Let k = 2n. A circuit is built by giving the output of an n-bit binary counter as input to an n-to-2n bit decoder. This circuit is equivalent to a A) k-bit binary up counter. B) k-bit binary down counter. C) k-bit ring counter. D) k-bit Johnson counter.

Official answer: C) k-bit ring counter.. Explanation: An n-bit binary counter has 2n = k distinct states (0, 1, , k-1). An n-to-2n decoder takes n inputs and activates exactly one of its 2n = k output lines corresponding to the binary value of the input. As the n-bit binary counter cycles through all its k counts, the decoder activates output line 0, then line 1, , up to line k-1, and then wraps around to line 0. At any point in time, exactly one of the k lines is active (HIGH), and this active state rotates sequentially across all k outputs. This behavior matches a k-bit ring counter, which consists of k flip-flops with a single circulating active bit (`1`). Therefore, (C) is the correct option.

Digital Logic · Combinational and Sequential Circuits

Official IIT answer key · IIT Kharagpur · Audited Aug 2026

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GATE CSE 2014 Set 2 Q7 · MCQ · 1 mark

Let $k = 2^n$ . A circuit is built by giving the output of an $n$ -bit binary counter as input to an $n$ -to- $2^n$ bit decoder. This circuit is equivalent to a

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Combinational and Sequential Circuits

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How GATE tests Combinational and Sequential Circuits

GATE extensively tests combinational and sequential circuit design through both analytical tracing and formal synthesis. In combinational circuits, questions focus on multiplexer tree implementations, decoder-to-memory interfacing, and Boolean function realization using universal MUX/decoder blocks. In sequential logic, the focus is on synchronous counter state tracing, excitation equation derivation (for D, T, and JK flip-flops), non-standard sequence flip-flop counting, ripple counter frequency division, and FSM equivalence minimization.

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Digital Logic · Combinational and Sequential Circuits

Official IIT answer key · IIT Kharagpur · Audited Aug 2026

Take topic test

GATE CSE 2014 Set 2 Q7 · MCQ · 1 mark

Let $k = 2^n$ . A circuit is built by giving the output of an $n$ -bit binary counter as input to an $n$ -to- $2^n$ bit decoder. This circuit is equivalent to a

Select an option to check

Key concept

Combinational and Sequential Circuits

Step-by-step verified solution

Examiner trap & misconception

Topic notes

How GATE tests Combinational and Sequential Circuits

Full Combinational and Sequential Circuits guide →

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Independent practice explanation verified by the GateAI team. GATE is conducted by the IITs; question text follows the official paper.

Let $k = 2^n$ . A circuit is built by giving the output of an $n$ -bit binary counter as input to an $n$ -to- $2^n$ bit decoder. This circuit is equivalent to a

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Let $k = 2^n$ . A circuit is built by giving the output of an $n$ -bit binary counter as input to an $n$ -to- $2^n$ bit decoder. This circuit is equivalent to a

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Let k=2nk = 2^nk=2n. A circuit is built by giving the output of an nnn-bit binary counter as input to an nnn-to-2n2^n2n bit decoder. This circuit is equivalent to a

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Let k=2nk = 2^nk=2n. A circuit is built by giving the output of an nnn-bit binary counter as input to an nnn-to-2n2^n2n bit decoder. This circuit is equivalent to a

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More on Combinational and Sequential Circuits

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Let $k = 2^n$ . A circuit is built by giving the output of an $n$ -bit binary counter as input to an $n$ -to- $2^n$ bit decoder. This circuit is equivalent to a

Let $k = 2^n$ . A circuit is built by giving the output of an $n$ -bit binary counter as input to an $n$ -to- $2^n$ bit decoder. This circuit is equivalent to a