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GATE ECE 2010 | Question-49

Common Data for Questions 48 and 49 :
Consider the common emitter amplifier shown below with the following circuit parameters: $\beta=100, g_{m}=0.3861 \mathrm{~A / V}, \mathrm{r}_{\mathrm{o}}=\infty, \mathrm{r}_{\mathrm{x}}=259 \; \Omega, \mathrm{R}_{\mathrm{s}}=1 \; \mathrm{k} \Omega, \mathrm{R}_{\mathrm{B}}=93 \; \mathrm{k} \Omega, \mathrm{R}_{\mathrm{C}}=250 \; \Omega, \mathrm{R}_{\mathrm{t}}=1 \; \mathrm{k} \Omega, \mathrm{C}_{1}=\infty$ and $\mathrm{C}_{2}=4.7 \mu \mathrm{F}.$


The lower cut-off frequency due to $\mathrm{C}_{2}$ is

  1. $33.9 \mathrm{~Hz}$
  2. $27.1 \mathrm{~Hz}$
  3. $13.6 \mathrm{~Hz}$
  4. $16.9 \mathrm{~Hz}$
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