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In a common emitter amplifier, using output resistance of 5000 Ω and input resistance of 2000 Ω, if the peak value of input signal voltage is 10 mV and β = 50, then peak value of output voltage is

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$5 \times 10^{- 6} V$

$12.5 \times 10^{- 4} V$

$1.25 V$

$125 V$

answer is C.

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Detailed Solution

$A_{V} = \frac{∆ V_{0}}{∆ V_{i}} = β \frac{R_{0}}{R_{i}} s o ∆ V_{0} = β \frac{R_{0}}{R_{i}} ∆ V_{i}$

$∆ V_{0} = 50 \frac{5000}{2000} \times 10 \times 10^{- 3} = 1.25$

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