The frequency of the oscillator circuit shown in the figure below is _______(in kHz, rounded off to two decimal places).
Given: \( R = 1 \, k\Omega; R_1 = 2 \, k\Omega; R_2 = 6 \, k\Omega; C = 0.1 \, \mu F \)
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For phase shift oscillators, the frequency depends on the equivalent resistance of the network and the capacitance. Make sure to use the correct equivalent resistance formula to calculate the frequency.
For a phase shift oscillator, the frequency \( f \) is given by the formula:
\[
f = \frac{1}{2\pi \sqrt{R_{{eq}} \cdot C}}
\]
where \( R_{{eq}} \) is the equivalent resistance of the oscillator circuit, given by:
\[
R_{{eq}} = \frac{R_1 \cdot R_2}{R_1 + R_2} + R_3
\]
Substituting the values:
\[
R_{{eq}} = \frac{2 \, k\Omega \cdot 6 \, k\Omega}{2 \, k\Omega + 6 \, k\Omega} + 1 \, k\Omega = \frac{12}{8} \, k\Omega + 1 \, k\Omega = 1.5 \, k\Omega + 1 \, k\Omega = 2.5 \, k\Omega
\]
Now, substitute \( R_{{eq}} = 2.5 \, k\Omega \) and \( C = 0.1 \, \mu F \) into the frequency formula:
\[
f = \frac{1}{2\pi \sqrt{2.5 \, k\Omega \cdot 0.1 \, \mu F}}
\]
\[
f = 1.50 \, {kHz}
\]
Thus, the frequency of the oscillator is approximately \( 1.50 \, {kHz} \).
