(C) The input current $i_i$ is given by $i_i = \frac{V_i}{R_i}$,where $V_i = 100 \ mV = 0.1 \ V$ and $R_i = 2000 \ \Omega$.$i_i = \frac{0.1}{2000} = 5

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(C) The input current $i_i$ is given by $i_i = \frac{V_i}{R_i}$,where $V_i = 100 \ mV = 0.1 \ V$ and $R_i = 2000 \ \Omega$.$i_i = \frac{0.1}{2000} = 5 \times 10^{-5} \ A$.The output current $i_o$ is given by $i_o = \beta \times i_i$.$i_o = 50 \times 5 \times 10^{-5} = 250 \times 10^{-5} = 2.5 \times 10^{-3} \ A$.The output voltage $V_o$ is given by $V_o = i_o \times R_o$,where $R_o = 5000 \ \Omega$.$V_o = 2.5 \times 10^{-3} \times 5000 = 12.5 \ V$.Thus,the peak value of the output voltage is $12.5 \ V$.

Class 12 Physics Semiconductor Electronics Junction Transistor

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In a common emitter amplifier,the output resistance is $5000 \Omega$ and the input resistance is $2000 \Omega$. If the peak value of input signal voltage is $100 mV$ and $\beta=50$,then the peak value of output voltage is

AP EAMCET 2018 All AP EAMCET

A
$5 \times 10^{-6} \ V$
B
$12.5 \times 10^{-4} \ V$
C
$12.5 \ V$
D
$1.25 \ V$

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Semiconductor Electronics

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Junction Transistor

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In a common emitter amplifier,the output resistance is $5000 \Omega$ and the input resistance is $2000 \Omega$. If the peak value of input signal voltage is $100 mV$ and $\beta=50$,then the peak value of output voltage is

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