Semiconductor Devices Test

Result

Semiconductor Devices Test - 104

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Weekly Quiz Competition

Question 1
4 / -1

In the circuit given below, V(t) is the sinusoidal voltage source, voltage drop V_AB (t) across the resistance R is

A
Is half wave rectified

B
Is full wave rectified

C
Has the same peak value in the positive and negative half cycles

D
Has different peak values during positive and negative half cycle
Question 2
4 / -1

The peak voltage in the output of a half-wave diode rectifier fed with a sinusoidal signal without filter is10 V. The D.0 component of the output voltage is

A

B
10 / πV

C
10 V

D
20 / πV

Solution
Question 3
4 / -1

Select the outputs Y of the combination of gates shown below for inputs A = 1, B = 0; A = 1, B = 1 and A = 0, B = 0 respectively

A
(0 1 1)

B
(0 0 1)

C
(1 0 0)

D
(1 1 1)
Question 4
4 / -1

In the following circuits PN–junction diodes D₁, D₂ and D₃ are ideal for the following potentials of A and B. The correct increasing order of resistance between A and B will be

A
(i) < (ii) < (iii)

B
(iii) < (ii) < (i)

C
(ii) = (iii) < (i)

D
(i) = (iii) < (ii)

Solution
Question 5
4 / -1

Find V_AB

A
10 V

B
20 V

C
30 V

D
None of these

Solution
Question 6
4 / -1

A diode is connected to 220 V (rms) A.C in series with a capacitor as shown in figure. The voltage across thecapacitor is

A
220 V

B
110 V

C
311.1 V

D

Solution
Question 7
4 / -1

A
0.25 A

B
0.45 A

C
0.56 A

D
0.64 A

Solution
Question 8
4 / -1

The contribution in the total current flowing through a semiconductor due to electrons and holes are 3/4 and 1/4 respectively. If the drift velocity of electrons is 5/2 times that of holes at this temperature, then the ratio of concentration of electrons and holes is

A
6 : 5

B
5 : 6

C
3 : 2

D
2 : 3

Solution
Question 9
4 / -1

The following configuration of gate is equivalent to

A
NAND

B
XOR

C
OR

D
None of these

Solution
Question 10
4 / -1

Figure gives a system of logic gates. From the study of truth table it can be found that to produce a high output (1) at R, we must have

A
X = 0, Y = 1

B
X = 1, Y = 1

C
X = 1,Y = 0

D
X = 0, Y = 0

Solution