In n – p – n transistor amplifier, the collector current is 9 mA. If 90% of the electrons emitted reach the collector, then:
Answer & explanation
Correct answer: option 3
Here, $I_C=9mA=\frac{90}{100}I_E$
$∴ I_E =10mA$
As $I_E = I_B + I_C$
$∴ I_B = I_E − I_C = 10 mA – 9 mA = 1 mA$
$α=\frac{I_C}{I_E}=\frac{9mA}{10mA}=0.9$
$β=\frac{I_C}{I_B}=\frac{9mA}{1mA}=9$