MCQ Questions for Class 12 Engineering Physics Semiconductor Electronics: Materials,Devices And Simple Circuits Quiz 10

CBSE Questions for Class 12 Engineering Physics Semiconductor Electronics: Materials,Devices And Simple Circuits Quiz 10 - MCQExams.com

CBSE Questions for Class 12 Engineering Physics Semiconductor Electronics: Materials,Devices And Simple Circuits Quiz 10 - MCQExams.com

Practice Class 12 Engineering Physics Quiz Questions and Answers

Practice Class 12 Engineering Physics Quiz Questions and Answers

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The following configuration of gate equivalent to :

0%
NAND
0%
XOR
0%
OR
0%
None of these

Explanation

The given output equation can also be written as

$Y=\left( A+B \right) \cdot \left( \bar { A } +\bar { B } \right)$ (Demorgan's theorem)

$\Rightarrow Y=A\bar { A } +A\bar { B } +B\bar { A } +B\bar { B }$

$\Rightarrow Y=0+A\bar { B } +\bar { A } B+0$

$\Rightarrow Y=\bar { A } B+A\bar { B }$

This is the expression for XOR gate.

The combination of gates shown in the figure below produces.

0%
NOR gate
0%
OR gate
0%
AND gate
0%
XOR gate

Explanation

Here, the output of these gate

$Y = \overline {\overline {A} - \overline {B}}$
But according to Boolean algebra,

$\overline {A}\cdot \overline {B} = \overline {A + B}$

$\therefore Y = \overline {\overline {A + B}} = A + B$
or

$Y = A + B$
It meas the whole circuit of gates behaves as OR gate.

In the circuit diagram, the current through the Zener diode is :

0%
$10 mA$
0%
$3.33 mA$
0%
$6.67 mA$
0%
$0 mA$

Explanation

The voltage drop across

${ R }_{ 2 }$ is

${ V }_{ { R }_{ 2 } }={ V }_{ z }=10 V$

The current through

${ R }_{ 2 }$ is

${ I }_{ { R }_{ 2 } }=\dfrac { { V }_{ { R }_{ 2 } } }{ { R }_{ 2 } } =\dfrac { 10 }{ 1500 } =0.667\times { 10 }^{ -2 }A$

$=6.67\times { 10 }^{ -3 }A=6.67 mA$

The voltage drop across

${ R }_{ 1 }$ is

${ V }_{ { R }_{ 1 } }=15 V - { V }_{ { R }_{ 2 } }=15 V-10 V=5 V$

The current through

${ R }_{ 1 }$ is

${ I }_{ { R }_{ 1 } }=\dfrac { { V }_{ { R }_{ 1 } } }{ { R }_{ 1 } } =\dfrac { 5V }{ 500\Omega } ={ 10 }^{ -3 }A=10\times { 10 }^{ -3 }=10 mA$

The current through Zener diode is

${ I }_{ z }={ I }_{ { R }_{ 1 } }-{ I }_{ { R }_{ 2 } }=\left( 10-6.67 \right) mA=3.33 mA$

Identify the mismatched pair from the following

0%
zener diode : voltage regulator
0%
germanium doped with phosphorous : n-type semiconductor
0%
semiconductor : band gap > 3 eV
0%
p-n junction diode : rectifier
0%
silicon doped with aluminium : p-type semiconductor

Explanation

For semiconductor band gap is

$<3eV$ so given option is incorrect.

When zener dioode is operated in breakdown region, voltge across it remains constant even if current through it changes by large amount.

Germanium when doped with phosphorous forms n-type semiconductor and silicon when doped with aluminum forms p-type.

p-n junction can be used as half wave or full wave rectifier.

The circuit gives the output as that of:

0%
AND gate
0%
OR gate
0%
NAND gate
0%
NOR gate
0%
NOT gate

Explanation

$Y=\overline{\overline{A\cdot B}+\overline{C}}=\overline{\overline{A\cdot B}}+\overline{\overline{C}}$

$=A\cdot B\cdot C=$ AND gate
Output is high only when all inputs are high as that of AND gate.

