Question

Given that \(R1=20\text{k}\Omega\), \(C1=2\text{nF}\), \(R2=20\text{k}\Omega\), \(C2=2\text{nF}\), find the approximate resonant frequency of a Wein bridge oscillator.

• A. \(4\text{kHz}\)
• B. \(3\text{kHz}\)
• C. \(25\text{kHz}\)
• D. \(15\text{kHz}\)

Hint:

To perform fast manual calculations during exams, approximate $\frac{1}{2\pi}$ as roughly $0.159$: \[ f_r = \frac{0.159}{R \cdot C} \] Given $RC = 40\,\mu\text{s}$: \[ f_r = \frac{0.159}{40 \times 10^{-6}} = \frac{159000}{40} = 3975\,\text{Hz} \approx 4\,\text{kHz} \] This shortcut provides the correct answer within seconds.

Answer 1

The Correct Option is A

Concept: A Wien Bridge Oscillator is a standard electronic oscillator that generates low-distortion sinusoidal waves at audio frequencies. The feedback network consists of a series RC combination ($R_1, C_1$) in one arm and a parallel RC combination ($R_2, C_2$) in the adjacent arm. The resonant frequency ($f_r$) of the bridge network is the specific frequency at which the phase shift through the feedback circuit is exactly $0^\circ$.

Step 1: Identifying the Resonant Frequency Formula
The general equation governing the frequency of oscillation for a Wien bridge circuit is given by the formula: \[ f_r = \frac{1}{2\pi \sqrt{R_1 R_2 C_1 C_2}} \] When the components in both arms are chosen to be perfectly symmetrical such that $R_1 = R_2 = R$ and $C_1 = C_2 = C$, the formula simplifies directly to: \[ f_r = \frac{1}{2\pi R C} \]

Step 2: Substituting the Given Numerical Values
From the problem description, we have:
• Resistance: $R = R_1 = R_2 = 20\text{k}\Omega = 20 \times 10^3 \, \Omega$
• Capacitance: $C = C_1 = C_2 = 2\text{nF} = 2 \times 10^{-9} \, \text{F}$ Since $R_1=R_2$ and $C_1=C_2$, we safely deploy the simplified equation: \[ f_r = \frac{1}{2 \cdot \pi \cdot (20 \times 10^3) \cdot (2 \times 10^{-9})} \]

Step 3: Calculating the Denominator
Combine the values inside the product: \[ R \times C = (20 \times 10^3) \times (2 \times 10^{-9}) = 40 \times 10^{-6} \, \text{seconds} \] Now evaluate the full denominator value incorporating $2\pi$: \[ 2 \times \pi \times (40 \times 10^{-6}) = 80\pi \times 10^{-6} \approx 80 \times 3.14159 \times 10^{-6} \approx 251.327 \times 10^{-6} \]

Step 4: Solving for Frequency ($f_r$)
\[ f_r = \frac{1}{251.327 \times 10^{-6}} = \frac{10^6}{251.327} \approx 3978.87 \, \text{Hz} \] Converting the answer from Hertz ($\text{Hz}$) to kilo-Hertz ($\text{kHz}$): \[ f_r \approx \frac{3978.87}{1000} \, \text{kHz} \approx 3.98 \, \text{kHz} \] Rounding to the nearest integer choice yields approximately $4\text{kHz}$.