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Arithmetic Circuits MCQs

Practice Arithmetic Circuits MCQs with four options, answers and explanations for focused Electronics & Telecommunication Engineering revision.

Question 1
Digital ElectronicsArithmetic Circuits

A 4-bit binary adder uses identical ripple-carry full-adder stages. If each stage requires 12 ns for its carry output to become valid after carry-in changes, approximately what is the worst-case carry propagation time through all four stages?

  • A12 ns
  • B24 ns
  • C36 ns
  • D48 ns
Report an error
Correct Answer: D. 48 ns
Explanation: In a ripple-carry adder, a carry may propagate serially through all four stages. The approximate worst-case carry delay is 4 × 12 ns = 48 ns.
Reference: Standard digital electronics: ripple-carry adder delay
Question 2
Digital ElectronicsArithmetic Circuits

The principal purpose of carry look-ahead logic in a binary adder is to?

  • AReduce carry propagation delay
  • BConvert binary to Gray code
  • CStore the carry permanently
  • DReduce the number of sum bits
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Correct Answer: A. Reduce carry propagation delay
Explanation: Carry look-ahead forms carry signals from generate and propagate terms in parallel, reducing the delay associated with a ripple-carry chain.
Reference: Standard digital electronics: carry look-ahead
Question 4
Digital ElectronicsArithmetic Circuits

For a 4-bit carry look-ahead group with bit propagate terms P0,P1,P2,P3, which product indicates that an input carry can propagate through all four bit positions?

  • AP0+P1+P2+P3
  • BP0P1P2P3
  • CP0⊕P1⊕P2⊕P3
  • DP0P3 only
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Correct Answer: B. P0P1P2P3
Explanation: A carry propagates through the entire group only if every bit position propagates it, so the group-propagate condition is the product P0P1P2P3.
Reference: Standard digital electronics: carry look-ahead group propagate
Question 5
Digital ElectronicsArithmetic Circuits

In unsigned binary subtraction A−B, a borrow is required at a bit position when the effective subtrahend at that position is?

  • AGreater than the available minuend bit after considering borrow-in
  • BAlways zero
  • CEqual to the minuend in every case
  • DIndependent of borrow-in
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Correct Answer: A. Greater than the available minuend bit after considering borrow-in
Explanation: A borrow is needed whenever the bit being subtracted, including any incoming borrow effect, exceeds the available minuend value at that position.
Reference: Standard digital electronics: binary subtraction
Question 6
Digital ElectronicsArithmetic Circuits

A 4-bit unsigned saturating adder is designed to clamp any result above its maximum representable value. What output should it produce when 1110₂ and 0101₂ are added?

  • A0011
  • B1111
  • C10011
  • D0000
Report an error
Correct Answer: B. 1111
Explanation: 14+5=19, which exceeds the 4-bit unsigned maximum of 15. A saturating adder clamps the result to 1111 rather than wrapping around.
Reference: Standard digital electronics: saturating arithmetic