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Solved Examples · Example 1

Q.A Hartley Oscillator has L1=2 mHL_1 = 2\,\text{mH}, L2=4 mHL_2 = 4\,\text{mH} and C=10 nFC = 10\,\text{nF}. Determine the frequency of oscillation.

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[!TLDR]

The Hartley tank inductance is L=L1+L2=6 mHL = L_1 + L_2 = 6\,\text{mH}, which resonates with C=10 nFC = 10\,\text{nF} to give f≈20.544 kHzf \approx 20.544\,\text{kHz}.

A Hartley oscillator, one of the LC oscillators in the Karnataka 2nd PUC Electronics course, fixes its frequency with a tank circuit in which a tapped inductor resonates with a capacitor. Because the two inductor sections L1L_1 and L2L_2 are in series across the tank, their inductances add, L=L1+L2L = L_1 + L_2 (mutual inductance neglected here). The oscillation frequency equals the tank resonant frequency

f=12πLC.f = \frac{1}{2\pi\sqrt{LC}}.

Step 1 — total inductance: L=2 mH+4 mH=6×10−3 HL = 2\,\text{mH} + 4\,\text{mH} = 6\times10^{-3}\,\text{H}.

Step 2 — substitute: f=12π(6×10−3)(10×10−9)=12π6×10−11f = \dfrac{1}{2\pi\sqrt{(6\times10^{-3})(10\times10^{-9})}} = \dfrac{1}{2\pi\sqrt{6\times10^{-11}}}.

Step 3 — evaluate: 6×10−11≈7.746×10−6\sqrt{6\times10^{-11}} \approx 7.746\times10^{-6}, so f≈16.284×7.746×10−6≈2.054×104 Hzf \approx \dfrac{1}{6.284\times7.746\times10^{-6}} \approx 2.054\times10^{4}\,\text{Hz}.

[!ANSWER]

f≈20.544 kHzf \approx 20.544\,\text{kHz}.

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