Calculation of thermal noise when resistors are connected in parallel

Q22. At a room temperature of 300K, calculate the thermal noise generated by two resistors of 10KΩ and 30 KΩ when the bandwidth is 10 KHz and the resistors are connected in parallel.

Option:

30.15*10^-3
8.23*10^-23
11.15*10^-7
26.85*10^-7

Explanation:

Noise voltage V_n= √(4R KTB)
Where, K = 1.381×10^-23 J/K, joules per Kelvin, the Boltzmann constant
B is the bandwidth at which the power P_n is delivered.
T noise temperature
R is the resistance
Noise voltage by resistors when connected in parallel is
V_n= √{4R KTB}
Here for resistors to be in parallel,
1/R=1/R₁ + 1/R₂
= 1/10K + 1/30K
= 0.1333
R= 7.502KΩ
V_n= √{4* 7.502*10³* 1.381×10^-23 * 300* 10*10³}
= √124.323*10^-14
= 11.15*10^-7