Blog Log in Get Started Free

NEET Practice Questions (MCQs) with Answers & Solutions

Practice free NEET NEET multiple-choice questions online with instant answers and detailed explanations. No login required.

All Physics Chemistry Botany Zoology

Language English हिंदी

● ● ● ● ●

In a PNP transistor the base is the N-region. Its width relative to the P-region is

Smaller

Larger

Same

Not related

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(a) The base is always thin

● ● ● ● ●

A common emitter amplifier is designed with NPN transistor ( $α$ = 0.99). The input impedance is 1 K $Ω$ and load is 10 K $Ω$ . The voltage gain will be

9.9

990

9900

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$β = \frac{α}{1 - α} = \frac{0.99}{(1 - 0.99)} = 99$

Resistance gain = $\frac{10 \times 10^{3}}{10^{3}} = 10$

$\Rightarrow$ Voltage gain = $99 \times 10 = 990$ .

● ● ● ● ●

The most commonly used material for making transistor is

Copper

Silicon

Ebonite

Silver

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(b) Silicon is commonly used in transistor as it is a cheap semi-conductor.

● ● ● ● ●

The part of a transistor which is heavily doped to produce a large number of majority carriers is-

Base

Emitter

Collector

None of these

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(b) Emitter is heavily doped

● ● ● ● ●

For a transistor, the current amplification factor is 0.8. The transistor is connected in common emitter configuration. The change in the collector current when the base current changes by 6 mA is

6 mA

4.8 mA

24 mA

8 mA

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(c)

$α = 0.8 \Rightarrow β = \frac{0.8}{(1 - 0.8)} = 4 Also β = \frac{∆ i_{c}}{∆ i_{b}} \Rightarrow ∆ i_{c} = β \times ∆ i_{b} = 4 \times 6 = 24 mA$

● ● ● ● ●

In a common base amplifier circuit, calculate the change in base current if that in the emitter current is 2 mA and $α$ = 0.98

0.04 mA

1.96 mA

0.98 mA

2 mA

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(a)

$∆ i_{c} = α ∆ i_{e} = 0.98 \times 2 = 1.96 mA ∴ ∆ i_{b} = ∆ i_{e} - ∆ i_{c} = 2 - 1.96 = 0.04 mA$

● ● ● ● ●

For a transistor, in a common emitter arrangement, the alternating current gain $β$ is given by

$β = {(\frac{∆ I_{c}}{∆ I_{B}})}_{V_{C}}$

$β = {(\frac{∆ I_{B}}{∆ I_{C}})}_{V_{C}}$

$β = {(\frac{∆ I_{c}}{∆ I_{E}})}_{V_{C}}$

$β = {(\frac{∆ I_{E}}{∆ I_{C}})}_{V_{C}}$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(a) For common emitter transistor : $β = {(\frac{∆ I_{c}}{∆ I_{B}})}_{V_{C}}$

● ● ● ● ●

The relation between $α$ and $β$ parameters of current gains for a transistors is given by

$α = \frac{β}{1 - β}$

$α = \frac{β}{1 + β}$

$α = \frac{1 - β}{β}$

$α = \frac{1 + β}{β}$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(b) $i_{e} = i_{b} + i_{c} \Rightarrow \frac{i_{e}}{i_{c}} = \frac{i_{b}}{i_{c}} + 1 \Rightarrow \frac{1}{α} = \frac{1}{β} + 1 \Rightarrow α = \frac{β}{(1 + β)}$

● ● ● ● ●

In the CB mode of a transistor, when the collector voltage is changed by 0.5 volt. The collector current changes by 0.05 mA. The output resistance will be

10 k $Ω$

20 k $Ω$

5 k $Ω$

2.5 k $Ω$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(a) Here $∆ V_{c} = 0.5 V, ∆ i_{c} = 0.05 mA = 0.05 \times 10^{- 3} A$

Output resistance is given by

$R_{out} = \frac{∆ V_{c}}{∆ i_{c}} = \frac{0.5}{0.05 \times 10^{- 3}} = 10^{4} Ω = 10 kΩ$

● ● ● ● ●

Consider an NPN transistor amplifier in the common-emitter configuration. The current gain of the transistor is 100. If the collector current changes by 1 mA, what will be the change in emitter current?

1.1 mA

1.01 mA

0.01 mA

10 mA

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(b) Current gain $β = \frac{∆ i_{c}}{∆ i_{b}} \Rightarrow ∆ i_{b} = \frac{1 \times 10^{- 3}}{100} = 10^{- 5} A = 0.01 mA$

By using, $∆ i_{e} = ∆ i_{b} + ∆ i_{c} \Rightarrow ∆ i_{e} = 1.01 + 1 = 1.01 mA$

Ready to ace NEET?

Free access · No credit card required

Browse All Questions Create Free Account

Frequently Asked Questions

Yes. You can attempt every NEET question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.