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For a transistor connected in common emi...

For a transistor connected in common emitter mode, the voltage drop across the collector is 2V and beta is 50. If `R_(c)` is `2 kOmega`, the base current is `axx10^(-5) A`. What is the value of a ?

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The correct Answer is:
2
Voltage drop across collector `= I_cR_c`
`:. 2=I_c xx(2xx10^3) ` or `I_c = 10^(-3)A`
`I_b =I_c/beta=(10^(-3))/50=2xx10^(-5)A`
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