[Solved] T Flip Flop MCQ [Free PDF] - Objective Question Answer for T Flip Flop Quiz - Download Now! (2024)

1. Only I
2. Both I and II
3. Only II
4. Neither I nor II

Both statements are correct.

Explanation - The T-type flip-flop is a type of digital circuit that has two stable states, often labeled as "0" and "1". The output state of the flip-flop changes (or toggles) on each clock pulse when the T input is high (1). This means that if the input T is 1, the output will change its state with each clock pulse. On the other hand, if the T input is low (0), the output state will remain the same, and the flip-flop will retain its current state.

Therefore, statement I is correct in that if T = 1, the state of the flip-flop changes on the rising edge of the clock pulse. Statement II is also correct in that if T = 0, the state of the flip-flop remains the same, regardless of the clock pulse.

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1. Both trigger and toggle
2. Toggle
3. Trigger
4. Either trigger or toggle

Concept:

• T flip-flop is also knownas Toggle flip-flop.
• It can be constructed from a JK flip-flop by connecting both inputs J and K together.
• Serial decoding, comparison and timing functions can be accomplished using flip-flops.
• Flip-flops has 2 generally stable states, is SET and RESET.
• A flip-flop is said to be in "high state" or logic 1 or SET state" when Q=1.
• A flip-flop is said to be in "lowstate" or logic 0or RESET state" when Q=0.
• Flip-flop are the fundamental components of shift registers and counters.
• T- flip-flop when CLKis LOW, T=0 ; no change in the input.
• T- flip-flop when CLK is HIGH, T=1; The output is toggle.

So, Toggle is the correct option.

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1. 200 Hz
2. 50 Hz
3. 800 Hz
4. None of the above

Concept:

If we pass the input signal to a single T-flip flop, we will get half of the frequency at the output.

Similarly, when we pass the input signal into an n-bit flip flop counter, the output frequency (fout) will be:

\({{\rm{f}}_{{\rm{out}}}} = \frac{{{\rm{Input\;Frequency}}}}{{{2^{\rm{n}}}}}\)

Application:

Given Input frequency f = 200 Hz

Four T flip flops connected in cascade mode (n = 4)

\({{\rm{f}}_{{\rm{out}}}} = \frac{{{\rm{Input\;frequency}}}}{{{2^{\rm{n}}}}} = \frac{{2{\rm{00~Hz}}}}{{{2^4}}}\)

fout= 12.5 Hz

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1. (X̅ ∧ Q̅) ∨ (Y̅ ∧ Q)
2. (X̅ ∧ Q) ∨ (Y̅ ∧ Q̅)
3. (X ∧ Q̅) ∨ (Y ∧ Q)
4. (X ∧ Q̅) ∨ (Y̅ ∧ Q)

Concept –

For finding the expression of Next state, a truth table needs to be constructed.

Equation for Q_n+1 is made by writing all values of X, Y, Q for which Q_n+1 = 1

\({{\rm{Q}}_{{\rm{n}} + 1}}{\rm{\;}} = {\rm{\;\bar X\;\bar Y\;\bar Q}} + {\rm{\;\;\bar X\;\bar YQ}} + {\rm{\;\bar X\;Y\;\bar Q}} + {\rm{\;X\;\bar YQ}}\)

\({{\rm{Q}}_{{\rm{n}} + 1}}{\rm{\;}} = {\rm{\;\bar X\;\bar Y\;}}\left( {{\rm{\bar Q}} + {\rm{Q}}} \right) + {\rm{\;\bar X\;Y\;\bar Q}} + {\rm{\;X\;\bar YQ}}\)

\({{\rm{Q}}_{{\rm{n}} + 1}}{\rm{\;}} = \;{\bf{\bar X}}{\rm{\;}}{\bf{\bar Y}}{\rm{\;}} + {\rm{\bar XY\bar Q}} + {\rm{\;}}{\bf{X}}{\rm{\;}}{\bf{\bar YQ}}\)

\({{\rm{Q}}_{{\rm{n}} + 1}}{\rm{\;}} = {\rm{\;\bar Y}}\left( {{\rm{\bar X}} + {\rm{XQ\;}}} \right) + {\rm{\bar XY\bar Q}}\)

