Q. In a common emitter transistor amplifier, the output resistance is 500 k$\Omega$ and the current gain $\beta$=49. If the power gain of the amplifier is ${5 \times 10}^{6}$, the input resistance is
A) 325$\ \Omega$
B) 165 $\Omega$
C) 198 $\Omega$
D) 240 $\Omega$
Explanation
(d) Given, $R_{o} = \ $500 k$\Omega$, $\beta\ $= 49 and $P$ = ${5 \times 10}^{6}$
We have $\ \ \ \ \ \ \ P = {\beta\ }^{2}\ \frac{R_{o}}{R_{i}}$
${5 \times 10}^{6} = \frac{(49)^{2} \times 500}{R_{i}}$
$R_{i} = \frac{49 \times 49 \times 500}{{5 \times 10}^{6}} =$240 $\Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.