JEE Main 2019 — Physics solution

JEE Main2019Physics

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

A
$0.33 k Ω, 1.5$
B
$0.33 k Ω, 300$
C
$0.67 k Ω, 200$
D
$0.67 k Ω, 300$

Solution

$β = \frac{l _{C}}{l _{B}} = 200 R_{i} = \frac{10 \times 10 ^{- 3}}{15 \times 10 ^{- 6}} = 0.67 K Ω V o l t a g e G ain = β (\frac{R _{0}}{R _{i}}) = 300$

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