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Boolean relation at the output stage - Y for the following circuit is :  
Option: 1 \bar{A}+\bar{B}
Option: 2 A+B
Option: 3 A\cdot B  
Option: 4 \bar{A}\cdot \bar{B}
 

 

 

 

 

The first part of the given circuit represents or gate whose output is A+B

The second part of the circuit is NOT gate whose input is A+B and output is  \overline{A+B}=\bar A.\bar B

So the correct option is 4.

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Posted by

vishal kumar

The circuit shown below is working as a 8 V dc regulated voltage source . When 12 V is used as input, the power dissipated (in mW) in each diode is ; (considering both zener diodes are identical ) _____.  
Option: 1 40
Option: 2 30
Option: 3 50
Option: 4 20
 

i=\frac{(12-8)}{(200+200)}A=\frac{4}{400}=10^{-2}A

Power \; loss\; in\; each\; diaode=(4)(10^{-2})W=40mW

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Posted by

avinash.dongre

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Which of the following gives a reversible operation?  
Option: 1

Option: 2

Option: 3 

Option: 4 
 

 

 

 

For reversible operation, NOT gate is used. If an input isAthen output=\bar{A}. The following circuit represents the NOT gate

So Option (2) is correct.

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Posted by

Ritika Jonwal

The current i in the  network is :
Option: 1 0.3A
Option: 2 0\: A
Option: 3 0.2\: A
Option: 4 0.6\: A
 

Both diodes are in reverse biased 

 

So new circuits can be drawn as

 

 

So I=\frac{9}{30}=\frac{3}{10}=0.3\; A

Hence the correct option is (1).

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Posted by

avinash.dongre

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In the given circuit, the value of Y is :
Option: 1 toggles between 0 and 1
Option: 2 1
Option: 3 0
Option: 4 will not execute
 

 

In NAND gate, when 

\\y=\overline{A.\overline{A.B}}\\Given\\A=1,\ B=0\\\Rightarrow A.B=0\\ \overline{A.B}=1\ and \ A.\overline{A.B}=1\ \\ \ therefore\ \overline{A.\overline{A.B}}=0

Hence the correct option is (3)

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Posted by

vishal kumar

Both the diodes used in the circuit shown are assumed to be ideal and have negligible resistance when these are forward biased. Built-in potential in each diode is 0.7V. For the input voltages shown in the figure, the voltage (in Volts) at a point A is ______________    
Option: 1 12
Option: 2 10
Option: 3 40
Option: 4 24

 

 

 

 

We want to find VE 

By applying kvl in AEFBA

12.7 - 0.7 - VE = 0.

implies

VE  = 12V

Hence the correct option is (1).


 

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Posted by

avinash.dongre

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A transistor is connected in common emitter circuit configuration, the collector supply voltage is 10 V and the voltage drop across a resistor of 1000 \Omega in the collector circuit is 0.6 V. If the current gain factor (\beta ) is 24, then the base current is _______ \mu A. (Round off to the Nearest Integer)
 

\begin{aligned} &V_{C C}=10 \mathrm{~V} \\ &I_{C} R_{C}=0.6 \mathrm{~V} \\ &I_{C}(1000)=0.6 \end{aligned}

I_{c}=6 \times 10^{-4} \mathrm{~A}

\begin{aligned} B=\frac{I_{C}}{I_{B}} &=24 \\ I_{B} &=\frac{6}{24} \times 10^{-4} . \\ &=0.25 \times 10^{-4} \\ I_{B} &=25 \mu \mathrm{A} \end{aligned}

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Posted by

vishal kumar

Find the truth table for the function \mathrm{Y}$ of $\mathrm{A}$ and $\mathrm{B} represented in the following figure.

Option: 1 \begin{array}{|c|c|c|} \hline \mathrm{A} & \mathrm{B} & \mathrm{Y} \\ \hline 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 1 & 0 \\ \hline \end{array}
Option: 2 \begin{tabular}{|c|c|c|} \hline $\mathrm{A}$ & $\mathrm{B}$ & $\mathrm{Y}$ \\ \hline 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \\ 1 & 1 & 1 \\ \hline \end{tabular}
 
Option: 3 \begin{tabular}{|c|c|c|} \hline $\mathrm{A}$ & $\mathrm{B}$ & $\mathrm{Y}$ \\ \hline 0 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \\ \hline \end{tabular}
 
Option: 4 \begin{tabular}{|c|c|c|} \hline $\mathrm{A}$ & $\mathrm{B}$ & $\mathrm{Y}$ \\ \hline 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 1 \\ \hline \end{tabular}

Y=A \cdot B+\bar{B}

\begin{array}{|c|c|c|} \hline A & B & Y=A \cdot B+\bar{B} \\ \hline 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \\ 1 & 1 & 1 \\ \hline \end{array}

The correct option is (2)

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vishal kumar

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Statement I:
By doping silicon semiconductor with pentavalent material,the electrons density increases.
Statement II :
The n-type semiconductor has net negative charge.
In the light of the above statements,choose the most appropriate answer from the options given below:
Option: 1 Both statement I and Statement II are false.
Option: 2  statement I  is true but Statement II is false.
Option: 3 Statement I is false but Statement II is true.
Option: 4 Both Statement I and Statement II are true.

Statement I
By doping silicon semiconductors with pentavalent material.
n_{e}> > n_{h}
Statement II
The n-type semiconductor has a net negative charge
Both the statements are true
Hence, the correct option is (4)

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Posted by

vishal kumar

Identify the logic operation carried out by the given circuit :

Option: 1 OR
Option: 2 AND
Option: 3 NAND
Option: 4 NOR


Z= \bar{A}.\bar{B} \\ or \\ Z= \overline{A+B}
\therefore The logic operation carried out in the above circuit is equivalent to NOR Gate
The correct option is (4)

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Posted by

vishal kumar

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