In a common - emitter amplifier- output resistance is 5000 - and input resitance is 2000 - If peak value of signal voltage is 10 mV and -50- then the peak value of output voltage isA- 1-25 VB- 5 - 10-6 VC- 125 VD- 2-5 - 10-4 V

The correct option is A 1.25 VVoltage amplification (A_v)is A_v = βR_0R_i = 50 ×50002000 A_v=V_0V_max V_0/10 × 10^ 3 = 125 (∵ V_max=10 × 10^ 3 ) Output Vol ...

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Standard XII

Physics

P-N Junction

In a common -...

Question

In a common - emitter amplifier, output resistance is $5000 Ω$ and input resitance is $2000 Ω$ . If peak value of signal voltage is $10 m V$ and $β = 50,$ then the peak value of output voltage is

5 \times 10^{- 6} V

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2.5 \times 10^{- 4} V

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1.25 V

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125 V

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Solution

The correct option is C $1.25 V$
Voltage amplification $(A_{v})$ is
$A_{v} = - β \frac{R_{0}}{R_{i}}$
$= - 50 \times \frac{5000}{2000}$

$A_{v} = \frac{V_{0}}{V_{m a x}}$
$\frac{V_{0}}{10 \times 10^{- 3}} = - 125$ $(∵ V_{m a x} = 10 \times 10^{- 3})$
Output Voltage $V_{0} = - 1.25 V$

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

The correct option is C 1.25 VVoltage amplification (Av)is Av=−βR0Ri =−50×50002000 Av=V0Vmax V010×10−3=−125 (∵Vmax=10×10−3) Output Voltage V0=−1.25V

In a common - emitter amplifier, output resistance is $5000 Ω$ and input resitance is $2000 Ω$ . If peak value of signal voltage is $10 m V$ and $β = 50,$ then the peak value of output voltage is

The correct option is C $1.25 V$
Voltage amplification $(A_{v})$ is
$A_{v} = - β \frac{R_{0}}{R_{i}}$
$= - 50 \times \frac{5000}{2000}$

$A_{v} = \frac{V_{0}}{V_{m a x}}$
$\frac{V_{0}}{10 \times 10^{- 3}} = - 125$ $(∵ V_{m a x} = 10 \times 10^{- 3})$
Output Voltage $V_{0} = - 1.25 V$