The solution is already provided and my question is how did they calculate the temperature? How come it is 290K? Please provide a detailed explanation.

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The solution is already provided and my question is how did they calculate the temperature? How come it is 290K? Please provide a detailed explanation.

69. The noise output of a resistor is
amplified by a noiseless amplifier
having a gain of 60 and a bandwidth of
20 kHz. A meter connected at the
output of the amplifier reads 1 mV RMS.
If the bandwidth of the amplifier is
reduced to 5 kHz, its gain remaining
constant, what does the meter read
now?
A. 0.5 mV
B. 0.5 uV
C.
5.0 mV
D. 5.0 uV
Transcribed Image Text:69. The noise output of a resistor is amplified by a noiseless amplifier having a gain of 60 and a bandwidth of 20 kHz. A meter connected at the output of the amplifier reads 1 mV RMS. If the bandwidth of the amplifier is reduced to 5 kHz, its gain remaining constant, what does the meter read now? A. 0.5 mV B. 0.5 uV C. 5.0 mV D. 5.0 uV
= V4KTBWR
V.²
R =
Vn
(1x 10-3
4(1.38x10)(290)(20x10°)
R = 3.123 G
with BW = 5 kHz
V, = )(290)(5x10°)
4(1.38x10-2
Vn = 0.5 mV
Transcribed Image Text:= V4KTBWR V.² R = Vn (1x 10-3 4(1.38x10)(290)(20x10°) R = 3.123 G with BW = 5 kHz V, = )(290)(5x10°) 4(1.38x10-2 Vn = 0.5 mV
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