The circuit is equivalent to

0%
NOR gate
0%
AND gate
0%
NAND gate
0%
OR gate

$A$ and

$B$ are the two points on a uniform ring of radius

$r$ . The resistance of the rings is

$R$ and

$\angle AOB = \theta$ as shown in the figure. The equivalent resistance between points

$A$ and

$B$ is _____

0%
$\dfrac {R(2\pi - \theta)}{4\pi}$
0%
$\dfrac {R\theta}{2\pi}$
0%
$R\left (1 - \dfrac {\theta}{2\pi}\right )$
0%
$\dfrac {R}{4\pi^{2}} (2\pi - \theta)\theta$

Explanation

Consider the ring as two parts as two resistances joined in parallel between two points

$A$ and

$B$ , then two resistance would be

$R_{1} = \dfrac {2\pi r} \cdot r\theta = \dfrac {R}{2\pi}\theta$
and

$R_{2} = \dfrac {R}{2\pi r} r (2\pi - \theta)$

$= \dfrac {R}{2\pi} (2\pi - \theta)$
Now, equivalent or effective resistance between

$A$ and

$B$

$R_{eq} = \dfrac {R_{1}\times R_{2}}{R_{1} + R_{2}}$

$\Rightarrow R_{eq} = \dfrac {\dfrac {R}{2\pi}\theta \times \dfrac {R}{2\pi}(2\pi - \theta)}{\dfrac {R}{2\pi}[\theta + 2\pi - \theta]}$

$= \left [\dfrac {\dfrac {R^{2}\theta (2\pi - \theta)}{4\pi^{2}}}{\dfrac {2\pi R}{2\pi}}\right ]$

$= \dfrac {R^{2}\theta (2\pi - \theta)}{4\pi^{2}} \times \dfrac {2\pi}{2\pi R}$

$= \dfrac {R\theta (2\pi - \theta)}{4\pi^{2}}$ .

Name the gate, which represents the Boolean expression

$Y = \overline{A\cdot B}$

0%
NAND
0%
AND
0%
NOT
0%
NOR

Explanation

Boolean equation for NAND gate is

$Y = \overline{A.B}$

The logic gates giving output '

$1$ ' for the inputs of '

$1$ ' and '

$0$ ' are

0%
AND and OR
0%
OR and NOR
0%
NAND and NOR
0%
NAND and OR
0%
AND and NOR

Explanation

Boolean expression of OR gate

$Y=A+B$
and Boolean expression of NAND gate

$Y=\overline { A\cdot B }$
i.e., the logic gate giving output

$1$ for the inputs of

$1$ and

$0$ are NAND and OR.

The inputs A, B and C to be given in order to get an output

$Y =1$ from the following circuit are

0%
0, 1, 0
0%
1, 0, 0
0%
1, 0, 1
0%
1, 1, 0
0%
0, 0, 1

Explanation

$(A+B) \cdot C=1$

$A\cdot C+B \cdot =1$

$A\cdot C= 1$ or

$B\cdot C = 1$

$A=1, B= 0, C=1$
So, (1, 0, 1) is correct option.

Consider the circuit, the current through the Zener diode is

0%
$20\ mA$
0%
$10\ mA$
0%
$15\ mA$
0%
$40\ mA$

Explanation

Voltage across Zener diode is constant

$i_{R_2}=\dfrac{20}{2000}=0.01 A$

$i_{R_1}=\dfrac{(40-20)}{2000}=0.01 A$

For which one of the following input combinations, the given logic circuit gives the output

$Y = 1$ ?

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

Explanation

The logic circuit is
The output,

$Y = \overline{\overline {\overline {A} \cdot B} + \overline {C}} = \overline {\overline {\overline {A} \cdot B} \cdot \overline {C}} = \overline {A} \cdot B\cdot C$

$[\overline {A + B} = \overline {A} \cdot \overline {B}$ , Demorgan's law]
when

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

$Y = 1 [\because \overline {A} = 1]$ .

