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.
a. `30.15 \times 10^{-3} `
b. `8.23 \times 10^{-23}`
c. `11.15 \times 10^{-7}`
d. `26.85 \times 10^{-7}`
Option C `11.15 \times 10^{-7}` is the Correct answer.
Explanation:
Noise voltage Vn = `\sqrt{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
Vn = `\sqrt{4R KTB}`
Here for resistors to be in parallel,
= `\frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}}`
= `\frac{1}{10K} + \frac{1}{30K} `
= 0.1333 R
= 7.502KΩ
Vn = `\sqrt{4 \times 7.502 \times 10^{3} \times 1.381 \times 10^{-23} \times300 \times 10 \times 10^{3}}`
= `\sqrt{124.323 \times 10^{-14}}`
= `11.15 \times 10^{-7}`
0 تعليقات