The input resistance of a transistor is `1000 Omega` on charging its base current by `10 muA`, the collector current increases by 2 mA. If a load resistance of `5 kOmega` is used in the circuit, the voltage gain of the amplifier is

Options

A. 100
B. 500
C. 1000
D. 1500

Answer

C

Explanation

Current gain, `beta=((DeltaI_C)/(DeltaI_B))_(V_(CE))` and voltage gain, `A_V=beta(R_("out"))/(R_("in"))`
Here, `R_("in")=1000 Omega, DeltaI_B=10 muA = 10^(-5)A`
`R_("out")=5 kOmega=5xx10^3 Omega`
`DeltaI_C=2 mA = 2xx10^(-3)A`
`beta=(2xx10^(-3))/10^(-5)=200`
Hence, `A_V=(200xx5xx10^3)/1000=1000`