In the adjoining circuit of logic gate, the output

$Y$ becomes zero if the inputs are

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

Explanation

The output

$Y = \overline {(A\cdot B)\cdot C} = \overline {(A \cdot B)} + \overline {\overline {C}}$

$= \overline {(A\cdot B)} + C$
So,

$Y = 0$ if

$A = 1, B = 1$ and

$C = 0$ .

A system of logic gates is shown in the figure. From the study of truth table it can be found that to produce a high output (1) at

$R$ , we must have

0%
$X=0,Y=1$
0%
$X=1,Y=1$
0%
$X=1,Y=0$
0%
$X=0,Y=0$

Explanation

The truth table can be written as

So,

$X=1,Y=0$ gives output

$R=1$

The following configuration of gates is equivalent to

0%
XOR
0%
NAND
0%
OR
0%
None of these

Explanation

The output

$Y$ of the following gates

$Y =\overline {\overline {X}}\overline {\overline {A + B}} = A + B$
This is the Boolean expression for

$OR$ gate.

The following logic circuit represents

0%
NAND gate with output $O=\overline { X } +\overline { Y }$
0%
NOR gate with output $O=\overline { X+Y }$
0%
NAND gate with output $O=\overline { X-Y }$
0%
NOR gate with output $O=\overline { X } .\overline { Y }$

Explanation

The given logic circuit is the combination of OR and NOT gates. So it should be the NOR gate. Besides, the NAND gate is the combination of AND and NOT gates.

The output of NOR gate will be

$O=\overline{X+Y}$ and output of the NAND gate will be

$O=\overline{XY}=\overline X+\overline Y$ (by de Morgan's law)

The following truth table is for which gate ?

A	B	Y=A + B
0	0	0
0	1	1
1	0	1
1	1	1

0%
AND
0%
OR
0%
NOR
0%
NAND

Explanation

R.E.F image

$F = A+B$

$A$ $B$	$A+B$
$0$ $0$	$0$
$0$ $1$	$1$
$1$ $0$	$1$
$1$ $1$	$1$

Select the output

$Y$ of the combination of gates shown in figure for inputs

$A = 1, B = 0; A = 1, B = 1$ and

$A = 0, B = 0$ respectively.

0%
$(0, 1, 1)$
0%
$(1, 0, 1)$
0%
$(1, 1, 1)$
0%
$(1, 0, 0)$

The arrangement shown in figure performs the logic function of

0%
OR gate
0%
XNOR gate
0%
AND gate
0%
NAND gate

The diagram of a logic gate circuit is given below. The output

$Y$ of the circuit is represented by

0%
$A\cdot (B + C)$
0%
$A\cdot (B\cdot C)$
0%
$A + (B\cdot C)$
0%
$A + (B + C)$

Explanation

Output of upper OR gate

$= A + B$
Output of lower OR gate

$= A + C$
Net output

$Y = (A + B)(A + C)$

$= AA + AC + BA + BC$

$= A(1 + C) + BA + BC [\because AA = A]$

$= A + BA + BC [\because 1 + C = 1]$

$= A(1 + B) + BC [\because 1 + B = 1]$

$= A + BC = A + (B\cdot C)$ .

For given logic diagram, output

$F = 1$ , then inputs are:

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

Explanation

When

$A = 0, B = 1$ and

$C = 0$ , then output becomes

$1$ .

The following circuit represents.

0%
OR gate
0%
XOR gate
0%
AND gate
0%
NAND gate

Explanation

Output of upper AND gate

$=\bar{A}B$
Output of lower AND gate

$=A\bar{B}$
Thus, output of OR gate

$=\bar{A}B+A\bar{B}$
This is Boolean expression for XOR gate.

Identify the gate used in the following diagram.

0%
AND
0%
OR
0%
NAND
0%
NOR

A photodiode is symbolised by the graphic.