\({{\rm{Q}}_{{\rm{n}} + 1}}{\rm{\;}} = {\bf{\bar Y}}\;{\bf{\bar X}} + {\rm{\bar YQ}} + {\bf{\bar XY\bar Q}}\)

\({{\rm{Q}}_{{\rm{n}} + 1}}{\rm{\;}} = {\rm{\bar X}}.{\rm{\bar Y}} + {\rm{\bar X}}\overline {.{\rm{Q}}} + {\rm{\bar YQ}}\)

Consensus theorem: ab + a̅c + ac = ab + a̅c

In a = Q̅, b = Y̅, c = Z̅

\({{\rm{Q}}_{{\rm{n}} + 1}}{\rm{\;}} = {\rm{\bar X\bar Q}} + {\rm{\bar YQ\;\;\;\;}}\)

Q_n+1 = (X̅ ∧ Q̅) ∨ (Y̅ ∧ Q)

Important Point:

. → AND → ∧

+ → OR → ∨ )

Minimization could have been done using K - Map

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1. T₁ = Q₁Q₀, T₀ = Q̅₁Q̅₀
2. T₁ = Q̅₁Q₀, T₀ = Q̅₁ + Q̅₀
3. T₁ = Q₁ + Q₀, T₀ = Q̅₁ + Q̅₀
4. T₁ = Q̅₁Q₀, T₀ = Q₁ + Q₀

Concept:

Output of T flip flop will change when T = 1 and remain same when T = 0

Excitation table for T flip flop

From this table,

T1= Q̅1Q0

T0= Q̅1+ Q̅0

Important Point:

T₀→ NAND Gate (function)

Tips:

If unable to write the function then construct the K- Map of twovariable withQ1 andQ0as input

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T Flip Flop Question 6

1. 200 Hz
2. 50 Hz
3. 800 Hz
4. None of the above

T Flip Flop Question 6 Detailed Solution

Concept:

If we pass the input signal to a single T-flip flop, we will get half of the frequency at the output.

Similarly, when we pass the input signal into an n-bit flip flop counter, the output frequency (fout) will be:

\({{\rm{f}}_{{\rm{out}}}} = \frac{{{\rm{Input\;Frequency}}}}{{{2^{\rm{n}}}}}\)

Application:

Given Input frequency f = 200 Hz

Four T flip flops connected in cascade mode (n = 4)

\({{\rm{f}}_{{\rm{out}}}} = \frac{{{\rm{Input\;frequency}}}}{{{2^{\rm{n}}}}} = \frac{{2{\rm{00~Hz}}}}{{{2^4}}}\)

fout= 12.5 Hz

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T Flip Flop Question 7

1. (X̅ ∧ Q̅) ∨ (Y̅ ∧ Q)
2. (X̅ ∧ Q) ∨ (Y̅ ∧ Q̅)
3. (X ∧ Q̅) ∨ (Y ∧ Q)
4. (X ∧ Q̅) ∨ (Y̅ ∧ Q)

T Flip Flop Question 7 Detailed Solution

Concept –

For finding the expression of Next state, a truth table needs to be constructed.

Equation for Q_n+1 is made by writing all values of X, Y, Q for which Q_n+1 = 1

\({{\rm{Q}}_{{\rm{n}} + 1}}{\rm{\;}} = {\rm{\;\bar X\;\bar Y\;\bar Q}} + {\rm{\;\;\bar X\;\bar YQ}} + {\rm{\;\bar X\;Y\;\bar Q}} + {\rm{\;X\;\bar YQ}}\)

\({{\rm{Q}}_{{\rm{n}} + 1}}{\rm{\;}} = {\rm{\;\bar X\;\bar Y\;}}\left( {{\rm{\bar Q}} + {\rm{Q}}} \right) + {\rm{\;\bar X\;Y\;\bar Q}} + {\rm{\;X\;\bar YQ}}\)

\({{\rm{Q}}_{{\rm{n}} + 1}}{\rm{\;}} = \;{\bf{\bar X}}{\rm{\;}}{\bf{\bar Y}}{\rm{\;}} + {\rm{\bar XY\bar Q}} + {\rm{\;}}{\bf{X}}{\rm{\;}}{\bf{\bar YQ}}\)

