Consider the circuit given below: The value of input resistance Rin is______kΩ.
102.38
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Solution
The correct option is A 102.38 Base emitter loop 500kIB+0.7+1k(IB+IC)=12 IC=100IB IE=101IB from here IB=18.80μA rπ=β×VTIC=VTIB=2618.8 kΩ=1.383 kΩ Rin=rπ+(1+β)RE =1.383k+101×1k=102.383 kΩ