NEET Practice Questions (MCQs) with Answers & Solutions

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The emitter-base junction of a transistor is …… biased while the collector-base junction is ……. biased

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Explanation

(d) The emitter base junction is forward biased while collector base junction is reversed biased.

If α = 0.98 and current through emitter ie = 20 mA, the value of β is 

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Explanation

(b)

β=α1-α=0.981-0.98=49

In a PNP transistor working as a common-base amplifier, current gain is 0.96 and emitter current is 7.2 mA. The base current is 

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Explanation

(c) 

α=icie=0.96 and ie=7.2 mAic=0.96×ie=0.96×7.2=6.91 mA ie=ic+ib7.2=6.91+ibib=0.29 mA

 If l1l2l3 are the lengths of the emitter, base and collector of a transistor then 

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Explanation

(d) For a transistor : length of collector > length of emitter> length of the base.

In an NPN transistor circuit, the collector current is 10 mA. If 90% of the electrons emitted reach the collector, the emitter current (iE) and base current (iB) are given by 

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Explanation

(d) 

ic=90100×iE10=0.9×iE=11mAAlso iE=iB+iciB=11-10=1mA

In a common emitter transistor, the current gain is 80. What is the change in collector current, when the change in base current is 250 μ

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Explanation

(a) Current gain β=icibic=β×ib=80×250 μA

Least doped region in a transistor 

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Explanation

(b) In transistor, base is least doped.

The transistors provide good power amplification when they are used in 

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Explanation

(b) Power gain is maximum in common emitter configuration as the current gain and voltage gain are medium but the output is the inverse of the input.

The transfer ratio of a transistor is 50. The input resistance of the transistor when used in the common-emitter configuration is 1 KΩ. The peak value for an A.C input voltage of 0.01 V peak is 

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Explanation

(d) 

β=50, Ri=1000Ω, Vi=0.01Vβ=icib and ib=ViRi=0.01103=10-5A

Hence, ic=50×10-5 A=500μA

For a transistor the parameter β = 99. The value of the parameter α is

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Explanation

(b) α=β1+β=991+99=0.99

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