Q190 / 200 · MHT CET - Medical 2026 (25 April Shift 1) · Physics
If the current gain in common-base configuration ($\alpha$) of a transistor increases from 0.95 to 0.96, the percentage change in current gain in common-emitter configuration ($\beta$) is approximately
- A$1\%$
- B$10\%$
- Correct answerC$26\%$
- D$50\%$
Solution
The relation between common-emitter current gain $\beta$ and common-base current gain $\alpha$ is given by $\beta = \dfrac{\alpha}{1 - \alpha}$.
For the initial state, $\alpha_1 = 0.95$.
$\beta_1 = \dfrac{0.95}{1 - 0.95} = \dfrac{0.95}{0.05} = 19$
For the final state, $\alpha_2 = 0.96$.
$\beta_2 = \dfrac{0.96}{1 - 0.96} = \dfrac{0.96}{0.04} = 24$
The change in current gain is $\Delta\beta = \beta_2 - \beta_1 = 24 - 19 = 5$.
The percentage change in $\beta$ is $\dfrac{\Delta\beta}{\beta_1} \times 100 = \dfrac{5}{19} \times 100 \approx 26.31\%$.
Thus, the percentage change is approximately $26\%$.
Answer: $26\%$
For the initial state, $\alpha_1 = 0.95$.
$\beta_1 = \dfrac{0.95}{1 - 0.95} = \dfrac{0.95}{0.05} = 19$
For the final state, $\alpha_2 = 0.96$.
$\beta_2 = \dfrac{0.96}{1 - 0.96} = \dfrac{0.96}{0.04} = 24$
The change in current gain is $\Delta\beta = \beta_2 - \beta_1 = 24 - 19 = 5$.
The percentage change in $\beta$ is $\dfrac{\Delta\beta}{\beta_1} \times 100 = \dfrac{5}{19} \times 100 \approx 26.31\%$.
Thus, the percentage change is approximately $26\%$.
Answer: $26\%$
Hence, the correct answer is option: 'C'.