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Solved Examples · Example 3

Q.The frequency of Hartley oscillator is 20 kHz20\,\text{kHz}. The capacitor used in the feedback network has a value of 576 pF576\,\text{pF}. If one of the inductors value is 100 mH100\,\text{mH}, calculate the value of the other inductor.

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[!TLDR]

The tank needs L≈109.91 mHL \approx 109.91\,\text{mH}; subtracting L1=100 mHL_1 = 100\,\text{mH} gives L2≈9.91 mHL_2 \approx 9.91\,\text{mH}.

In a Hartley oscillator the two inductor sections in series give L=L1+L2L = L_1 + L_2, and the frequency is f=12πLCf = \dfrac{1}{2\pi\sqrt{LC}}. First find the total tank inductance required, then subtract the known section:

L=1(2πf)2 C,L2=L−L1.L = \frac{1}{(2\pi f)^2\,C}, \qquad L_2 = L - L_1.

Step 1 — total inductance: with f=20×103 Hzf = 20\times10^{3}\,\text{Hz} and C=576×10−12 FC = 576\times10^{-12}\,\text{F}, (2πf)2≈1.579×1010(2\pi f)^2 \approx 1.579\times10^{10}, so L=11.579×1010×576×10−12≈109.91×10−3 HL = \dfrac{1}{1.579\times10^{10}\times576\times10^{-12}} \approx 109.91\times10^{-3}\,\text{H}.

Step 2 — subtract the known inductor: L2=(109.91−100)×10−3 HL_2 = (109.91 - 100)\times10^{-3}\,\text{H}.

[!ANSWER]

L2≈9.91×10−3 H=9.91 mHL_2 \approx 9.91\times10^{-3}\,\text{H} = 9.91\,\text{mH}.

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