In an n- p-n transistor cirrect , the collectorcurrect is `10m A if 90 %` of the electrons reach the collector.
A
the emitter currect will be `9mA`
B
the base currect will be `1 mA`
C
the emitter currect will be `11 mA`
D
the base currect will be `- 1 mA`
Text Solution
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The correct Answer is:
B, C
`1_(e) = 10mA` `90%` of eklectrons emitted produce a cellector currect of `10mA` . The base currect `I_(b) = 10% of aU_(c) = (10)/(100) xx 10 = 1mA` Now I_(b) =I_(c) = 1 + 10 = 11mA`
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