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Question

Consider the circuit given below:
The value of input resistance Rin is _____
  1. 102.38

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Solution

The correct option is A 102.38

500kIB+0.7+1k(IB+IC)=12

IC=100IB

IE=101 IB


from here IB=18.80 μA

rπ=β×VTIC=VTIB=2618.8kΩ=1.383 kΩ

Rin=rπ+(1+β)RE

=1.383k+101×1k=102.383 kΩ


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