In the circuit shown in the figure, the input voltage `V_(i)` is `20V,V_(BE)=0` and `V_(CE)=0`. The values of `I_(B),I_(C)` and `beta` are given by:

Options

A. `I_(B) = 20muA, I_(C)=5mA, beta=250`
B. `I_(B)=25muA, I_(C) = 5mA, beta=200`
C. `I_(B)=40muA, I_(C)=10 mA, beta=250`
D. `I_(B) = 40 muA, I_(C)=5mA, beta=125`

Answer

D

Explanation

Given, `V_(BE)=0V, V_(CE)=0V` and `V_(i)=20V`

Applying Kirchoff's law to the base-emitter loop, we get
`V_(i) = I_(B)R_(B) + V_(BE)`
Substituting the values, we get
`20 = I_(B) xx (500 xx 10^(3))+ 0`
`rArr I_(B) = (20)/(500 xx 10^(3)) = 0.04 xx 10^(-3)`
`=40 xx 10^(-6)=40muA`.........(i)
Similarly, `V_(CC) = I_(C)R_(C)+V_(CE)`
Substituting the given values, we get
`20 = I_(C) xx (4 xx 10^(3))+0`
`rArr =I_(C) = 20/(4 xx 10^(3))= 5 xx 10^(-3)=5mA`........(ii)
current gain is given as `beta=I_(C)/I_(B)`
Subsituting the value of `I_(B)` and `I_(C)` from Eqs. (i) and (ii), we get
`rArr beta=(5 xx 10^(-3))/(40 xx 10^(-6))=0.125 xx 10^(3)=125`