AIIMS2018Physics-Electronics

AIIMS 2018 Physics Transistor Circuits MCQ Question

Type: MCQ-numerical-Medium-Class 12

An N-P-N transistor is connected in common emitter configuration in which collector supply is 9 V9\text{ V} and the voltage drop across the load resistance of 1000 Ω1000\ \Omega connected in the collector circuit is 1 V1\text{ V}. If current amplification factor is (25/26)(25/26), If the internal resistance of the transistor is 200 Ω200\ \Omega, then which of the following options is correct.

A

IB=2.04×103 AI_B = 2.04 \times 10^{-3}\text{ A}

B

IB=4.04×103 AI_B = 4.04 \times 10^{-3}\text{ A}

C

IB=1.04×103 AI_B = 1.04 \times 10^{-3}\text{ A}

D

IB=5.04×103 AI_B = 5.04 \times 10^{-3}\text{ A}

Correct Answer

Option C

Detailed Explanation

The collector current ICI_C is determined by the voltage drop across the load resistor RLR_L divided by its resistance, expressed as IC=VdropRLI_C = \frac{V_{drop}}{R_L}. Substituting the given values yields a collector current of 12 mA, which corresponds to option C. Options A (8 mA), B (10 mA), and D (15 mA) do not satisfy the equation based on the provided voltage drop and resistance, indicating they are incorrect. Understanding this relationship is crucial for analyzing transistor circuits and their behavior in amplifying signals.

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