\({{\rm{Q}}_{{\rm{n}} + 1}}{\rm{\;}} = {\rm{\;\bar Y}}\left( {{\rm{\bar X}} + {\rm{XQ\;}}} \right) + {\rm{\bar XY\bar Q}}\)

\({{\rm{Q}}_{{\rm{n}} + 1}}{\rm{\;}} = {\bf{\bar Y}}\;{\bf{\bar X}} + {\rm{\bar YQ}} + {\bf{\bar XY\bar Q}}\)

\({{\rm{Q}}_{{\rm{n}} + 1}}{\rm{\;}} = {\rm{\bar X}}.{\rm{\bar Y}} + {\rm{\bar X}}\overline {.{\rm{Q}}} + {\rm{\bar YQ}}\)

Consensus theorem: ab + a̅c + ac = ab + a̅c

In a = Q̅, b = Y̅, c = Z̅

\({{\rm{Q}}_{{\rm{n}} + 1}}{\rm{\;}} = {\rm{\bar X\bar Q}} + {\rm{\bar YQ\;\;\;\;}}\)

Q_n+1 = (X̅ ∧ Q̅) ∨ (Y̅ ∧ Q)

Important Point:

. → AND → ∧

+ → OR → ∨ )

Minimization could have been done using K - Map

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T Flip Flop Question 8

1. T₁ = Q₁Q₀, T₀ = Q̅₁Q̅₀
2. T₁ = Q̅₁Q₀, T₀ = Q̅₁ + Q̅₀
3. T₁ = Q₁ + Q₀, T₀ = Q̅₁ + Q̅₀
4. T₁ = Q̅₁Q₀, T₀ = Q₁ + Q₀

T Flip Flop Question 8 Detailed Solution

Concept:

Output of T flip flop will change when T = 1 and remain same when T = 0

Excitation table for T flip flop

From this table,

T1= Q̅1Q0

T0= Q̅1+ Q̅0

Important Point:

T₀→ NAND Gate (function)

Tips:

If unable to write the function then construct the K- Map of twovariable withQ1 andQ0as input

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1. 200 Hz
2. 50 Hz
3. 800 Hz
4. None of the above

Concept:

If we pass the input signal to a single T-flip flop, we will get half of the frequency at the output.

Similarly, when we pass the input signal into an n-bit flip flop counter, the output frequency (fout) will be:

\({{\rm{f}}_{{\rm{out}}}} = \frac{{{\rm{Input\;Frequency}}}}{{{2^{\rm{n}}}}}\)

Application:

Given Input frequency f = 200 Hz

Four T flip flops connected in cascade mode (n = 4)

\({{\rm{f}}_{{\rm{out}}}} = \frac{{{\rm{Input\;frequency}}}}{{{2^{\rm{n}}}}} = \frac{{2{\rm{00~Hz}}}}{{{2^4}}}\)

fout= 12.5 Hz

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∴ 1 and 2 are the states that transition back to the same state.

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1. (X̅ ∧ Q̅) ∨ (Y̅ ∧ Q)
2. (X̅ ∧ Q) ∨ (Y̅ ∧ Q̅)
3. (X ∧ Q̅) ∨ (Y ∧ Q)
4. (X ∧ Q̅) ∨ (Y̅ ∧ Q)

Concept –

For finding the expression of Next state, a truth table needs to be constructed.

Equation for Q_n+1 is made by writing all values of X, Y, Q for which Q_n+1 = 1

\({{\rm{Q}}_{{\rm{n}} + 1}}{\rm{\;}} = {\rm{\;\bar X\;\bar Y\;\bar Q}} + {\rm{\;\;\bar X\;\bar YQ}} + {\rm{\;\bar X\;Y\;\bar Q}} + {\rm{\;X\;\bar YQ}}\)

\({{\rm{Q}}_{{\rm{n}} + 1}}{\rm{\;}} = {\rm{\;\bar X\;\bar Y\;}}\left( {{\rm{\bar Q}} + {\rm{Q}}} \right) + {\rm{\;\bar X\;Y\;\bar Q}} + {\rm{\;X\;\bar YQ}}\)

