MCQEasyJEE 2023Transistors (BJT: Amplifier & Switch)

JEE Physics 2023 Question with Solution

In an n-p-n common emitter (CE) transistor, the collector current changes from 5mA5 \, \text{mA} to 16mA16 \, \text{mA} for the change in base current from 100μA100 \, \mu A and 200μA200 \, \mu A, respectively. The current gain of the transistor is:

  • A

    110110

  • B

    0.90.9

  • C

    210210

  • D

    99

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: Collector current changes from 5mA5 \, \text{mA} to 16mA16 \, \text{mA}, and base current changes from 100μA100 \, \mu A to 200μA200 \, \mu A.

Find: The current gain β\beta of the CE transistor.

For a common emitter transistor,

β=ΔICΔIB\beta = \frac{\Delta I_C}{\Delta I_B}

Here,

ΔIC=16mA5mA=11mA\Delta I_C = 16 \, \text{mA} - 5 \, \text{mA} = 11 \, \text{mA}

and

ΔIB=200μA100μA=100μA\Delta I_B = 200 \, \mu A - 100 \, \mu A = 100 \, \mu A

Now substitute the values:

β=11mA100μA=11×103100×106=110\beta = \frac{11 \, \text{mA}}{100 \, \mu A} = \frac{11 \times 10^{-3}}{100 \times 10^{-6}} = 110

Therefore, the current gain is 110110. The solution working gives 110110, which matches option A even though the solution labels the correct option as C.

Common mistakes

  • Using ICIB\frac{I_C}{I_B} instead of ΔICΔIB\frac{\Delta I_C}{\Delta I_B} is incorrect here because the question asks for current gain from the change in currents. Use differences first, then take the ratio.

  • Ignoring unit conversion between mA and μA\mu A gives a wrong answer. Convert both currents to the same unit before dividing.

  • Choosing option C only because the solution says so is a source inconsistency. The numerical working clearly gives 110110, so the defensible option is A.

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