Question

In a transistor $10^8$ electrons enter at the emitter in $10^{-4}$ s, out of which 2% electron go to the base, the current transfer ratio in common base configuration is

• A. 98
• B. 2
• C. 0.98
• D. 0.2

Answer 1

Answer: (C)

0.98

Answer 2

$I_{b}=2 \%$ of $I_{e},$ $I_{c}=98 \%$ of $I_{e},$ $\alpha=?$ $\alpha=\frac{I_{c}}{I_{e}}=\frac{98 \% \text { of } I_{e}}{I_{e}}$ $=98 \%=0.98$