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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?` A. `29 K Omega`B. `82 K Omega`C. `108 K Omega`D. `55 K Omega` |
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Answer» Correct Answer - B `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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