KCET · Physics · Semiconductors
In a transistor, the collector current varies by \( 0.49 \mathrm{~mA} \) and emitter current varies by \( 0.50 \mathrm{~mA} \).
Current gain \( \beta \) measured is
- A 19
- B \( 150 \)
- C \( 99 \)
- D \( 100 \)
Answer & Solution
Correct Answer
(A) 19
Step-by-step Solution
Detailed explanation
Given, collector current variation, \( \Delta I_{\mathrm{c}}=0.49 \mathrm{~mA} \) emittercurrent variation, \( \Delta I_{\mathrm{E}}=0.50 \mathrm{~mA} \)
Now, current gain,
\[
\begin{array}{l}
\beta=\frac{\Delta I_{C}}{\Delta I_{B}}=\frac{\Delta I_{C}}{\Delta I_{E}-\Delta I_{C}}=\frac{0.49 \mathrm{~mA}}{(0.50-0.49) \mathrm{mA}} \\
\beta=\frac{0.49 \mathrm{~mA}}{0.01 \mathrm{~mA}}=49
\end{array}
\]
Therefore, current gain \( \beta \) is \( 49 \).
Now, current gain,
\[
\begin{array}{l}
\beta=\frac{\Delta I_{C}}{\Delta I_{B}}=\frac{\Delta I_{C}}{\Delta I_{E}-\Delta I_{C}}=\frac{0.49 \mathrm{~mA}}{(0.50-0.49) \mathrm{mA}} \\
\beta=\frac{0.49 \mathrm{~mA}}{0.01 \mathrm{~mA}}=49
\end{array}
\]
Therefore, current gain \( \beta \) is \( 49 \).
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