Browse Questions

# The emitter current in a transistor is $1mA$ If $\alpha =0.95$ value of base current is :

$(a)\;0.05 \;mA \\ (b)\;0.95\;mA \\ (c)\;10\;mA \\ (d)\;1.5 \;mA$

$\alpha =\large\frac{I_C}{I_B}$
$I_E=I_B+I_C$
$I_C=I_E-I_B$
$0.95=\large\frac{1 mA-I_B}{I_B}$
$\qquad= \large\frac{1 m A}{I_B} -\frac{I_B}{I_B}$
$1.95 =\large\frac{1mA}{I_B}$$-1$
$1.95=\large\frac{1mA}{I_B}$
$I_B= \large\frac{1mA}{1.95}$
$I_B= 0.05 \;mA$
Hence a is the correct answer.