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In the following common emitter configuration an $n p n$ transistor with current gain $\beta=100$ is used. The output voltage of the amplifier will be
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Verified Answer
The correct answer is:
$1.0 \mathrm{~V}$
We know that the output voltage is given
by, $\quad v_0=v_i \times \beta \times \frac{R_L}{R_{B E}}$
Here, $v_i=1 \mathrm{mV}, \beta=100$
$R_L=10 \mathrm{k} \Omega, R_{B E}=1 \mathrm{k} \Omega$
$$
\therefore \quad v_0=1 \times 10^{-3} \times 100 \times \frac{10}{1}=1.0 \mathrm{~V}
$$
by, $\quad v_0=v_i \times \beta \times \frac{R_L}{R_{B E}}$
Here, $v_i=1 \mathrm{mV}, \beta=100$
$R_L=10 \mathrm{k} \Omega, R_{B E}=1 \mathrm{k} \Omega$
$$
\therefore \quad v_0=1 \times 10^{-3} \times 100 \times \frac{10}{1}=1.0 \mathrm{~V}
$$
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