Semiconductor Electronics: Materials, Devices and Simple Circuits Test 38

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Semiconductor Electronics: Materials, Devices and Simple Circuits Test 38

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Weekly Quiz Competition

Question 1
1 / -0

In the given circuit, the binary inputs at $$A$$ and $$B$$ are both $$1$$ in one case and both $$0$$ in the next case. The respective outputs at $$Y$$ in these two cases will be

A
$$1,1$$

B
$$0,0$$

C
$$0,1$$

D
$$1,0$$

Solution

$$Y=\overline { AB+\overline { A } \overline { B } } $$
for $$A=1,B=1$$ $$\Rightarrow$$ $$Y=0$$ and for $$A=0,B=0$$ $$\Rightarrow$$ $$Y=0$$
Question 2
1 / -0

Directions For Questions

Base your answers to the following questions on the diagram below, which represents a germanium semiconductor device.

...view full instructions

In the diagram, section A represents:

A
N-type germanium

B
P-type germanium

C
Anode

D
Diode

Solution

In the given problem, section A has excess of electrons and section B has excess of holes. Hence, section A represents N-type Germanium and section B represents P-Type Germanium.
Question 3
1 / -0

The diagram of a logic circuit is given below. The output A of the circuit is represented by :

A
$$P \cdot (R+Q)$$

B
$$P \cdot (R\cdot Q)$$

C
$$P+ (R \cdot Q)$$

D
$$P+ (R +Q)$$

Solution

$$\textbf{Hint: OR gate add the inputs and AND gate multiplies the input}$$

$$\textbf{Step1: Find output of OR gate}$$
Output of upper OR gate $$=P+R$$
Output of lower OR gate$$ = P+Q$$

$$\textbf{Step2: Find output A of the circuit}$$
$$\therefore$$ New output $$ A=(P+R).(P+Q)$$
$$= P.P+P.Q+R.P+R.Q$$
$$=P(1+Q)+R.P+R.Q$$ ($$\because $$ $$1+Q =1$$ and $$P.P =P$$)
$$\therefore A=P+R\cdot P+R\cdot Q$$
$$\Rightarrow$$$$P(1+R)+R\cdot Q$$ now $$1+R = 1$$
$$\Rightarrow$$$$P+R\cdot Q$$
$$\textbf{Hence option C correct}$$
Question 4
1 / -0

The truth table given is for which gate?
A B Y
$$0$$ $$0$$ $$1$$
$$0$$ $$1$$ $$1$$
$$1$$ $$0$$ $$1$$
$$1$$ $$1$$ $$0$$

A
XOR

B
OR

C
AND

D
NAND

Solution

The given truth table is of NAND gate.
Question 5
1 / -0

For a common emitter circuit amplifier, the load resistance of the output circuit is $$500$$ times the resistance of the input. circuit. If $$\alpha = 0.98$$, then, current gain is :

A
$$0.98$$

B
$$50$$

C
$$98$$

D
$$49$$

Solution

Given, $$\alpha = 0.98$$
$$\beta = \dfrac{\alpha}{1-\alpha} = \dfrac{0.98}{1-0.98} = 49$$
Question 6
1 / -0

Find the corresponding truth table for the function Z of X and Y represented by given figure is :

A

B

C

D
None of these

Solution

The output in terms of input is
$$Z=X.(X+Y)$$
the truth table for which is represented for option (a).
Question 7
1 / -0

The following truth table corresponds to the logic gate :
A B Y
0 0 0
0 1 1
1 0 1
1 1 1

A
NAND

B
AND

C
XOR

D
OR

Solution

This is truth table for OR logic gate because for
OR gate, $$X=A+B$$.
Question 8
1 / -0

To get an output Y$$=1$$ from circuit of adjoining figure, the input must be :

A
A$$=0$$, B$$=1$$, C$$=0$$

B
A$$=1$$, B$$=0$$, C$$=0$$

C
A$$=1$$, B$$=0$$, C$$=1$$

D
A$$=1$$, B$$=1$$, C$$=0$$

Solution

$$Y=1$$, if A$$=1$$, B$$=0$$ and C$$=1$$ because A and B are input of OR gate and the output of A and B and as well as C are the inputs of AND gate.
Question 9
1 / -0

The resistance of mercury at $$4.2\ K$$ is

A
Infinity

B
Greater than at lab temperature

C
Same as that of lab temperature

D
Almost zero

Solution

Onnes discovered to his own amazement that below 4.2 K (about -269°C), the resistance of mercury suddenly dropped to zero.

Question 10

1 / -0

The truth table for the following logic circuit is

$$\begin{vmatrix} A & B & Y \\ 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{vmatrix}$$

$$\begin{vmatrix} A & B & Y \\ 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 1 \end{vmatrix}$$

$$\begin{vmatrix} A & B & Y \\ 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{vmatrix}$$

$$\begin{vmatrix} A & B & Y \\ 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \end{vmatrix}$$

Solution

The truth table for the given logic circuit is

$$A$$	$$B$$	$$\overline { A } $$	$$\overline { B } $$	$$\overline { A } \cdot B$$	$$A\overline { B } $$	$$Y=\overline { A } B+A\overline { B } $$
$$0$$	$$0$$	$$1$$	$$1$$	$$0$$	$$0$$	$$0$$
$$0$$	$$1$$	$$1$$	$$0$$	$$1$$	$$0$$	$$1$$
$$1$$	$$0$$	$$0$$	$$1$$	$$0$$	$$1$$	$$1$$
$$1$$	$$1$$	$$0$$	$$0$$	$$0$$	$$0$$	$$0$$

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Semiconductor Electronics: Materials, Devices and Simple Circuits Test 38

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Semiconductor Electronics: Materials, Devices and Simple Circuits Test 38

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SHARING IS CARING

Question 1

$$1,1$$

$$0,0$$

$$0,1$$

$$1,0$$

Solution

Question 2

N-type germanium

P-type germanium

Anode

Diode

Solution

Question 3

$$P \cdot (R+Q)$$

$$P \cdot (R\cdot Q)$$

$$P+ (R \cdot Q)$$

$$P+ (R +Q)$$

Solution

Question 4

XOR

OR

AND

NAND

Solution

Question 5

$$0.98$$

$$50$$

$$98$$

$$49$$

Solution

Question 6

None of these

Solution

Question 7

NAND

AND

XOR

OR

Solution

Question 8

A$$=0$$, B$$=1$$, C$$=0$$

A$$=1$$, B$$=0$$, C$$=0$$

A$$=1$$, B$$=0$$, C$$=1$$

A$$=1$$, B$$=1$$, C$$=0$$

Solution

Question 9

Infinity

Greater than at lab temperature

Same as that of lab temperature

Almost zero

Solution

Question 10

$$\begin{vmatrix} A & B & Y \\ 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{vmatrix}$$

$$\begin{vmatrix} A & B & Y \\ 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 1 \end{vmatrix}$$

$$\begin{vmatrix} A & B & Y \\ 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{vmatrix}$$

$$\begin{vmatrix} A & B & Y \\ 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \end{vmatrix}$$

Solution

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