Which expression represents Johnson-Nyquist thermal noise power in a resistor?

• A. N = k T / B
• B. N = B / (k T)
• C. N = k / (T B)
• D. N = k T B

Answer: (D)

Thermal noise in a resistor comes from random motion of charge carriers, and its power across a given bandwidth is flat with respect to frequency. The power per unit bandwidth is kT, where k is Boltzmann’s constant and T is the absolute temperature. When you look over a bandwidth B, you collect that power across B, giving N = k T B. This makes intuitive sense: higher temperature means more vigorous thermal fluctuations, and a wider bandwidth lets more of those fluctuations express themselves as detectable power. The units line up too: kT has units of energy (joules), and multiplying by bandwidth (cycles per second) yields joules per second, i.e., watts.

So the Johnson-Nyquist noise power in a resistor, observed in a bandwidth B, is N = k T B. The other forms would misrepresent how noise scales with temperature or bandwidth, or would violate unit consistency.