Semiconductor Electronics: Materials, Devices and Simple Circuits Test 42

Result

Semiconductor Electronics: Materials, Devices and Simple Circuits Test 42

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

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.

A
$$(0, 1, 1)$$

B
$$(1, 0, 1)$$

C
$$(1, 1, 1)$$

D
$$(1, 0, 0)$$

Solution

Each gate in the above circuit is a NAND gate which gives output as 0 only when both the inputs are 1 while gives 0 in rest of the cases.
Truth table of the circuit is shown above which implies that option D is correct.
Question 2
1 / -0

Identify the gate used in the following diagram.

A
AND

B
OR

C
NAND

D
NOR

Solution

Truth table:-
A B output
1 1 1
0 0 0
1 0 0
Truth table corresponds to an AND gate.
Question 3
1 / -0

A pure semiconductor is __________.

A
an extrinsic semiconductor

B
an intrinsic semiconductor

C
p-type semiconductor

D
n-type semiconductor

Solution

A pure semiconductor is "an intrinsic semiconductor"
Question 4
1 / -0

With increase in temperature the conductivity of?

A
Metals increases and of semiconductor decreases

B
Semiconductors increases and of metals decreases

C
In both metals and semiconductors increases

D
In both metal and semiconductor decreases

Solution

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.
Question 5
1 / -0

Diffusion is the process of.

A
Rarefaction of particles

B
Movement of particles through a semipermeable membrane

C
Movement of particles from higher concentration to lower concentration

D
Accumulation of particles on a solid surface

Solution

Diffusion is a physical process in which molecules move from area of higher concentration to area of lower concentration.
Question 6
1 / -0

A photodiode is symbolised by the graphic.

A

B

C

D

Solution
Question 7
1 / -0

A wire connected in the left gap of a meter bridge balance is $$10\Omega$$ resistance in the right gap to a point, which divides the bridge wire in the ratio $$3:2$$. If the length of the wire is $$1$$. The length of one ohm wire is?

A
$$0.057$$m

B
$$0.067$$m

C
$$0.37$$m

D
$$0.134$$m

Solution

Given,
$$R:S=3:2$$
$$S=10\Omega$$
Ratio, $$\dfrac{R}{S}=\dfrac{3}{2}$$
$$R=\dfrac{3}{2}\times 10=15\Omega$$
The length of $$1\Omega $$ wire is given by
$$l=\dfrac{1}{15}$$
$$l=0.067m$$
The correct option is B.
Question 8
1 / -0

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?

A
$$\displaystyle\frac{2}{3}$$mA

B
$$1$$mA

C
$$10$$mA

D
$$\displaystyle\frac{3}{2}$$mA

Solution

$$\because$$ 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.
Question 9
1 / -0

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 $$62.5\ W$$?

A
$$15\Omega$$

B
$$10\Omega$$

C
$$5\Omega$$

D
$$25\Omega$$

Solution

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.
Question 10
1 / -0

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

A
Hole

B
Majority carrier

C
Minority carrier

D
Electron

Solution