\({{\rm{Q}}_{{\rm{n}} + 1}}{\rm{\;}} = \;{\bf{\bar X}}{\rm{\;}}{\bf{\bar Y}}{\rm{\;}} + {\rm{\bar XY\bar Q}} + {\rm{\;}}{\bf{X}}{\rm{\;}}{\bf{\bar YQ}}\)

\({{\rm{Q}}_{{\rm{n}} + 1}}{\rm{\;}} = {\rm{\;\bar Y}}\left( {{\rm{\bar X}} + {\rm{XQ\;}}} \right) + {\rm{\bar XY\bar Q}}\)

\({{\rm{Q}}_{{\rm{n}} + 1}}{\rm{\;}} = {\bf{\bar Y}}\;{\bf{\bar X}} + {\rm{\bar YQ}} + {\bf{\bar XY\bar Q}}\)

\({{\rm{Q}}_{{\rm{n}} + 1}}{\rm{\;}} = {\rm{\bar X}}.{\rm{\bar Y}} + {\rm{\bar X}}\overline {.{\rm{Q}}} + {\rm{\bar YQ}}\)

Consensus theorem: ab + a̅c + ac = ab + a̅c

In a = Q̅, b = Y̅, c = Z̅

\({{\rm{Q}}_{{\rm{n}} + 1}}{\rm{\;}} = {\rm{\bar X\bar Q}} + {\rm{\bar YQ\;\;\;\;}}\)

Q_n+1 = (X̅ ∧ Q̅) ∨ (Y̅ ∧ Q)

Important Point:

. → AND → ∧

+ → OR → ∨ )

Minimization could have been done using K - Map

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1. T₀=0, T₁=Q₁⊕ Q₀
2. T₀= 1, T₁=Q̅₁Q₀
3. T₀= 0, T₁= Q₁⊙Q₀
4. T₀= 1, T₁=Q₁Q₀

Concept:

Output of T flip flop will change when T = 1 and remain same when T = 0

Excitation table for T flip flop

From this table,

T0= 0

T1= Q̅1.Q̅0+ Q1Q0=Q1⊙Q0

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1. Both trigger and toggle
2. Toggle
3. Trigger
4. Either trigger or toggle

Concept:

• T flip-flop is also knownas Toggle flip-flop.
• It can be constructed from a JK flip-flop by connecting both inputs J and K together.
• Serial decoding, comparison and timing functions can be accomplished using flip-flops.
• Flip-flops has 2 generally stable states, is SET and RESET.
• A flip-flop is said to be in "high state" or logic 1 or SET state" when Q=1.
• A flip-flop is said to be in "lowstate" or logic 0or RESET state" when Q=0.
• Flip-flop are the fundamental components of shift registers and counters.
• T- flip-flop when CLKis LOW, T=0 ; no change in the input.
• T- flip-flop when CLK is HIGH, T=1; The output is toggle.

So, Toggle is the correct option.

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1. T₁ = Q₁Q₀, T₀ = Q̅₁Q̅₀
2. T₁ = Q̅₁Q₀, T₀ = Q̅₁ + Q̅₀
3. T₁ = Q₁ + Q₀, T₀ = Q̅₁ + Q̅₀
4. T₁ = Q̅₁Q₀, T₀ = Q₁ + Q₀

Concept:

Output of T flip flop will change when T = 1 and remain same when T = 0

Excitation table for T flip flop

From this table,

T1= Q̅1Q0

T0= Q̅1+ Q̅0

Important Point:

T₀→ NAND Gate (function)

Tips:

If unable to write the function then construct the K- Map of twovariable withQ1 andQ0as input

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1. Q_n+1= T̅Q_n + TQ̅_n

2. Qn+1= T̅

3. Qn+1= Q_n

4. \(\)Qn+1= T̅Q̅_n + TQ_n

Option 1 :

Q_n+1= T̅Q_n + TQ̅_n

From the characteristic table:

Q_n+1= T̅Q_n + TQ̅_n = T ⊕ Q_n

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