If an amplifier has peak output voltage of 10 V across a load resistor of 1 k- - If the input power to amplifier is 100 - W - the Power gain will be

Given that nbsp; nbsp; P_i = 100 μ watt nbsp; and nbsp; nbsp; nbsp; nbsp; V_0 nbsp; nbsp; = 10 V peak to peak nbsp; ⇒ V_0 = 5 cosω t n ...

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Standard XII

Physics

Power

If an amplifi...

Question

If an amplifier has peak output voltage of 10 V across a load resistor of 1 $k Ω$ . If the input power to amplifier is 100 $μ W$ , the Power gain will be

225

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250

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500

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125

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Solution

The correct option is D 125
Given that $P_{i} = 100 μ$ watt
and $V_{0}$ = 10 V peak to peak
$\Rightarrow V_{0}$ = 5 $cos ω t$
$R_{L} = 1 K Ω$

then $P_{0} = \frac{V_{r m s}^{2}}{R_{L}}$
= 12.5 mW

so, power gain = $\frac{P_{0}}{P_{i}} = \frac{12.5 m W}{0.1 m W}$ = 125

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If an amplifier has peak output voltage of 10 V across a load resistor of 1 kΩ . If the input power to amplifier is 100 μW , the Power gain will be

The correct option is D 125Given that Pi=100μ watt and V0 = 10 V peak to peak ⇒V0 = 5 cosωt RL=1KΩ then P0=V2rmsRL = 12.5 mW so, power gain = P0Pi=12.5mW0.1mW = 125

If an amplifier has peak output voltage of 10 V across a load resistor of 1 $k Ω$ . If the input power to amplifier is 100 $μ W$ , the Power gain will be

The correct option is D 125
Given that $P_{i} = 100 μ$ watt
and $V_{0}$ = 10 V peak to peak
$\Rightarrow V_{0}$ = 5 $cos ω t$
$R_{L} = 1 K Ω$

then $P_{0} = \frac{V_{r m s}^{2}}{R_{L}}$
= 12.5 mW

so, power gain = $\frac{P_{0}}{P_{i}} = \frac{12.5 m W}{0.1 m W}$ = 125