Question

(C4.002.P01) (b) The Wien's bridge is shown in Figure 2 has a series RC in one am and a parallel combination in adjoining am. It is designed tomasure frequency. EKT 21274 PRINCIPLE OF MEASUREMENT AND INSTRUMENTATION ASSIGNMENT CHAPTER 4 SEMESTER II SESSION 2019 2020 CA Detector Figure 2 The impedance of series RC is given by ZI -R1-Cl and the admittance of a parallel combination is given Y3 - 1/R3+jC3, derive an equation to compute frequency, (5 Marks) i) Given that f -3981, RI - 400, R2 = 1000, R3 = 300 and C1-203. Assume that the bridge is balanced, calculate CI, C and R4. (5 Marks)

Answer 1

b).Yes

Given circuit is wien's bridge , it is designed to measure frequency of a given system.

i).

When detector detects zero deflection then this is called balanced bridge

During balanced condition

$Z_1Z_4=Z_2Z_3$

$Z_1=R_1-\frac {j}{\omega C_1}, \ Z_4=R_4, \ Z_3=\frac{R_3}{1+j\omega C_3 R_3}, \ Z_2= R_2$

$(R_1-\frac {j}{\omega C_1})R_4=\frac{R_3}{1+j\omega C_3 R_3}\times R_2$

$(1+j\omega C_1R_1)({1+j\omega C_3 R_3})R_4=j\omega C_1R_2R_3$

Separate real and imaginary parts

Real parts

$(1-{\omega^2}C_1C_3R_3R_1)R_4=0$

$\omega = \frac{1}{\sqrt{C_1C_3R_1R_3}}$

$f = \frac{1}{2 \pi \sqrt{C_1C_3R_1R_3}}$

Imaginary parts

$\frac{C_3}{C_1}+\frac{R_1}{R_3}=\frac{R_2}{R_4}$

ii).

Given

$f=398Hz$

$C_1=2C_3$

We know

$f = \frac{1}{2 \pi \sqrt{C_1C_3R_1R_3}}$

$f = \frac{1}{2 \pi \sqrt{2C_3^2R_1R_3}}$

$398 = \frac{1}{2 \pi \sqrt{2C_3^2\times 400\times 800}}$

$C_3=4.99858\times 10^{-7} \approx 5 \times 10^{-7}F$

$C_1=2C_3=10^{-6}$

and from the relation

$\frac{C_3}{C_1}+\frac{R_1}{R_3}=\frac{R_2}{R_4}$

$\frac{C_3}{2C_3}+\frac{400}{800}=\frac{1000}{R_4}$

$\frac{1}{2}+\frac{1}{2}=\frac{1000}{R_4}$

$R_4=1000$

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