Question

The base current of a transistor is $105\mu A$ and the collector current is $2.05mA$, $\alpha$ of the transistor is

• A. 0.951
• B. 9.51
• C. 95.1
• D. 0.0951

Answer 1

The correct option is A 0.951

Given that:
$I_b=105\mu A=1.05\times {10}^{-4}A$
$I_c=2.05mA=2.05\times {10}^{-3}A$

For transistors,
$\beta =\frac{I_c}{I_b}$
$\beta =\frac{2.05\times {10}^{-3}}{1.05\times {10}^{-4}}$
$\beta =19.52$

Therefore,
$\alpha =\frac{\beta }{1+\beta }$
$\alpha =\frac{19.52}{1+19.52}$
$\alpha =\frac{19.52}{20.52}=0.951$