Q136 / 180 · AIIMS 2012 · Physics

For a common-emitter transistor, input current is $5 \mu \mathrm{A}, \beta=100$ circuit is operated at load resistance of $10 \mathrm{k} \Omega$, then voltage across collector emitter will be

  • Correct answer
    A
    $5 \mathrm{~V}$
  • B
    $10 \mathrm{~V}$
  • C
    $12.5 \mathrm{~V}$
  • D
    $7.5 \mathrm{~V}$

Solution

Here, $I_B=5 \mu \mathrm{A}=5 \times 10^{-6} \mathrm{~A}$ $\beta=100$ $R_L=10 \mathrm{k} \Omega=10 \times 10^3 \Omega$ As $\beta=\frac{I_C}{I_B}$ or $I_C=(100)\left(5 \times 10^{-6} \mathrm{~A}\right)=5 \times 10^{-4} \mathrm{~A}$ The voltage across collector emitter is $V_{C E}=R_L I_C=\left(10 \times 10^3 \Omega\right)\left(5 \times 10^{-4} \mathrm{~A}\right)=5 \mathrm{~V}$

Hence, the correct answer is option: 'A'.

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