# 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/R_{1} + 1/R_{2}

= 1/10K + 1/30K

= 0.1333

R= 7.502KΩ

V_{n}= √{4* 7.502*10^{3}* 1.381×10^{-23} * 300* 10*10^{3}}

= √124.323*10^{-14}

= 11.15*10^{-7}