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A change of 8.0mA in the emitter current...

A change of `8.0mA` in the emitter current brings a change of 7.9mA in the collector current. How much change in the base current is required to have the same change 7.9mA in the collector current?Find the values of `(alpha)`and `(beta)`.

Text Solution

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We have
`I_E=I_B+I_C`
`or,(Delta)I_E=(Delta)I_B+(Delta)I_C`
Form the question,when (Delta)I_E=8.0mA,(Delta)I_C=7.9mA.`
Thus,
`(Delta)I_B=8.0mA-7.9mA=0.1mA.`
So a change of 0.1mA in the base current is required to have a change of 7.9mA in the collector current.
`(alpha)=I_(C)/I_(E)=(Delta)I_(C)/(Delta)I_(E)`
`7.9mA/8.0mA~~0.99.`
`(beta)=I_(C)/I_(B)=(Delta)I_(C)/(Delta)I_(B)`
`7.9mA/0.1mA=79.`
Check if these values of (alpha) and (beta)satisfy the equation
`(beta)=(alpha)/(1-(alpha)).`
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