshooting.

Given Data / Assumptions:

• CLEAR is inactive when HIGH and active when LOW.
• Mode selects S1 and S0 determine operation per the standard truth table: 00 = hold, 01 = shift right, 10 = shift left, 11 = parallel load.
• Clocked operation: the selected action occurs on the appropriate clock transition.

Concept / Approach:
When CLEAR is HIGH, the device is not being forced into reset. The effective operation is then solely dictated by S1 and S0. The conventional truth table for universal shift registers assigns S1 = 1 and S0 = 1 to the parallel load function, meaning that the parallel data present at the D inputs is transferred into the internal flip-flops and then presented at the parallel outputs on the next clock event.

Step-by-Step Solution:

Verification / Alternative check:
Consult any standard 74xx194/195 mode table: the row with S1S0 = 11 shows “Parallel Load.” A lab check by tying CLEAR HIGH, S1 = S0 = HIGH, and stepping the clock while changing the parallel inputs will show the outputs updating to the input word each clock.

Why Other Options Are Wrong:

Common Pitfalls:
Confusing active-LOW CLEAR with an active-HIGH reset; misreading the mode table and swapping the meaning of S1/S0; forgetting that the load happens on the clock event, not immediately upon setting S1/S0.

Final Answer:
Parallel inputs are loaded and appear at the parallel outputs