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Physics MCQs for NEET — Practice Questions with Answers

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The most commonly used material for making transistor is

Copper

Silicon

Ebonite

Silver

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Explanation

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

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

Base

Emitter

Collector

None of these

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Explanation

(b) Emitter is heavily doped

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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

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Explanation

(c)

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

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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

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Explanation

(a)

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

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

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

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

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

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

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Explanation

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

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

$α = \frac{β}{1 - β}$

$α = \frac{β}{1 + β}$

$α = \frac{1 - β}{β}$

$α = \frac{1 + β}{β}$

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Explanation

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

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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 $Ω$

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Explanation

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

Output resistance is given by

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

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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

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

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In a common base amplifier the phase difference between the input signal voltage and the output voltage is

$π / 4$

$π / 2$

$π$

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Explanation

(a) In CB amplifier Input and output voltage signal are in same phase.

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In NPN transistor the collector current is 10 mA. If 90% of electrons emitted reach the collector, then

Emitter current will be 9 mA

Emitter current will be 11.1 mA

Base current will be 0.1 mA

Base current will be 0.01 mA

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Explanation

(b) As 90% of the electrons reach the collector, then

$\Rightarrow$ 90% of emitter current = 10 mA

$\Rightarrow$ emitter current = $\frac{10}{90} \times 100 = 11.1 m A$

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