An NPN transistor is used in common emitter configuration as an amplifier with `1k Omega` load resistance. Signal voltage of 10 mV is applied a across the base-emitter. This produces a 3 mA change in the collector current and `15 mu A` change in the base currect of the amplifier. The input resistance and voltage gain are :

Options

A. `0.67 k Omega, 200`
B. `0.33 k Omega, 300`
C. `0.67 k Omega, 300`
D. `0.33 k Omega, 1.5`

Answer

C

Explanation

input resistance `= r_(1)=(10 mV)/(15xx10^(-6))~~ 0.67xx10^(3) Omega=6.0xx10^(2)Omega=0.67 K Omega`
Voltage gain `= +B (Rc)/(r) =+ ( 3 mA)/(15 mu A)xx(1K Omega)/(0.67 K Omega) = + 300` Volt