• # question_answer An NPN transistor is used in common emitter configuration as an amplifier with $1k\Omega$load resistance. Signal voltage of 10 mV is applied across the base-emitter. This produces a 3 mA change in the collector current and $15\mu A$ change in the base current of the amplifier. The input resistance and voltage gain are: [JEE Main 9-4-2019 Morning] A) $0.33k\Omega ,1.5$              B) $0.67k\Omega ,200$C) $0.33k\Omega ,300$             D) $0.67k\Omega ,300$

input current $=15\times {{10}^{6}}$ output current $=3\times {{10}^{3}}$ resistance output = 1000 ${{V}_{input}}=10\times {{10}^{3}}$ Now ${{V}_{input}}={{r}_{input}}\times {{i}_{input}}$ $10\times {{10}^{-3}}={{r}_{input}}\times 15\times {{10}^{-6}}$ ${{r}_{input}}=\frac{2000}{3}=0.67K\Omega .$ voltage gain =$=\frac{{{V}_{output}}}{{{V}_{input}}}=\frac{1000\times 3\times {{10}^{-3}}}{10\times {{10}^{-3}}}=300$ Option [d]