In a solid, an unfilled vacancy in an electronic energy level is called.

0%
Hole
0%
Majority carrier
0%
Minority carrier
0%
Electron

What will be the current flowing through the

$6K\Omega$ resistor in the circuit shown, where the breakdown voltage of the zener is

$6$ V?

0%
$\displaystyle\frac{2}{3}$ mA
0%
$1$ mA
0%
$10$ mA
0%
$\displaystyle\frac{3}{2}$ mA

Explanation

$\because$ Zener break done

$=6$ V
Sp potential across

$4K\Omega =6$ V
and potential across

$6K\Omega =(10-6)=4$ V
Current through the

$6K\Omega =\displaystyle\frac{4}{6000}A\Rightarrow \displaystyle\frac{2}{3000}A=\frac{2}{3}$ mA.

Which logic gate is represented by the following logic gates?

0%
NOR
0%
NAND
0%
AND
0%
OR

A pure semiconductor is __________.

0%
an extrinsic semiconductor
0%
an intrinsic semiconductor
0%
p-type semiconductor
0%
n-type semiconductor

Which of following gates produces output of

$1$ ?

Explanation

For (A), NAND Gate, output will be

$Y = \overline {1.1} = \bar 1 = 0$

For (B), NOR Gate, output will be

$Y = \overline{0+0} = \bar 0 = 1$

For (C), AND Gate, output will be

$Y= {1.0} = 0$

For (D), OR Gate, output will be

$Y = {0+0} = 0$

Hence, option (B) is the correct one.

A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.0

$^{\circ}$ C. What is the change in the diameter of the hole when the sheet is heated to 227

$^{\circ}$ C? Coefficient of linear expansion of copper is 1.70 x 10

$^{-5}$

0%
$1.44 \times 10^{-2}$ cm
0%
$2.44 \times 10^{-3}$ cm
0%
$1.44\times 10^{-2}$ mm
0%
$2.44 \times 10^{-3}$ mm

Explanation

$D_2=4.24cm$

Initial Area of the hole

$(A_0)=\pi r^2=22/7(4.24/2)^2\\=4.494\pi cm^2$

Initial temp.

$(T_1)=27°=27+273=300K$

Final temp.

$(T_2)=227°=227+273=500K$

Coefficientof linear exp.

$(O)=1.7\times 10^{-5}/°C$

Superficial expansion

$(b)=2\times$ Linear expansion

$=2\times1.7\times10^{-5}°C\\=3.4\times 10^{-5}°C$

Use formula

$A=A_0(1+b\Delta T)\\A=4.49\pi[1+3.4\times10^{-5}\times(500-300)]\\ A=4.494\pi[1+0.0068]\\A=4.525cm^2=\pi D 2^2/$\\B2^2=4.523\times4\\ D2=4.2544cm$

Change in diameter

$=\Delta D=D_2-D_1\\=4.2544-4.24\\=0.0144cm$

With increase in temperature the conductivity of?

0%
Metals increases and of semiconductor decreases
0%
Semiconductors increases and of metals decreases
0%
In both metals and semiconductors increases
0%
In both metal and semiconductor decreases

Explanation

As the temperature increases, more electrons get the energy to jump from Conduction band to valence band, and thereby increases the conductivity of the semiconductor.

In metal, resistance increases with increase in temperature hence conductivity decreases.

So, option

$(B)$ is correct.

If the load resistance decreases in a zener regulator, the zener current

0%
Decreases
0%
Stays the same
0%
Increases
0%
Equals the source voltage divided by the series resistance

Explanation

The load current is given by

$I_L=\dfrac{V_z}{R_L}$ and the zener current is given by

$I_z=I_s-I_L$ . Thus, if the load resistance decreases, load current increases. This in turn will reduce the zener current

Diffusion is the process of.

0%
Rarefaction of particles
0%
Movement of particles through a semipermeable membrane
0%
Movement of particles from higher concentration to lower concentration
0%
Accumulation of particles on a solid surface

The logic circuit shown in the figure represents characteristic of which logic gate?

