In the following common emitter configuration an `NPN` transistor with current gain `beta=100` is used. The output voltage of the amlifier will be

Options

A. 10mV
B. 0.1V
C. 1.0V
D. 10V

Answer

C

Explanation

We known that output voltage is given by,
`v_0=v_ixxbetaxx(R_L)/(R_(BE))`
Here, `v_i=1mV, beta = 100`
`R_L=10kOmega, R_(BE)=1kOmega`
`therefore v_0=1xx10^(-3)xx100xx10/1=1.0V` .