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# In the use of transistor as an amplifier $I_f \alpha =\large\frac{I_c}{I_e}$ and $\beta =\large\frac{I_c}{I_b}$ where $I_c$- collector current $I_b$- base current $I_e$- emitter current then

$(a)\;\beta= \large\frac{1+\alpha}{\alpha} \\ (b)\;\beta = \frac{\alpha}{1+\alpha} \\ (c)\;\beta =\frac{1- \alpha}{\alpha} \\ (d)\;beta = \frac{\alpha}{1- \alpha}$
Can you answer this question?

$I_c= I_c+I_b$
$\therefore \large\frac{I_e}{I_c} =\frac{I_c +I_b}{I_c}$
$\qquad= 1+ \large\frac{I_b}{I_c}$
=> $\large\frac{1}{\alpha} =1+\large\frac{1}{\beta}$
$\therefore \beta=\large\frac{\alpha}{1- \alpha}$
Hence d is the correct answer.
answered Jan 28, 2014 by