JEE Main & Advanced JEE Main Paper (Held on 09-4-2019 Morning)

  • 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) \[0.67k\Omega ,300\]

    Correct Answer: D

    Solution :

    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]


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