Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

The transfer characterstic curve of a transistor, having input and output resistance $100\Omega$ and $100K\Omega$ respectively, is shown in the figure. The Voltage and Power gain are respectively:

Option 1)
$5\times10^{4},2.5\times10^{6}$

Option 2)
$5\times 10^{4},5\times10^{5}$

Option 3)
$5\times10^{4},5\times10^{6}$

Option 4)
$2.5\times10^{4},2.5\times10^{6}$

Answers (1)

best_answer

$V gain$ $=\left ( \frac{\triangle I_{C}}{\triangle I_{B}} \right )\frac{R_{out}}{R_{in}}$

$=\left ( \frac{5\times 10^{-3}}{100\times10^{-6}} \right )\times10^{3}$

$=\frac{1}{20}\times10^{6}=5\times10^{4}$

$P gain$ = $\left ( \frac{\triangle I_{c}}{\triangle I_{b}} \right )\cdot V gain$

$=\left ( \frac{5\times10^{-3}}{100\times 10^{-6}} \right )\cdot 5\times10^{4}$

$\Rightarrow 2.5\times10^{6}$

$\Rightarrow V gain =5\times10^{4}, P gain=2.5\times10^{6}$

Option 1)

$5\times10^{4},2.5\times10^{6}$

Option 2)

$5\times 10^{4},5\times10^{5}$

Option 3)

$5\times10^{4},5\times10^{6}$

Option 4)

$2.5\times10^{4},2.5\times10^{6}$

Posted by

solutionqc

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

NTA JEE Main 2026 session 1 admit card likely today; download link generated but not active yet

anam-khan

JEE Main 2026: IIT Guwahati CSE cut-off drops; BTech biosciences still least competitive

anam-khan

JEE Main Admit Card 2026 (OUT) LIVE: Session 1 hall ticket link active now; NTA announces shift timings

Latest Question