Question

For a transistor, \( \alpha \) and \( \beta \) are given as \( \alpha = \frac{I_C}{I_E} \) and \( \beta = \frac{I_C}{I_B} \). Then the correct relation between \( \alpha \) and \( \beta \) will be:

• A. \( \alpha = \frac{1 - \beta}{\beta} \)
• B. \( \beta = \frac{\alpha}{1 - \alpha} \)
• C. \( \alpha \beta = 1 \)
• D. \( \alpha = \frac{\beta}{1 - \beta} \)

Hint:

The relationship between \( \alpha \) and \( \beta \) for a transistor can be derived by expressing the currents in terms of each other and using the current conservation equation.

Answer 1

The Correct Option is B

Step 1: We know the relationships: \[ \alpha = \frac{I_C}{I_E}, \quad \beta = \frac{I_C}{I_B}. \] Step 2: The total current is conserved in the transistor, so: \[ I_E = I_C + I_B. \] Step 3: From the definition of \( \alpha \), we can express \( I_C \) as: \[ I_C = \alpha I_E. \] Substituting this into the equation for \( \beta \): \[ \beta = \frac{\alpha I_E}{I_B}. \] Step 4: Since \( I_E = I_B + I_C \), we can express \( I_B \) in terms of \( I_E \) and \( \alpha \): \[ I_B = \frac{I_E}{1 - \alpha}. \] Step 5: Substituting this into the expression for \( \beta \), we get: \[ \beta = \frac{\alpha}{1 - \alpha}. \]