Question

In a 4-bit synchronous binary counter using JK flip-flops, what will be the state of the counter after 15 clock pulses, if it starts at 0000?

• A. 1111
• B. 0000
• C. 1010
• D. 0101

Hint:

Each pulse adds one; starting from 0, after 15 pulses the count is 15, which is 1111 in 4 bits.

Answer 1

The Correct Option is A

Step 1: A 4-bit synchronous binary counter counts up by one on every clock pulse. The JK flip-flops are wired so that the count follows the normal binary sequence 0000, 0001, 0010, and so on.

Step 2: The counter starts at the value 0000, which is decimal \(0\).

Step 3: After \(n\) clock pulses the state equals the starting value plus \(n\), taken modulo \(16\) because 4 bits can hold values \(0\) to \(15\). Here \(n = 15\).

Step 4: \[0 + 15 = 15\] Since \(15 < 16\), there is no wrap around yet. Decimal \(15\) in 4 bits is \(1111\).

Step 5: So after 15 pulses the counter shows 1111. It would roll back to 0000 only on the 16th pulse.