0%
OR gate
0%
NOR gate
0%
AND gate
0%
NAND gate

Explanation

$C=A'+B'$

By De-Morgan's theorem,

$(AB)'=A'+B'\\ \therefore C=(AB)'$

Therefore, NAND gate.

What is true about the breakdown voltage in a zener diode?

0%
It decreases when current increases.
0%
It destroys the diode.
0%
It equals the current times the resistance.
0%
It is approximately constant.

A heater is designed to operate with a power of

$1000\ W$ in a

$100\ V$ line. It is connected in combination with a resistance of

$10\Omega$ and a resistance

$R$ , to a

$100\ V$ mains as shown in the figure. What will be the value of

$R$ so that the heater operates with a power of

$62.5\ W$ ?

0%
$15\Omega$
0%
$10\Omega$
0%
$5\Omega$
0%
$25\Omega$

Explanation

Given that,

Power rating of a heater,

$P=1000\ W$

Voltage rating of heater,

$V=100\ V$

$\therefore$ Resistance of heater,

$R= \dfrac{V^2}P= \dfrac{100\times100}{1000}=10 \Omega$

Given that, Power at which heater operates ,

$P'=62.5\ W$

Hence, Voltage drop across the heater,

$V'=\sqrt{RP'}=\sqrt{10\times62.5}=25 \ V$

Since the voltage drop across the heater is

$25\ V$ hence voltage drop across

$10\ Ω$ resistor is

$, V_r= (100-25) = 75\ V$

Current in the circuit,

$I=\dfrac{V_r}R= \dfrac{75}{10}=7.5\ A$

Further, the current divides into two parts.

Let

$I_1$ be the current that passes through the heater.

$\therefore 25=I_1\times 10\implies I_1=2.5\ A$

Hence, current through resistance

$R$ is,

$i=7.5-2.5=5\ A$

Applying Ohm's law across

$R$ , we get

$\therefore iR=25 \implies R=\dfrac{25}5=5\Omega$

Hence, option

$(C)$ is correct.

The current flowing through the zener diode in figure is:

0%
$20\,mA$
0%
$25\,mA$
0%
$15\,mA$
0%
$5\,mA$

Explanation

Using Kirchoff's current law:

$\dfrac{{10 - 5}}{{500}} = {I_1} + \dfrac{5}{{1000}}$

$\dfrac{5}{{500}} - \dfrac{5}{{1000}} = {I_1}$

$\dfrac{{10 - 5}}{{1000}} = {I_1}$

$\boxed{5\,mA = {I_1}}$

The graph given represents the I-V characteristics of a Zener diode. Which part of the characteristic curve is most relevant for its operation as a voltage regulator?

0%
$ab$
0%
$bc$
0%
$cd$
0%
$de$

Explanation

Hence ans should be

$de$ bez zenve diode acts as voltage congulator is reverse biasing where in chore voltage tange called zener breatdown voltage

Zener diodes with breakdown voltage ranging over 2V- 200 V are commercially available. Breakdown voltage of a zener diode

0%
Increases with increasing doping concentration
0%
Decreases with increasing doping concentration
0%
Does not depend on doping concentration
0%
May increases or decreases with increasing doping concentration

When the source voltage increases in a zener regulator, which of these currents remains approximately constant?

0%
Series current
0%
Zener current
0%
Load current
0%
Total current

There are two holes O1 and O2 in a tank of height H.The water emerging from O1 and O2 strikes the ground at the same points, as shown in fig.then.

0%
$H = {h_1} + {h_2}$
0%
$H = {h_1} - {h_2}$
0%
$H = {h_1} {h_2}$
0%
$H = {h_1}/ {h_2}$

Which of the following is a correct statement about current carrying conductor?

