1.

What is the base resistance R_(B) in the circuit as shown in figure , if beta_(d.c.) = 90, V_(BE) = 0.7 V, V_(CE) =4V?

Answer»

`29 K OMEGA`
`82 K Omega`
`108 K Omega`
`55 K Omega`

SOLUTION :`V_(CE)=I_(c)xx2 xx10^(3)+9=4`
`I_(c)(5xx10^(-3))/(2) rArr2.5 xx 10^(-3)`
`beta=90 (I_(C))/(I_(B))`
`I_(beta)=(2.5 xx10^(-3))/(90)`
`V_(BE)=-I_(B)R_(B) +3`
`0.7 + (2.5 xx 10^(-3))/(50) R_(B) =3`
`R_(B)=(2.3xx90)/(2.5xx10^(-3))`
`=(2.3 xx 10^(-3))/(25) xx 10^(4)`
`rArr (2070)/(25) xx 10^(3)rArr 82 K Omega`


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