Modern Physics 6 Question 7

7. An npn transistor is used in common emitter configuration as an amplifier with 1kΩ load resistance. Signal voltage of 10mV is applied across the base-emitter. This produces a 3 mA change in the collector current and 15μA change in the base current of the amplifier. The input resistance and voltage gain are

(a) 0.67kΩ,200

(b) 0.33kΩ,1.5

(c) 0.67kΩ,300

(d) 0.33kΩ,300

(Main 2019, 9 April I)

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Solution:

  1. Given, load resistance, RL=1kΩ

Input voltage, Vin =10mV=10×103V

Base current, ΔIB=15μA=15×106A

Collector current, ΔIC=3mA

Input resistance,

Rin=VinΔIB=10×10315×106=0.67kΩ

and voltage gain =β×RLRin =ΔIC×RLΔIB×Rin 

=3mA15μA×1kΩ0.67kΩ=3×10315×1061×1030.67×103=1000×3×315×2=300(0.672/3)

Alternate Solution

Voltage gain =Voutput Vinput =RL×ΔUCVin 

=1×103×3×10310×103=300



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