0%
On increasing temperature drifts speed of free electron decreases
0%
On increasing temperature drift speed of free electron increases
0%
On increasing temperature, drift speed of free electron remains same.
0%
On increasing P.D. average kinetic energy of electron decreases ( Neglected heating effect of electric current)

Explanation

$vd =\dfrac{et}{m} E$

$t \downarrow$ as

$T \uparrow ]$

As the temperature increases, the thermal agitation of the electrons increases thereby, increasing the number of collisions. Hence, drift velocity of the electrons decreases.

In the circuit shown in figure, the diode is ideal. The potential difference between A and B is?

0%
$V/4$
0%
V
0%
Zero
0%
$V/2$

Explanation

Diode short circuits the resistance of

$2\Omega$ .

$\therefore$ Voltage across points A and B will be

$V_1=\dfrac{\dfrac{2}{3}}{\dfrac{2}{3}+2}\times V=\dfrac{V}{4}$ .

The correct output for the given circuit is:

Explanation

$0-{T}_{0}$

$C=(\overline { 1+1 } .1)=0$

${T}_{0}-2{T}_{0}$

$C=(\overline { 1+1 } .1)=0$

$2{T}_{0}-\infty$

$C=(\overline { 1+0 } .1)=0$
1073266_992736_ans_1fcafd564d7b4dee8fd3b1ad34d32e8e.png

Read the following diagram and tell which of the following statements are true:

0%
In circuit 1 lamp does not glow but in circuit 2 lamp glows
0%
In circuit 1 as well as 2 lamp does not glow
0%
In circuit 1 lamp glows but in 2 lamp does not glow
0%
In both circuit lamp glows

Germanium and silicon junction diodes are connected in parallel. These are connected in series with a resistance

$R$ , a milliammeter

$(mA)$ and a key

$(K)$ as shown in Fig. When key

$(K)$ is closed a current begins to flow in the milliammeter. The potential drop across the germanium diode is then

0%
$0.3 V$
0%
$0.7 V$
0%
$1.1 V$
0%
$12 V$

Explanation

In Fig. germanium diode is reverse biased and silicon diode is forward biased. Therefore, there will be no current in the branch of germanium diode. The potential barrier of silicon diode is

$0.7 V$ . Therefore, for conduction minimum potential difference across silicon is

$0.7 V$ .
Maximum potential difference across resistance,

$R = 12 - 0.7 = 11.3 V.$

For the plate voltage, the plate current in a triode valve is maximum when the potential of :

0%
The gird is positive and plate is negative
0%
The grid is zero and plate is positive
0%
The grid is negative and plate is positive
0%
The grid is positive and plate is positive

Explanation

When grid is given potential more electrons will cross the grid to reach the positive plate

$p$ . Hence current increases.
1741182_1012508_ans_1cfa4fd8a680412ba6eaacc5c5b5b1df.png

The electrical conductivity of a semiconductor increases when emf radiation of wavelength shorter the

$2480$ nm is incident in it. The band gap (in eV) for the semiconductor is?

0%
$0.9$
0%
$0.7$
0%
$0.5$
0%
$1.1$

In a p-n junction diode, not connected to any circuit

0%
The potential is the same everywhere
0%
The p-type side is at a higher potential than the n-type side
0%
There is an electric field at the junction directed from the n-type side to the p-type side
0%
There is an electric field at the junction directed from the p-type side to the n-type side

Explanation

In pn function diode p side of diode (+ve side) attract

$-ve$ charge and due to induct ion opposite charge

$+ve$ charge appraises on n side which produce potential barrierand we know that charge separation produce electric fields from inside to p side of diode.

Consider two

$npn$ transistors as shown in figure. If

$0$ volts corresponds to false and

$5$ volts correspond to true then the output at

$C$ correspond to true then the output at

$C$ corresponds to:

0%
$A\ NAND\ B$
0%
$A\ OR\ B$
0%
$A$ and $B$
0%
$A\ NOR\ B$

$\overline { X }$

$\overline { X }$

$P=\overline { X } +Y$

$Q=\overline { X+Y }$

$R=\overline { P+Q }$