joric
Newbie level 3
The thermal noise power of a channel with a bandwidth of 400 kHz is 1.776 x 10–15 watts, what is its operating temperature in ℃?
N = kTB
N = thermal noise in watts
k = Boltzmann’s constant = 1.38 x 10-23
T = Temperature in kelvin x + 273
T = N / kB
N = 1.776 x 10^-15
k = 1.38 x 10^-23
B = 4 x 10^5
T = 1.776 x 10^-15 / 1.38 x 10^-23 * 4 x 10^5
= 1.776 x 10^-15 / 5.2 x 10^-18
= 0.341538461 x 10^3
= 341.538k
341.538 -273 = 68.538 ℃
Is the answer correct?
N = kTB
N = thermal noise in watts
k = Boltzmann’s constant = 1.38 x 10-23
T = Temperature in kelvin x + 273
T = N / kB
N = 1.776 x 10^-15
k = 1.38 x 10^-23
B = 4 x 10^5
T = 1.776 x 10^-15 / 1.38 x 10^-23 * 4 x 10^5
= 1.776 x 10^-15 / 5.2 x 10^-18
= 0.341538461 x 10^3
= 341.538k
341.538 -273 = 68.538 ℃
Is the answer correct?