1.

For a transistor in common emitter configuration, the voltage drop across the load of 1000 `Omega` is 0.5 V. If the value of `alpha ` for the transistor is 0. 98, then the base current will be approximately equal toA. 5 `muA`B. 8`muA`C. 10`muA`D. 15`muA`

Answer» Correct Answer - C
`beta=0.98/(1-0.98)=98/2 = 49`
`I_C=V/R = 0.5/1000 = 5xx10^(-4) A`
`therefore beta=I_C/I_B`
`therefore I_B=I_C/beta=(5xx10^(-4))/49 = (500xx10^(-6))/49=10 muA`


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