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- About the calculation of equivalent inductance when taking into account mutual induction in the analysis of Hartley Oscillator

Hi,

I'm watching a video about Hartley oscillator and I'm in trouble with a simple assumption: as stated at minute 5:50 if the two coils ##L_1## and ##L_2## are wound on the same core then taking into account the mutual inductance M he gets:

##L^ {'} _1 = L_1 + M##

##L^ {'} _2 = L_2 + M##

From my understanding it surely makes sense if we assume equals the currents through the two coils ##I_{L_1} = I_{L_2} = I##.

However I believe they are not the same from the analysis of the small signal equivalent circuit at least when we do *not neglect* the BJT input impedance ##h_{ie}##.

I'm watching a video about Hartley oscillator and I'm in trouble with a simple assumption: as stated at minute 5:50 if the two coils ##L_1## and ##L_2## are wound on the same core then taking into account the mutual inductance M he gets:

##L^ {'} _1 = L_1 + M##

##L^ {'} _2 = L_2 + M##

From my understanding it surely makes sense if we assume equals the currents through the two coils ##I_{L_1} = I_{L_2} = I##.

However I believe they are not the same from the analysis of the small signal equivalent circuit at least when we do *not neglect* the BJT input impedance ##h_{ie}##.

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