A npn transistor is connected in common emitter configuration in a given amplifier- A load resistance of 800 - is connected in the collector circuit and the voltage drop across it is 0-8 V- If the current amplification factor is 0-96 and the input resistance of the circuit is 192 - the voltage gain and the power gain of the amplifier will respectively be-A- 4-3-84B- 4-4 C- 4-3-69D- 3-69-3-84

The correct option is A 4, 3.84R_0 =800 Ω V_0 =0.8 V β=0.96 R_1 192 Ω Δ I_G =0.8/800=1mA Voltage gain A_V=βR_0/R_1=0.96 × 800/192=4 Power gain =β^2 R_0/R_1=0.9 ...

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Standard XII

Physics

Transistor as an Amplifier

A npn transis...

Question

A npn transistor is connected in common emitter configuration in a given ampliﬁer. A load resistance of $800 Ω$ is connected in the collector circuit and the voltage drop across it is 0.8V. If the current ampliﬁcation factor is 0.96 and the input resistance of the circuit is $192 Ω$ , the voltage gain and the power gain of the amplifier will respectively be:

4, 3.84

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4, 4

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4, 3.69

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Solution

The correct option is A 4, 3.84
$R_{0} = 800 Ω V_{0} = 0.8 V$
$β = 0.96 R_{1} 192 Ω$
$Δ I_{G} = \frac{0.8}{800} = 1 m A$
Voltage gain $A_{V} = β \frac{R_{0}}{R_{1}} = \frac{0.96 \times 800}{192} = 4$
Power gain $= β^{2} \frac{R_{0}}{R_{1}} = \frac{0.96 \times 0.96 \times 800}{192} = 3.84$

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The correct option is A 4, 3.84R0=800ΩV0=0.8V β=0.96R1192Ω ΔIG=0.8800=1mA Voltage gain AV=βR0R1=0.96×800192=4 Power gain =β2R0R1=0.96×0.96×800192=3.84

The correct option is A 4, 3.84
$R_{0} = 800 Ω V_{0} = 0.8 V$
$β = 0.96 R_{1} 192 Ω$
$Δ I_{G} = \frac{0.8}{800} = 1 m A$
Voltage gain $A_{V} = β \frac{R_{0}}{R_{1}} = \frac{0.96 \times 800}{192} = 4$
Power gain $= β^{2} \frac{R_{0}}{R_{1}} = \frac{0.96 \times 0.96 \times 800}{192} = 3.84$