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NEET PHYSICSEasy

An n-p-n transistor is connected in a common emitter configuration (see figure) in which collector voltage drop across load resistance (800 Ω800\ \Omega) connected to the collector circuit is 0.8 V0.8\text{ V}. The collector current is:

A

2 mA2\text{ mA}

B

0.1 mA0.1\text{ mA}

C

1 mA1\text{ mA}

D

0.2 mA0.2\text{ mA}

Step-by-Step Solution

The collector current (ICI_C) can be calculated using Ohm's law, given the voltage drop across the load resistance (VRLV_{R_L}) and the load resistance (RLR_L). Given: Voltage drop across load resistance, VRL=0.8 VV_{R_L} = 0.8\text{ V} Load resistance, RL=800 ΩR_L = 800\ \Omega Collector current, IC=VRLRLI_C = \frac{V_{R_L}}{R_L} IC=0.8800=11000 A=10−3 A=1 mAI_C = \frac{0.8}{800} = \frac{1}{1000}\text{ A} = 10^{-3}\text{ A} = 1\text{ mA}.

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