Q. The current gain of a transistor in common emitter mode is 49. The change in collector current and emitter current corresponding to the change in base current by 5.0 $\mu$A are
A) $\mathrm{\Delta}i_{C} = 245\ \mu A$, $\mathrm{\Delta}i_{E} =$ 250 $\mu A$
B) $\mathrm{\Delta}i_{C} = 252\ \mu A$, $\mathrm{\Delta}i_{E} =$ 145 $\mu A$
C) $\mathrm{\Delta}i_{C} = 125\ \mu A$, $\mathrm{\Delta}i_{E} =$ 250 $\mu A$
D) $\mathrm{\Delta}i_{C} = 252\ \mu A$, $\mathrm{\Delta}i_{E} =$ 230 $\mu A$
Explanation
(a) Current gain in common emitter mode of transistor
$\beta = \frac{\mathrm{\Delta}i_{C}}{\mathrm{\Delta}i_{B}}$
Or $\ \ \ \ \ \ \ \ \mathrm{\Delta}i_{C} = \ \beta\mathrm{\Delta}i_{B}$
Given, $\ \ \ \ \ \ \beta =$49, $\mathrm{\Delta}i_{B} = 5.0\ \mu A$
$\mathrm{\Delta}i_{C} = 49 \times 5.0 = 245\mu A$
For a transistor, emitter current is the sum of base current and collector current.
ie, $i_{E} = i_{C} + i_{B}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \mathrm{\Delta}i_{E} = {\mathrm{\Delta}i}_{C} + {\mathrm{\Delta}i}_{B}$
$\therefore$ $\mathrm{\Delta}i_{E} = 245 + 5.0 = 250\ \mu A$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.