Question

The diagram shows the biasing of an n-p-n transistor in common emitter configuration used in an amplifier.
The design of the transistor is such that 98% of the charge carriers passing through the emitter reach the collector. If base current changes from $50\mu A$ to $100\mu A$, then the corresponding change in the voltage across the load resistance will be

• A. 0.25 V
• B. 0.5 V
• C. 24.5 V
• D. 49.0 V

Answer 1

The correct option is C 24.5 V

$\beta =\frac{\Delta I_C}{\Delta I_B}=\frac{0.98I_E}{0.02I_E}=49$

$\therefore \Delta V$ (across $R_L$)$=\beta \Delta I_B.R_L=49\times 50\times {10}^{-6}\times {10}^4$

$=24.5V$