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In common emitter amplifier, input resistance is $1000 \Omega$, peak value of input signal voltage is $5 \mathrm{mV}$ and $\beta=60$. The peak value of output current is
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The correct answer is:
$3 \times 10^{-4} \mathrm{~A}$
$r_{L}=1000 \Omega$
$v_{i}=5 \mathrm{mV}=5 \times 10^{-3} \mathrm{~V}$
$\beta=60$
$I_{\text {input }}=\frac{5 \times 10^{-3}}{1000}=5 \times 10^{-6} \mathrm{~A}$
$\therefore I_{\text {output }}=60 \times 5 \times 10^{-6} \mathrm{~A}=300 \times 10^{-6} \mathrm{~A}$
$=3 \times 10^{-4} \mathrm{~A}$
$v_{i}=5 \mathrm{mV}=5 \times 10^{-3} \mathrm{~V}$
$\beta=60$
$I_{\text {input }}=\frac{5 \times 10^{-3}}{1000}=5 \times 10^{-6} \mathrm{~A}$
$\therefore I_{\text {output }}=60 \times 5 \times 10^{-6} \mathrm{~A}=300 \times 10^{-6} \mathrm{~A}$
$=3 \times 10^{-4} \mathrm{~A}$
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