If the equivalent inductance in a Hartley oscillator is doubled, the frequency is:
Reduced by √2
The Hartley oscillator is an LC (inductor-capacitor) oscillator used in radio frequency applications. The oscillation frequency is given by the formula:
\(f = \frac{1}{2\pi\sqrt{L_{\text{eq}}C}}\)
where \(f\) is the frequency of oscillation, \(L_{\text{eq}}\) is the equivalent inductance, and \(C\) is the capacitance.
In this problem, it is given that the equivalent inductance \(L_{\text{eq}}\) is doubled. Let's see the effect on the frequency:
- Initial frequency, \(f_1 = \frac{1}{2\pi\sqrt{L_{\text{eq}}C}}\)
- After doubling the inductance, the new equivalent inductance becomes \(2L_{\text{eq}}\).
- The new frequency, \(f_2 = \frac{1}{2\pi\sqrt{2L_{\text{eq}}C}}\)
Let's calculate \(f_2\) in terms of \(f_1\):
\(f_2 = \frac{1}{2\pi\sqrt{2L_{\text{eq}}C}} = \frac{1}{\sqrt{2}} \cdot \frac{1}{2\pi\sqrt{L_{\text{eq}}C}} = \frac{1}{\sqrt{2}} f_1\)
This shows that the frequency \(f_2\) is reduced by a factor of \(\sqrt{2}\) compared to \(f_1\).
Hence, the frequency is reduced by \(\sqrt{2}\) when the equivalent inductance is doubled. Therefore, the correct answer is:
Reduced by √2
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