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An NPN transistor is used in common emmitter configuration as an amplifire with $1k\Omega$ load resistance. Signal voltage of $10\; mV$ is applied across the base-emitter. This produces a $3 \; mA$ change in the collector current and $15\: \mu A$ change in the base current of the amplifier. The input resistance and voltage gain are :

Option 1)
$0.33\: k\Omega ,1.5$ Drift along river flow $x= d\left ( \frac{u}{v} \right )$

Option 2)
$0.67\: k\Omega ,300$

Option 3)
$0.67\: k\Omega ,200$

Option 4)
$0.33\: k\Omega ,300$

Answers (1)

best_answer

$R_{L}=1K\Omega$

$\Delta I_{B}=15\mu A$

$\Delta I_{C}=3\: m A$

$V in =10 \; mv$

$\frac{Vin}{\Delta I_{B}}=r\; in$

$r\; in=\frac{10mV}{15\times 10^{-6}}$

$r in=0.67\; K\Omega$

$\Delta v=-\beta \frac{R_{L}}{r_{i}}$

$\beta=\frac{\Delta I_{C}}{\Delta I_{B}}$

$\Delta v=-\left ( \frac{3 mA}{15\mu A} \right )\left ( \frac{1K\Omega }{2000}\times 3 \right )$

$\Delta v=-300$

Option 1)

$0.33\: k\Omega ,1.5$ Drift along river flow $x= d\left ( \frac{u}{v} \right )$

Option 2)

$0.67\: k\Omega ,300$

Option 3)

$0.67\: k\Omega ,200$

Option 4)

$0.33\: k\Omega ,300$

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