Free Objective Test 02 Practice Test - 11th and 12th

Question 1

Current gain in common base configuration is less than 1 because

I_{e} < I_{b}

I_{b} < I_{e}

I_{c} < I_{e}

I_{e} < I_{c}

SOLUTION

Solution : C

$\propto= \frac{I_{c}}{I_{e}} < 1$ or $I_{c} < I_{e}$

Question 2

Current gain $β_{A C}$ in common emitter mode of transistor is

β_{A C} = \frac{△ I_{C}}{△ I_{B}}, V_{C} = c o n s t a n t

β_{A C} = \frac{△ I_{B}}{△ I_{C}}, V_{C} = c o n s t a n t

β_{A C} = \frac{△ I_{C}}{△ I_{E}}, V_{C} = c o n s t a n t

β_{A C} = \frac{△ I_{E}}{△ I_{C}}, V_{C} = c o n s t a n t

SOLUTION

Solution : A

It is $β_{A C} = \frac{△ I_{C}}{△ I_{B}}, V_{C} = c o n s t a n t$
Collector voltage is to be kept a constant.

Question 3

Current gain of a transistor in common base mode is 0.95. Its value in common emitter mode is

A. 0.95

B. 1.5

C. 19

D. (19)

^{- 1}

SOLUTION

Solution : C

$β = \frac{\propto}{1 - \propto} = \frac{0.95}{1 - 0.95} = \frac{0.95}{0.05}$ = 19

Question 4

In a common emitter transistor amplifier $β$ = 60, $R_{0}$ = 5000 $Ω$ and internal resistance of a transistor is 500 $Ω$ . The voltage amplification of amplifier will be

A. 500

B. 460

C. 600

D. 560

SOLUTION

Solution : C

$A_{v} = β \frac{R_{o}}{R_{i}} = 60 \times \frac{5000}{500}$ = 600

Question 5

A transistor has a base current of 1 mA and emitter current 90 mA. The collector current will be

A. 90 mA

B. 1 mA

C. 89 mA

D. 91 mA

SOLUTION

Solution : C

$I_{c} = I_{e} - I_{b}$ = 90 - 1 = 89 mA.

Question 6

A p-n junction (D) shown in the figure can act as a rectifier. An alternating current source (V) is connected in the circuit. The current (I) in the resistor (R) can be shown by

A. a

B. b

C. c

D. d

SOLUTION

Solution : B

The given circuit works is half wave rectifier. In this circuit, we will get current through R when p-n junction diode is forward biased and no current when p-n junction is reversed biased. Thus the current through resistance will be shown by graph (b).

Question 7

How many NAND gates are used in an OR gate?

A. Four

B. Two

C. Three

D. Five

SOLUTION

Solution : C

For this purpose we use three NAND gate in manner as shown. The first two NAND gates are operated as NOT gates and their outputs are fed to the third. The resulting circuit is OR gate.

Question 8

A Si and a Ge diode has identical physical dimensions. The band gap in Si is larger than that in Ge. An identical reverse bias is applied across the diodes

A. The reverse current in Ge is larger than that in Si

B. The reverse current in Si is larger than that in Ge

C. The reverse current is identical in the two diodes

D. The relative magnitude of the reverse currents cannot be determined from the given data only

SOLUTION

Solution : C

When a diode is reverse biased, then the applied voltage supports the barrier voltage. Due to it, the reverse current is weak. It will be identical in two diodes.

Question 9

To get an output Y = 1 from the circuit shown in figure the inputs A, B and C must be respectively

A. 1, 0, 1

B. 1, 1, 0

C. 0, 1, 0

D. 1, 0, 0

SOLUTION

Solution : A

The output of OR gate is

            Y’ = (A + B)

The output and AND gate is

            Y = Y’ . C = (A + B) . C

If  A = 1, B = 0, C = 1, they Y = (1 + 0). 1 = 1

Question 10

In an unbiased p-n junction,

A. potential at p is equal to that at n

B. potential at p is +ve and that at n is –ve

C. potential at p is more than that at n

D. potential at p is less than that at n

SOLUTION

Solution : D

Option (d)
At the barrier, there are electron concentration on p side.

Question 11

The combination of the following gates produces

A. AND gate

B. NAND gate

C. NOR gate

D. OR gate

SOLUTION

Solution : A

Output of first NAND gate is, $Y^{'} = ¯ ¯¯¯¯¯¯¯¯¯ ¯ A . B$
Output of second NAND gate is,
$Y = {¯ ¯¯ ¯ Y}^{'} = ¯ ¯¯¯¯¯¯¯¯¯ ¯ ¯ ¯¯¯¯¯¯¯¯¯ ¯ A . B = A . B$
Thus AND gate is formed.

Question 12

The current I through 10 $Ω$ resistor in the circuit shown in figure.

A. 50 mA

B. 20 mA

C. 40 mA

D. 80 mA

E. 35 mA

SOLUTION

Solution : D

In the given circuit, junction diode $D_{1}$ if forward biased, will conduct current and junction diode $D_{2}$ is reverse biased, will not conduct current. Therefore, current
I = $\frac{2}{10 + 15}$ = 0.08 A = 80 mA.

Question 13

In the circuit as shown in figure, A and B represent two inputs and C represents the

A. OR gate

B. NOR gate

C. AND gate

D. NAND gate

SOLUTION

Solution : A

The circuit represents OR gate, as the output at C is 1, when either A or B or both A and B have input at level 1. But output at C is zero, when both A and B are at zero level and the Boolean expression is C = (A + B).

Question 14

A working transistor with its three legs marked P, Q and R is tested using a multi-meter. No conduction is found between P and Q. By connecting the common (negative) terminal of the multimeter to R and the other (positive) terminal to P or Q, some resistance is seen on the multi-meter. Which of the following is true for the transistor?

A. It is an n-p-n transistor with R as collector

B. It is an n-p-n transistor with R as base

C. It is an p-n-p transistor with R as collector

D. It is an p-n-p transistor with R as emitter

SOLUTION

Solution : B

When a multimeter is connected between P and Q there is no conduction between P and Q, then P and Q are of same type of semiconductor, ie, either both are on n-type or of p-type. It means emitter and collector of a transistor are P and Q. When R is connected to negative terminal of multimeter and positive terminal to P or Q, then emitter-base junction will conduct to some extent due to reverse biasing.

The similar is the case of collector – base junction. Thus the transistor is n-p-n transistor with R as base.

Question 15

In common base mode of transistor, the collector current is 5.488 mA for an emitter current of 5.60 mA. The value of the base current amplification factor $(β)$ will be

A. 48

B. 49

C. 50

D. 51

SOLUTION

Solution : B

$I_{b} = I_{e} - I_{c}$ = 5.60 - 5.488 = 0.112 mA
$β = \frac{I_{e}}{I_{b}} = \frac{5.488}{0.112}$ = 49.

Free Objective Test 02 Practice Test - 11th and 12th

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

Related ExamsObjective Test 02Objective Test 02Objective Test 02

Related Exams
Objective Test 02
Objective Test 02
Objective Test 02