# Understanding of coupling coefficient and power transfer efficiency

I am a bit confused about my derivation to get power efficiency from the coupling coefficient.

The coupling coefficient $$\k=M/\sqrt{L_1L_2}\$$. $$V_2=M\frac{dI_1}{dt}=L_2\frac{dI_2}{dt}\\ V_1=M\frac{dI_2}{dt}=L_1\frac{dI_1}{dt}\\$$ Therefore, $$M={V_2}/{\frac{dI_1}{dt}}={V_1}/{\frac{dI_2}{dt}}\\ L_1={V_1}/{\frac{dI_1}{dt}}\\ L_2={V_2}/{\frac{dI_2}{dt}}$$ Then I substitute them into $$\k\$$ to get: $$k=\sqrt{\frac{V_1dI_1}{V_2dI_2}}=\sqrt{\frac{V_2dI_2}{V_1dI_1}}\\ k^2=\frac{P_1}{P_2}=\eta_{12}=\frac{P_2}{P_1}=\eta_{21}$$ So I can say: if there are two coils that one is given power source and the other is receiving the power, then the two coils can transmit and feedback the power to each other with the efficiency of $$\k^2=\eta\$$. Therefore, the total power received in the receiver coil will be: $$P_2=P_1(\eta+\eta^3+\eta^5+\dots)=P_1\sum_{n=1}^\infty\eta^{2n-1}=P_1\frac{\eta}{1-\eta^2}$$ As you can see, the efficiency of the power transfer here becomes $$\\frac{\eta}{1-\eta^2}\$$.

Is this correct? Or am I having any misunderstanding here?

• True. I have missed the point that $I$ has $90^{\circ}$ phase lagging related to $V$. What I should consider is to put a load across the receiver conductor that is seen as a voltage source and a resistor series to the transmitter one so that I can evaluate the power transfer efficiency. Thank you. – ONLYA Dec 12 '20 at 16:13