In a common emitter amplifier, using output resistance of $5000\;{\rm{\Omega }}$ and input resistance of $2000\;{\rm{\Omega }}$, if the peak value of input signal voltage is 10 mV and ${\rm{\beta }} = 50$, then peak value of output voltage is

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

$12.50 \times {10^{ - 6}}{\rm{\;V}}$

NEET UG PHYSICS MCQ: SEMI CONDUCTOR DEVICES

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Application +4 | -1

In a common emitter amplifier, using output resistance of

5000\;\mathrm{\Omega }

and input resistance of

2000\;\mathrm{\Omega }

, if the peak value of input signal voltage is 10 mV and

\mathrm{\beta } = 50

, then peak value of output voltage is

View answer and explanation

Correct option: C

1.25 V

Option C (1.25 V) is correct because it is the calculated peak value of the output voltage using the given parameters in a common emitter amplifier.

Practice All Chapter MCQs →

I = A{T^2}{e^{q/{V_L}}}

{V_L}

\sqrt{n}.

n

p - n\;

{10^{17}}\mathrm{c}{\mathrm{m}^{ - 3}}

P

{10^{16}}

\mathrm{c}{\mathrm{m}^{ - 3}}

N

kT =

{n_i} = 1.4 \times {10^{10}}

\mathrm{c}{\mathrm{m}^{ - 3}}